Intro to Polar Coordinates
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Intro to Polar Coordinates 1. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (3, 120◦), p2 = (−6, 120◦), p3 = (−3, −120◦), p4 = (3, −300◦), p5 = (−3, 135◦) Solution: 90◦ 60◦ 120◦
30◦
p1
150◦
b
p4
p3 b
0◦ 180◦
p5
b
330◦ 210◦
p2 b
300◦
240◦ 270◦
2. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (2, −120◦), p2 = (−4, −120◦), p3 = (−4, 120◦), p4 = (4, −60◦ ), p5 = (−4, 135◦) Solution:
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pg. 6
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez 90◦ 60◦ 120◦
p2 b
30◦
150◦
0◦ 180◦
p1 b
210◦
p4
pp35 b
b
330◦
300◦ 240◦ 270◦
3. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (1, 45◦ ), p2 = (−2, 45◦), p3 = (−5, −45◦), p4 = (5, −225◦), p5 = (−5, 135◦) Solution:
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pg. 7
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez 90◦ 60◦ 120◦
p4
b
p3 30◦
150◦
p1 b
0◦ 180◦
p2 b
330◦ 210◦
p5
b
300◦ 240◦ 270◦
4. Convert the polar coordinates, (−4, 210◦) to cartisian coordinates Solution:
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pg. 8
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦
6 60◦
120◦
5
30◦ 150◦
(−4, 210◦) b
x = r cos θ y = r sin θ so...
4 3
(3.46, 2)
x = r cos θ = −4 cos(210◦) ≈ 3.46 0◦ -6
180◦
b
2 1
-5
-4
-3
-2
-1
1
2
3
4
5
6
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = −4 sin(210◦) ≈2
-2 -3 -4 -5
300◦ 240◦
-6 270◦
5. Convert the polar coordinates, (4, 210◦) to cartisian coordinates Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 4 cos(210◦) ≈ −3.46 0◦
2 1
-6
180◦
-5
-4
-3
-2
-1 -1
(4, 210◦) b
330◦ 210◦
y = r sin θ = 4 sin(210◦) ≈ −2
(−3.46, −2) b
-2 -3 -4 -5
300◦ 240◦
-6 270◦
6. Convert the polar coordinates, (5, 150◦) to cartisian coordinates
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pg. 9
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solution:
90◦
6 60◦
120◦
150◦
5
30◦
(5, 150◦) b
x = r cos θ y = r sin θ so...
4
(−4.33, 2.5)
3
b
x = r cos θ = 5 cos(150◦) ≈ −4.33 0◦
2 1
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = 5 sin(150◦) ≈ 2.5
-2 -3 -4 -5
300◦ 240◦
-6 270◦
7. Convert the polar coordinates, (6, −240◦) to cartisian coordinates Solution:
90◦
6
(6, −240◦)
(−3, 5.2)
60◦
120◦ b
b
30◦ 150◦
5
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 6 cos(−240◦) ≈ −3 0◦ -6
180◦
2 1
-5
-4
-3
-2
-1 -1
330◦ 210◦
y = r sin θ = 6 sin(−240◦) ≈ 5.2
-2 -3 -4 -5
300◦ 240◦
-6 270◦
8. Convert the polar coordinates, (−5, 0◦ ) to cartisian coordinates
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pg. 10
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = −5 cos(0◦) ≈ −5 0◦
(−5, 0◦) b
2 1
(−5, 0) b
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = −5 sin(0◦) ≈0
-2 -3 -4 -5
300◦ 240◦
-6 270◦
9. Convert the polar coordinates, (6, −180◦) to cartisian coordinates Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 6 cos(−180◦) ≈ −6 (−6, −0) 0◦
(6, −180◦) b
2 1
b
-6
180◦
-5
-4
-3
-2
-1 -1
330◦ 210◦
y = r sin θ = 6 sin(−180◦) ≈ −0
-2 -3 -4 -5
300◦ 240◦
-6 270◦
10. Convert the polar coordinates, (6, 180◦) to cartisian coordinates
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pg. 11
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 6 cos(180◦) ≈ −6 (−6, 0) 0◦
(6, 180◦) b
2 1
b
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = 6 sin(180◦) ≈0
-2 -3 -4 -5
300◦ 240◦
-6 270◦
11. Convert the cartesian coordinates, (-2, 3), to polar coordinates. Solution:
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pg. 12
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4
(r, θ) b
150◦
(−2, 3)
30◦
b
3 2 1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 3/ − 2 (note: eq has infinite solutions) ◦ θ ≈ −56.31 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−2)2 + (3)2 r2 =13
√ r = ± 13 ≈ ±3.61
note: if we use the angle −56.31◦, then we must choose the negative value of r, r ≈ −3.61, thus the polar cordinates (−3.61, −56.31◦) is one way to represent the point (−2, 3)
12. Convert the cartesian coordinates, (2, -3), to polar coordinates. Solution:
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pg. 13
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2
(r, θ)
330◦
b
-3
(2, −3) b
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = −3/2 (note: eq has infinite solutions) ◦ θ ≈ −56.31 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(2)2 + (−3)2 r2 =13
√ r = ± 13 ≈ ±3.61
note: if we use the angle −56.31◦, then we must choose the positive value of r, r ≈ 3.61, thus the polar cordinates (3.61, −56.31◦) is one way to represent the point (2, −3)
13. Convert the cartesian coordinates, (-5, -3), to polar coordinates. Solution:
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pg. 14
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2
(r, θ)
(−5, −3)
330◦
b
b
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = −3/ − 5 (note: eq has infinite solutions) θ ≈ 30.96◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−5)2 + (−3)2 r2 =34
√ r = ± 34 ≈ ±5.83
note: if we use the angle 30.96◦, then we must choose the negative value of r, r ≈ −5.83, thus the polar cordinates (−5.83, 30.96◦) is one way to represent the point (−5, −3)
14. Convert the cartesian coordinates, (-4, -1), to polar coordinates. Solution:
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pg. 15
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦ 180◦
-6
(r, θ) b
-5
-4
-3
(−4, −1) b
-2
-1
1
2
3
4
5
6
-1 -2
330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = −1/ − 4 (note: eq has infinite solutions) θ ≈ 14.04◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−4)2 + (−1)2 r2 =17
√ r = ± 17 ≈ ±4.12
note: if we use the angle 14.04◦, then we must choose the negative value of r, r ≈ −4.12, thus the polar cordinates (−4.12, 14.04◦) is one way to represent the point (−4, −1)
15. Convert the cartesian coordinates, (-6, 3), to polar coordinates. Solution:
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pg. 16
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4
(r, θ) b
(−6, 3)
30◦
b
150◦
3 2 1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 3/ − 6 (note: eq has infinite solutions) ◦ θ ≈ −26.57 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−6)2 + (3)2 r2 =45
√ r = ± 45 ≈ ±6.71
note: if we use the angle −26.57◦, then we must choose the negative value of r, r ≈ −6.71, thus the polar cordinates (−6.71, −26.57◦) is one way to represent the point (−6, 3)
16. Convert the cartesian coordinates, (-8, 4), to polar coordinates. Solution:
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pg. 17
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5
(r, θ)
(−8, 4)
b
b
4
30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 4/ − 8 (note: eq has infinite solutions) ◦ θ ≈ −26.57 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−8)2 + (4)2 r2 =80
√ r = ± 80 ≈ ±8.94
note: if we use the angle −26.57◦, then we must choose the negative value of r, r ≈ −8.94, thus the polar cordinates (−8.94, −26.57◦) is one way to represent the point (−8, 4)
17. Plot Points, for a Famous Graph Plot the following points and connect the points.
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pg. 18
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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angle theta r 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 345 360
5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00
pg. 19
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
angle theta r 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 345 360
Solution: 18. Plot Points, for a Famous Graph Plot the following points and connect the points.
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pg. 20
5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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angle theta r 0 0.00 15 1.29 30 2.50 45 3.54 60 4.33 75 4.83 90 5.00 105 4.83 120 4.33 135 3.54 150 2.50 165 1.29 180 0.00 195 -1.29 210 -2.50 225 -3.54 240 -4.33 255 -4.83 270 -5.00 285 -4.83 300 -4.33 315 -3.54 330 -2.50 345 -1.29 360 -0.00 5sin th
pg. 21
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
angle theta r 0 0.00 15 1.29 30 2.50 45 3.54 60 4.33 75 4.83 90 5.00 105 4.83 120 4.33 135 3.54 150 2.50 165 1.29 180 0.00 195 -1.29 210 -2.50 225 -3.54 240 -4.33 255 -4.83 270 -5.00 285 -4.83 300 -4.33 315 -3.54 330 -2.50 345 -1.29 360 -0.00 5sin th
Solution:
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pg. 22
Trigonometry Sec. 02 notes
MathHands.com M´ arquez
Introduction to Polar Graphs Main Idea Plot Points You should be very conformable plotting points in radians as well as degrees. Simply make a table of values with r’s and corresponding θ’s, then start plotting until you can make sense of the the graph. If connecting the points is uneasy, choose smaller angle intervals, i.e. instead of plotting and calculating the r every 30 degrees, plot every 5 degrees. Get to Know Your Calculator I am usually not a big calculator fan, but this may be a good time when we can use it appropriately to graph some of these functions. After all, there is limited beauty and creativity that occurs when plotting points. Get to know Famous Questions 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ? Get to know Famous Graphs Get to know Famous Graphs 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ?
r = sin θ
r = cos θ
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r = cos(2θ)
r = cos(3θ)
r = 5 + 3 sin θ
r = sin(2θ)
r = sin(4θ)
r = 3 + 5 sin θ
r = sin(3θ)
r = cos(5θ)
math hands
pg. 1
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
Introduction to Polar Graphs 1. Graph and Understand the graph of r = 5 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 2. Graph and Understand the graph of r = 5 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 3. Graph and Understand the graph of r = −3 cos θ
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pg. 2
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 4. Graph and Understand the graph of r = 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 5. Graph and Understand the graph of r = −6 sin θ
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pg. 3
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 6. Graph and Understand the graph of r = 5 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 7. Graph and Understand the graph of r = 5 sin θ. Limit the study of this graph to the values of θ ranging from −90◦ to 0◦ .
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pg. 4
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 8. Graph and Understand the graph of r = −3 cos θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 9. Graph and Understand the graph of r = 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 180◦.
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pg. 5
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 10. Graph and Understand the graph of r = −6 sin θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 11. Graph and Understand the graph of r = 5 + cos θ
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pg. 6
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 12. Graph and Understand the graph of r = 4 − 2 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 13. Graph and Understand the graph of r = −3 + 2 cos θ
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pg. 7
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 14. Graph and Understand the graph of r = 3 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 15. Graph and Understand the graph of r = 4 − 2 cos θ
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pg. 8
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 16. Graph and Understand the graph of r = −2 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 17. Graph and Understand the graph of r = 3 + 2 sin θ
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pg. 9
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 18. Graph and Understand the graph of r = 3 − 2 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 19. Graph and Understand the graph of r = −3 + 2 sin θ
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pg. 10
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 20. Graph and Understand the graph of r = 3 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 21. Graph and Understand the graph of r = 2 − 3 sin θ
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math hands
pg. 11
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 22. Graph and Understand the graph of r = −2 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 23. Graph and Understand the graph of r = 2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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pg. 12
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 24. Graph and Understand the graph of r = 2 − 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 60◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 25. Graph and Understand the graph of r = −2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
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math hands
pg. 13
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 26. Graph and Understand the graph of r = 5 + 5 sin θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 27. Graph and Understand the graph of r = 3 − 2 sin θ. Limit the study of this graph to the values of θ ranging from 135◦ to 225◦ .
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 28. Graph and Understand the graph of r = −3 + 2 sin θ. Limit the study of this graph to the values of θ ranging from 360◦ to 540◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 29. Graph and Understand the graph of r = 5 cos(2θ)
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 30. Graph and Understand the graph of r = 6 cos(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 31. Graph and Understand the graph of r = 5 sin(2θ)
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 32. Graph and Understand the graph of r = 6 sin(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 33. Graph and Understand the graph of r = 5 cos(3θ)
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 34. Graph and Understand the graph of r = 6 cos(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 35. Graph and Understand the graph of r = 5 sin(3θ)
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 36. Graph and Understand the graph of r = 4 sin(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 37. Graph and Understand the graph of r = 5 sin(2θ). Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 38. Graph and Understand the graph of r = 6 sin(4θ). Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 39. Graph and Understand the graph of r = 5 cos(3θ). Limit the study of this graph to the values of θ ranging from 0◦ to 180◦ .
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 40. Graph and Understand the graph of r = 6 cos(5θ). Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 41. Graph and Understand the graph of r2 = 25 sin2 θ √ Solution: Note, we can solve for r as r√= ± 25 sin2 θ... then we can plot each r separately, that is we graph √ r = 25 sin2 θ and then we graph r = − 25 sin2 θ
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
42. Graph and Understand the graph of r2 = 25 sin2 3θ √ 25 sin2 3θ... then we can plot each r separately, that is we graph Solution: Note, we can solve for r as r = ± √ √ 2 r = 25 sin 3θ and then we graph r = − 25 sin2 3θ 90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
43. Understanding The graph of r = 2 − 3 cos θ has an inner loop. What values of θ trace precisely this inner loop? i.e. ?◦ ≤ θ ≤?◦ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center
44. Understanding On the graph of r = 5 sin(3θ) (a) ’sin 3θ’ will take values between 1 and −1; never larger than one, never smaller than −1. Find the angles θ, at which sin 3θ take on its largest value 1, or is smallest value -1. (i.e solve sin 3θ = 1 and solve sin 3θ = −1 )
(b) Graph r = 5 sin(3θ) and mark the points obtained from part a). What can you say about these points? A) nothing B) they are the ’tip of the leaves’ C) they are not on the leaves. (c) Find the angles at which the graph goes through (0, 0) (d) Find the interval/s for theta that trace the second ’leaf’ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center, such leaf starts and ends at center.. 45. Understanding On the graph of r = −5 + 2 cos θ
(a) ’cos θ’ will take values between 1 and −1; never larger than one, never smaller than −1. What is the largest or the smallest that the quantity ”−5 + 2 cos θ” can become?
(b) Find the angles at which ”−5 + 2 cos θ” takes on its largest value -3, or is smallest value -7. (c) Graph r = −5 + 2 cos θ and mark the points obtained from part b). What can you say about these points? A) nothing B) its the furthest point from the origin (d) Find the angles at which the graph goes through (0, 0) (e) Find the interval/s for theta that trace ONLY the outer leaf.
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Trigonometry Sec. 03 notes
MathHands.com M´ arquez
Converting Equations Polar & Cartesian The IDEA In the last couple sections we’ve established, among other things, a ’dictionary’ to convert coordinate points from cartesian to polar and vise-varsa. In this section, we would like to exploit that same dictionary to convert equations. As a reminder, the dictionary has two components, ’from’ polar, and ’to’ polar. The goal in converting, in most cases, is to eliminate all x’s and y’s and replace them with appropriate expressions involving r’s and θ’s if we are translating to polar. When converting from polar, the idea is to eliminate the r’s and θ’s. We summarize the ’dictionary’ below. Translating
Polar
Cartesian
90◦ 120◦
y = r sin θ x = r cos θ
60◦ 30◦
150◦ b
b
(r, θ)
(x, y)
0◦
180◦
210◦
330◦
θ = tan−1 300◦
240◦
r2 = x2 + y
270◦
y x 2
Example: Converting Equation of a line to Polar Convert the following equation to polar coordinates: y=x Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
y y x tan θ θ
=x
(given)
=1
(div by x, setting up to use translating dictionary, tan θ = y/x)
=1 = 45◦
(used tan θ = y/x) (notice: this there are infinite possibilities for θ (ie −135 , 225 ...), this is just one of them.) ◦
◦
Thus, one way to translate the equation y = x to polar is to convert to θ = 45◦ Example: Converting Equation of a circle to Polar Convert the following equation to polar coordinates: x2 + y 2 = 25 Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
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Trigonometry Sec. 03 notes
MathHands.com M´ arquez x2 + y 2 = 25
(given)
2
2
(used the dictionary r = x2 + y 2 )
r = 25
at this point we have eliminated all x’s and y’s from the equation, thus the converting is complete. It is sometimes desirable to solve for r explicitly, thus we could also say r = 5 or r = −5, in fact all three of these equations, r2 = 25, r = 5, AND r = −5 all yield the same graph as x2 + y 2 = 25. Example: Converting the equation of a circle to Polar Convert the following equation to polar coordinates: x2 + 3x + 5 + y 2 − 4y + 2 = 25 Solution:
x2 + 3x + 5 + y 2 − 4y + 2 = 25 2
2
(given)
2
(x + y ) + 3(x) − 4(y) = 18
(just cleaning up, getting ready to use the dictionary)
(r ) + 3(r cos θ) − 4(r sin θ) = 18
(used the dictionary)
2
r + 3r cos θ − 4r sin θ = 18
(algebra)
Example: Converting the equation of a circle from Polar Convert the following equation of a circle to polar coordinates: r = 12 cos θ Solution: r = 12 cos θ r · r = r(12 cos θ)
r2 = 12(r cos θ) 2
2
x + y = 12x 2
2
2
y + x + − 12x = 0 2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
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(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Converting Equations Polar & Cartesian 1. Convert the following equation to polar coordinates: y = 2x
Solution: Note: the ’translating’ of this equation is not unique.
y = 2x y =2 x tan θ = 2 θ ≈ 63.435
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) ◦
(notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 63.435◦ 2. Convert the following equation to polar coordinates: 2 y= − x 3
Solution: Note: the ’translating’ of this equation is not unique. 2 y= − x 3 2 y = − x 3 2 tan θ = − 3 θ ≈ −33.69◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ −33.69◦ 3. Convert the following equation to polar coordinates: y=
2 x 5
Solution: Note: the ’translating’ of this equation is not unique.
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez 2 x 5 2 = 5 2 = 5 ≈ 21.801◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 21.801◦ 4. Convert the following equation to polar coordinates: y=
4 x 3
Solution: Note: the ’translating’ of this equation is not unique. 4 x 3 4 = 3 4 = 3 ≈ 53.13◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 4x to polar is to convert to θ ≈ 53.13◦ 5. Convert the following equation to polar coordinates: 4 y= − x 3 Solution: Note: the ’translating’ of this equation is not unique. 4 y= − x 3 4 y = − x 3 4 tan θ = − 3 θ ≈ −53.13◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = − 4x to polar is to convert to θ ≈ −53.13◦ 6. Convert the following equation of a circle to polar coordinates: 2x2 + 3x + 2y 2 + − 5y = 7
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution:
2x2 + 3x + 2y 2 + − 5y = 7 2 x2 + y 2 + 3(x) + − 5(y) = 7 2 r2 + 3(r cos θ) + − 5(r sin θ) = 7
(given) (ready to use the dictionary....) (used the dictionary)
7. Convert the following equation of a circle to polar coordinates: − 5x2 + 2x + − 5y 2 + 7y = 25 Solution:
− 5x2 + 2x + − 5y 2 + 7y = 25 − 5 x2 + y 2 + 2(x) + 7(y) = 25 − 5 r2 + 2(r cos θ) + 7(r sin θ) = 25
(given) (ready to use the dictionary....) (used the dictionary)
8. Convert the following equation of a circle to polar coordinates: 3 4x2 + x + 4y 2 + 1y = 9 2
Solution: 3 4x2 + x + 4y 2 + 1y = 9 2 3 4 x2 + y 2 + (x) + 1(y) = 9 2 3 2 4 r + (r cos θ) + 1(r sin θ) = 9 2
(given) (ready to use the dictionary....) (used the dictionary)
9. Convert the following equation of a circle to polar coordinates: 3x2 + 3x + 3y 2 + 5y = 4
Solution:
3x2 + 3x + 3y 2 + 5y = 4 3 x2 + y 2 + 3(x) + 5(y) = 4 3 r2 + 3(r cos θ) + 5(r sin θ) = 4
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(given) (ready to use the dictionary....) (used the dictionary)
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
10. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 9 25
Solution: y2 x2 + =1 9 25 (r cos θ)2 (r sin θ)2 + =1 9 25 ! cos2 θ sin2 θ 2 + =1 r 9 25
(given) (ready to use the dictionary....) (alebra)
11. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 16
Solution: y2 x2 + =1 4 16 (r cos θ)2 (r sin θ)2 + =1 4 16 ! cos2 θ sin2 θ + =1 r2 4 16
(given) (ready to use the dictionary....) (alebra)
12. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 49
Solution: y2 x2 + =1 4 49 (r cos θ)2 (r sin θ)2 + =1 4 49 ! cos2 θ sin2 θ =1 + r2 4 49
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(given) (ready to use the dictionary....) (alebra)
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
13. Convert the following equation of a circle to polar coordinates: r = 2 sin θ
Solution: r = 2 sin θ
(given)
r · r = r(2 sin θ)
(clever little idea, mult both sides by r)
r2 = 2(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 2y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 2y = 0
(optional step: complete the square)
2
x + y + − 2y + 1 = 1 2 x2 + y + − 1 = 1
(optional to step: complete the square) (Shows its a circle, shows center/radius)
14. Convert the following equation of a circle to polar coordinates: r = 5 sin θ
Solution: r = 5 sin θ
(given)
r · r = r(5 sin θ)
(clever little idea, mult both sides by r)
2
r = 5(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 5y = 0 25 25 x2 + y 2 + − 5y + = 4 4 2 25 5 = x2 + y + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
15. Convert the following equation of a circle to polar coordinates: r = 6 sin θ
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 6 sin θ r · r = r(6 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 6(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 6y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 6y = 0
(optional step: complete the square)
2
x + y + − 6y + 9 = 9 2 x2 + y + − 3 = 9
(optional to step: complete the square) (Shows its a circle, shows center/radius)
16. Convert the following equation of a circle to polar coordinates: r = 11 sin θ
Solution: r = 11 sin θ r · r = r(11 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 11y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 11y = 0 121 121 = x2 + y 2 + − 11y + 4 4 2 11 121 x2 + y + − = 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
17. Convert the following equation of a circle to polar coordinates: r = 8 cos θ
Solution: r = 8 cos θ
(given)
r · r = r(8 cos θ)
(clever little idea, mult both sides by r)
2
r = 8(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 8x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 8x = 0
(optional step: complete the square)
2
y + x + − 8x + 16 = 16 2 y 2 + x + − 4 = 16
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(optional to step: complete the square) (Shows its a circle, shows center/radius)
math hands
pg. 8
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
18. Convert the following equation of a circle to polar coordinates: r = 12 cos θ
Solution: r = 12 cos θ
(given)
r · r = r(12 cos θ)
(clever little idea, mult both sides by r)
r2 = 12(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 12x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 12x = 0
(optional step: complete the square)
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
(optional to step: complete the square) (Shows its a circle, shows center/radius)
19. Convert the following equation of a circle to polar coordinates: r = 5 cos θ
Solution: r = 5 cos θ
(given)
r · r = r(5 cos θ)
(clever little idea, mult both sides by r)
2
r = 5(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 5x = 0 25 25 y 2 + x2 + − 5x + = 4 4 2 25 5 = y2 + x + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
20. Convert the following equation of a circle to polar coordinates: r = 11 cos θ
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 11 cos θ r · r = r(11 cos θ) r2 = 11(r cos θ) 2
2
x + y = 11x 2
2
y + x + − 11x = 0 121 121 = y 2 + x2 + − 11x + 4 4 2 121 11 = y2 + x + − 2 4
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(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 10
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
Intro to Complex Numbers The Question In the last section we wondered what might be far to the right and to the left of the real number line. In doing so, we entertained the existence of the very special unreal number ∞. In this section, we continue on this path and now consider what lies above or below the real number line?
? ∞
−∞ -5
-4
Out of Audacity, a number is born
-3
-2
-1
0
1
2
3
4
5
?
Here is the account, directly from the world famous, the almost immortal, mathematician Euler himself.... ”I sat in that soft comfortable chair leaned back and enjoyed the million thoughts dancing inside my head. A blank paper, a pencil and my awesome coffee was all there was on the desk. On the paper the equation x2 = −1 The equation was screaming, enticing, talking trash, challenging me, saying ”you can’t solve me!” Hours went by faster than I would have liked. Days past by, weeks and months...There was no real number that would solve the equation. But the forces were greater. The inspiration divine. I would not be stopped.. and one day it happened. There was no real number solution, I had looked on the positive side on the negative side and all numbers between. Resolved to avoid defeat at all costs, I invented a number. From my own imagination, I gathered all my might, my courage, and my audacity, √ and I thought...I will create a number. I will call it i, and I will solve my problem by declaring i = −1. It’s my number so I can make it behave however I please, just as the artists paints the clouds at his whim... This solves the equation x2 = −1 and marks the birth of a grand elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.” Needless, to say, I have taken some artistic liberties with this account of events. In fact, traces of complex numbers or ’imaginary’ numbers can be found in 9th century’s Al-Khwarizmi’s Algebra text. During the next couple centuries these ideas made their way to Italy and France, as people were learning to solve degree 3 equations.√ By the turn of the 17th century Descartes coined the phrase ”imaginary” numbers, referring √ to numbers such as −1. At last, it was Euler, in the 18th century who named such number i, declaring i = −1. Thus... by definition of i;
i2 = −1 and, i is a solution to x2 = −1 In addition to solving the equation x2 =− 1, and this marks the birth of a grand and elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.
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Trigonometry Sec. 04 notes
MathHands.com M´ arquez
Where do all the i’s Live With the invent of i came other numbers such as 2i, 3i, −5i, or 2 + 3i. Generally, complex numbers are numbers that can be written in the form a + bi where a and b are real numbers. Now, in the previous sections we noted a visual representation of the real numbers using the real number line. Considering this, the natural question is where or how do we represent the complex numbers visually? Over the centuries, the most powerful and common way to represent the complex numbers is to place them as an extension of the real number line, extending it above and below to make what we commonly call the complex plane. In essence, this is done by placing an ’imaginary’ axis perpendicular to the real number line. Then we position every number a+ bi on the plane similarly to placing the ordered point (a, b) on the cartesian [xy]plane. came numbers such as 2i, 3i, 4i, . . . as well as numbers such as 2 + 3i. Here are some visual representations of a few complex numbers along with their position relative to the known real number line. 3i
The Complex Numbers, C
2i b
−4 + i
b
3 + 2i
3
4
1i 2 + 0i
R
b
-5
-4
-3
-2
-1
0
1
2
5
−1i −2i −3i b
4 − 3i
Negative Radicals With the invention of i we can now make sense of radicals (i.e. square roots) of √ negative real numbers. Consider √ the radical −1, the number whose square is −1. Recall when we first defined 4 we did so as ’the number whose number square is 4.’ But there are two such numbers 2 and −2. By default, we declared to radical to mean the positive number √ whose square is 4. We follow a similar logic here, as we are confronted with the same dilemma. If we define −1 as the number whose square is −1, we will find there are two possible choices, i and −i (see examples). By convention, we will define negative radical to be, the positive i rather than −i. Examples 1. Simplify
√ −4
√
√ −4 = i 4 = i2 = 2i
2. Simplify
√ −10
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√ √ −10 = i 10
math hands
(neg rad) (def rad) (CoLM)
(neg rad)
pg. 2
Trigonometry Sec. 04 notes 3. Simplify
4. Simplify
√ −15
MathHands.com M´ arquez √ √ −15 = i 15
(neg rad)
√ −x
Solution: Notice, we do not know the value of x. We don’t know if x is positive √ or negative. This means we don’t know if −x is positive or negative therefore we don’t know if the radical −x is positive or negative.
5. Adding in C
3 + 5i + 2 + 3i
Solution:
3 + 5i + 2 + 3i = = (3 + 2) + (5i + 3i)
(given) (ALA)
= 3 + 2 + (5 + 3)i
(DL)
= 5 + 8i
(AT)
6. Multiplying in C (3 + 5i)(2 + 3i)
Solution:
(3 + 5i)(2 + 3i) = = 3 · 2 + 5i · 2 + 3 · 3i + 5i · 3i = 6 + 10i + 9i + 15i2
= 6 + (10 + 9)i + 15i
2
(given) (FOIL) (BI) (DL)
2
(AT) (Def of i)
= 6 + 19i + −15 = −9 + 19i
(BI) (BI)
= 6 + 19i + 15i = 6 + 19i + 15 · −1
7. Multiplying in C (4 +− 5i)(2 + 3i)
Solution:
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2007-2009 MathHands.com
math hands
pg. 3
Trigonometry Sec. 04 notes
MathHands.com M´ arquez (4 +− 5i)(2 + 3i) =
(given)
= 4 · 2 + 5i · 2 + 4 · 3i + 5i · 3i −
−
= 8 + 10i + 12i + 15i −
−
2
= 8 + ( 10 + 12)i + 15i −
−
= 6 + 2i + 15i −
2
2
(FOIL) (BI) (DL) (BI)
= 6 + 2i + 15 · −1 = 6 + 2i + 15
(Def of i) (NNT)
−
= 21 + 2i
(BI)
8. Multiplying in C (4 + 3i)(2 + 3i)
Solution:
(4 + 3i)(2 + 3i) =
(given)
= 8 + 12i + 6i + 9i
2
= 8 + 12i + 6i + 9 · 1 −
(FOIL) (def i)
= 8 + 12i + 6i + 9
(BI)
= 1 + 18i
(BI)
−
−
9. Multiplying in C i7
Solution:
i7 = iiiiiii 2 2 2
(+Expo)
=i i i i
(+Expo)
= −1 · −1 · −1 · i = −1 · i
(Def of i) (BI)
= −i
(MT)
And Now, Divide We have now introduced the imaginary number, their standard form ’a+bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide.
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pg. 4
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3 5i − 3 5i − 3 = ·1 i+3 i+3 = =
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2007-2009 MathHands.com
5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
math hands
pg. 5
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
=
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 11 −3 − 11i = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
2i + 1 −i+1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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math hands
pg. 6
Trigonometry Sec. 04 notes
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 5i − 2 2i + 3 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
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2007-2009 MathHands.com
2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
math hands
pg. 7
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Intro to Complex Numbers 1. Plot the following points. (a) 2 + 3i (b) 5 + 1i (c) −4 + −3i
(d) −4i
Solution:
8i 7i 6i 5i 4i b
3i
2 + 3i
2i b
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5 + 1i
5
6
7
8
9 10 11 12
-2i
−4 + −3i b
-3i
0 + −4i
-4i b -5i -6i -7i -8i
2. Plot the following points. (a) −1 + 4i
(b) 2 + 2i
(c) 5 + −5i
(d) −7i
Solution:
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2007-2009 MathHands.com
math hands
pg. 8
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
8i 7i 6i 5i
−1 + 4i b 4i 3i b
2i
2 + 2i
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5
6
7
8
9 10 11 12
-2i -3i -4i -5i
5b + −5i
-6i
0 + −7i
-7i b -8i
Simplify
√ −90
Solution: √ −90
4. 3. Simplify
√ = i 90 √ = i 9 · 10 √ √ = i 9 10 √ = i · 3 10 √ = 3i 10
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −150
Solution:
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math hands
pg. 9
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −150
5. Simplify
√ = i 150 √ = i 25 · 6 √ √ = i 25 6 √ = i·5 6 √ = 5i 6
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −375
Solution: √ −375
6. Simplify
√ = i 375 √ = i 25 · 15 √ √ = i 25 15 √ = i · 5 15 √ = 5i 15
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −18
Solution: √ −18
7. Simplify
√ = i 18 √ =i 9·2 √ √ =i 9 2 √ = i·3 2 √ = 3i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −50
Solution:
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math hands
pg. 10
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −50
8. Simplify
√ = i 50 √ = i 25 · 2 √ √ = i 25 2 √ = i·5 2 √ = 5i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −8
Solution: √ −8
9. Simplify
√ =i 8 √ = i 4·2 √ √ =i 4 2 √ = i·2 2 √ = 2i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −125
Solution: √ −125
10. Simplify
√ = i 125 √ = i 25 · 5 √ √ = i 25 5 √ = i·5 5 √ = 5i 5
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ 3 −8
Solution: −2 since (−2)3 = −8 11. Simplify
√ 3 −125
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math hands
pg. 11
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: −5 since (−5)3 = −125 12. Simplify
√ 3 −1
Solution: −1 since (−1)3 = −1 13. Add 3i + 2 + 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 5i) = 2 + (3i + 4) + 5i = 2 + (4 + 3i) + 5i = (2 + 4) + (3i + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i
(ALA) (CoLA) (ALA) (DL) (BI)
14. Add 2i + 7 + 9i + 3 Solution:
by convention, we write in ’standard form’, ”a + bi” (7 + 2i) + (3 + 9i) = 7 + (2i + 3) + 9i = 7 + (3 + 2i) + 9i = (7 + 3) + (2i + 9i) = (7 + 3) + (2 + 9)i = 10 + 11i
(ALA) (CoLA) (ALA) (DL) (BI)
15. Add 3i − 2 + − 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (−2 + 3i) + (4 + −5i) = −2 + (3i + 4) + −5i
= −2 + (4 + 3i) + −5i = (−2 + 4) + (3i + −5i)
= (−2 + 4) + (3 + −5)i = 2 + −2i
(ALA) (CoLA) (ALA) (DL) (BI)
16. Add 30i + 11 + 5i + 5
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pg. 12
Trigonometry Sec. 04 exercises Solution:
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” (11 + 30i) + (5 + 5i) = 11 + (30i + 5) + 5i = 11 + (5 + 30i) + 5i = (11 + 5) + (30i + 5i) = (11 + 5) + (30 + 5)i = 16 + 35i
(ALA) (CoLA) (ALA) (DL) (BI)
17. Add i + 1 + i + 2 Solution:
by convention, we write in ’standard form’, ”a + bi” (1 + 1i) + (2 + 1i) = 1 + (1i + 2) + 1i = 1 + (2 + 1i) + 1i = (1 + 2) + (1i + 1i) = (1 + 2) + (1 + 1)i = 3 + 2i
(ALA) (CoLA) (ALA) (DL) (BI)
18. Add 3i + 2 + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 0i) = 2 + (3i + 4) + 0i = 2 + (4 + 3i) + 0i = (2 + 4) + (3i + 0i) = (2 + 4) + (3 + 0)i = 6 + 3i
(ALA) (CoLA) (ALA) (DL) (BI)
19. Add 3i + i Solution:
by convention, we write in ’standard form’, ”a + bi” (0 + 3i) + (0 + 1i) = 0 + (3i + 0) + 1i = 0 + (0 + 3i) + 1i = (0 + 0) + (3i + 1i) = (0 + 0) + (3 + 1)i = 0 + 4i
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math hands
(ALA) (CoLA) (ALA) (DL) (BI)
pg. 13
Trigonometry Sec. 04 exercises 20. Multiply 3i + 2
MathHands.com M´ arquez 5i + 4
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 5i + 4 = (2)(4) + (2)(5i) + (3i)(4) + (3i)(5i) = 8 + 10i + 12i + 15i
2
9i + 3
(BI)
= 8 + 10i + 12i + 15(−1)
(def i)
= (8 + −15) + (10i + 12i) = (8 + −15) + (10 + 12)i
(CoLA, ALA,BI) (DL)
= −7 + 22i
21. Multiply 2i + 7
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 2i + 7 9i + 3 = (7)(3) + (7)(9i) + (2i)(3) + (2i)(9i) 2
= 21 + 63i + 6i + 18i = 21 + 63i + 6i + 18(−1)
(BI) (def i)
= (21 + −18) + (63i + 6i) = (21 + −18) + (63 + 6)i
(CoLA, ALA,BI) (DL)
= 3 + 69i
22. Multiply 3i − 2
− 5i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i − 2 − 5i + 4 = (−2)(4) + (−2)(−5i) + (3i)(4) + (3i)(−5i) = −8 + 10i + 12i + −15i
2
= −8 + 10i + 12i + −15(−1) = (−8 + 15) + (10i + 12i)
= (−8 + 15) + (10 + 12)i = 7 + 22i
23. Multiply 30i + 11
5i + 5
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(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
math hands
pg. 14
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: by convention, we write in ’standard form’, ”a + bi” 30i + 11 5i + 5 = (11)(5) + (11)(5i) + (30i)(5) + (30i)(5i) = 55 + 55i + 150i + 150i
2
= 55 + 55i + 150i + 150(−1) = (55 + −150) + (55i + 150i)
= (55 + −150) + (55 + 150)i = −95 + 205i
24. Multiply i + 1
i+2
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
25. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i)
(BI) (def i)
= (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i
(CoLA, ALA,BI) (DL)
= 8 + −8i + 12i + −12i = 8 + −8i + 12i + −12(−1)
= 20 + 4i
26. Multiply 3i − 1
i−1
(FOIL)
2
(BI)
Solution:
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math hands
pg. 15
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i)
(DL)
= −2 + −4i
i+2
(BI) (def i) (CoLA, ALA,BI)
= (1 + −3) + (−1 + −3)i
27. Multiply i + 1
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
28. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i) = 8 + −8i + 12i + −12i
2
= 8 + −8i + 12i + −12(−1) = (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i = 20 + 4i
29. Multiply 3i − 1
i−1
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution:
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math hands
pg. 16
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i) = (1 + −3) + (−1 + −3)i = −2 + −4i
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
30. multiply i3 Solution: i3 = i2 · i
= −1 · i = −i
(JAE) (def of i) (MT)
31. multiply i4 Solution: i4 = i2 · i2 = −1 · −1 =1
(JAE) (def of i) (NotNot)
32. multiply i6 Solution: i6 = i4 · i2
= 1 · −1 = −1
(JAE) (see previous problem) (MiD)
3
33. (3i + 2)
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math hands
pg. 17
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution:
(3i + 2)3
(given) 3
= (3i)
2
+
3
2
3 · (3i) (2)
2
+
3(3i)(2)
+
3
(2)
(famous, PP3)
= 27i + 54i + 36i + 8
(BI)
= 27(−i) + 54(−1) + 36(i) + 8 = −46 + 9i
(BI) (BI)
3
34. (2i + 1)
Solution:
(2i + 1)3
(given) 3
= (2i) 3
2
+
3 · (2i) (1)
2
2
+
3(2i)(1)
+
3
(1)
(famous, PP3)
= 8i + 12i + 6i + 1
(BI)
= 8(−i) + 12(−1) + 6(i) + 1 = −11 + −2i
(BI) (BI)
3
35. (−2i + 1)
Solution:
3
(−2i + 1)
(given) 3
= (−2i)
3
2
+ 2
3 · (−2i) (1)
+
= − 8i + 12i − 6i + 1
= −8(−i) + 12(−1) + −6(i) + 1 = −11 + 2i
2
3(−2i)(1)
+
3
(1)
(famous, PP3) (BI) (BI) (BI)
3
36. (−2i + 3)
Solution:
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math hands
pg. 18
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 3
(−2i + 3)
(given) 3
= (−2i)
2
+
3
2
3 · (−2i) (3)
+
2
3(−2i)(3)
3
+
(3)
= − 8i + 36i − 54i + 27 = −8(−i) + 36(−1) + −54(i) + 27
(famous, PP3) (BI) (BI)
= −9 + −46i
(BI)
37. (−1i + 1)3 Solution:
3
(−1i + 1)
(given) 3
= (−1i)
+
3
2
2
3 · (−1i) (1)
+
2
3(−1i)(1)
3
+
(1)
= − i + 3i − 3i + 1 = −1(−i) + 3(−1) + −3(i) + 1
(famous, PP3) (BI) (BI)
= −2 + −2i
(BI)
38. (1i + 1)3 Solution:
3
(1i + 1)
(given) 3
= (1i) 3
39. Divide
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2007-2009 MathHands.com
+ 2
2
3 · (1i) (1)
+
2
3(1i)(1)
+
3
(1)
(famous, PP3)
= i + 3i + 3i + 1 = 1(−i) + 3(−1) + 3(i) + 1
(BI) (BI)
= −2 + 2i
(BI)
7i + 1 7i + 3
math hands
pg. 19
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
40. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
41. Divide
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2007-2009 MathHands.com
5i + 2 4i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
math hands
pg. 20
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
42. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
43. Divide
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3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
math hands
pg. 21
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
44. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
45. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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math hands
pg. 22
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
46. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
47. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 48. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 49. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 50. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2
(just add exponents) (setting up to use fact i4 = 1)
3
= (1) · i2 =i
(yipei-kae-yeh...)
2
= −1
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51. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
52. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
53. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
54. Compute w/ Complex Numbers Calculate and write in standard form. i−5
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Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
55. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
56. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
57. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
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math hands
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
pg. 26
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
58. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
59. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
60. Compute w/ Complex Numbers Calculate and write in standard form. i11
Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1)
= (1) · i
=i
(ez as sundays)
61. Compute w/ Complex Numbers Calculate and write in standard form. i−6
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MathHands.com M´ arquez
Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
62. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
= (1)8 · i
(used i4 = 1)
=i
(ez as sundays)
63. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
64. Compute w/ Complex Numbers Calculate and write in standard form.
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1 1 √ +√ i 2 2
math hands
2
pg. 28
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i = √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
(FOIL) (used i2 = −1)
65. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 3 1 + i 2 2
(FOIL) (used i2 = −1)
66. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
67. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
68. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
(getting ready to use A3 = i)
10
= (i)
(used A3 = i)
= −1
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(ez as Sundays...)
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pg. 30
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
69. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 70. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
71. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 72. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 73. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 74. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
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More onComplex Numbers The Idea Last section was ’out of this world’. We introduced the imaginary number, their standard form ’a + bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide. The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3
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MathHands.com M´ arquez 5i − 3 5i − 3 = ·1 i+3 i+3 = =
5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi = 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 − 11i −3 11 = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so.. 2i + 3 − 5i − 2 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
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MathHands.com M´ arquez
Instructions 1. Divide
7i + 1 7i + 3
Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
2. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
3. Divide
4i + 2 5i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
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Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
4. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
5. Divide
3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
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Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
6. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
7. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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2007-2009 MathHands.com
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MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
8. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
9. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 10. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 11. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 12. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2 3
= (1) · i2 =i
(just add exponents) (setting up to use fact i4 = 1) (yipei-kae-yeh...)
2
= −1
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13. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
14. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
15. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
16. Compute w/ Complex Numbers Calculate and write in standard form. i−5
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Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
17. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
18. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
19. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
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20. Compute w/ Complex Numbers Calculate and write in standard form. i2
Solution: i2 = i0 · i1 0 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
0
(used i4 = 1) (ez as sundays)
= (1) · i =i
21. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
22. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 = i4 ·i = (1)
−1
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
23. Compute w/ Complex Numbers Calculate and write in standard form. i11
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Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1) (ez as sundays)
= (1) · i =i
24. Compute w/ Complex Numbers Calculate and write in standard form. i−6
Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
25. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
8
(used i4 = 1) (ez as sundays)
= (1) · i =i
26. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
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Trigonometry Sec. 5
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27. Compute w/ Complex Numbers Calculate and write in standard form.
1 1 √ +√ i 2 2
2
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
=
(FOIL) (used i2 = −1)
28. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 1 3 + i 2 2
(FOIL) (used i2 = −1)
29. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
30. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
31. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
10
= (i)
= −1
(getting ready to use A3 = i) (used A3 = i) (ez as Sundays...)
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Trigonometry Sec. 5
MathHands.com M´ arquez
32. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 33. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
34. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 35. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 36. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 37. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
c
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Trigonometry Sec. 8
MathHands.com M´ arquez
1. Vector Dot Product (a) Compute the following DOT product: < 1, 2 > · < 3, 5 > Solution: < 1, 2 > · < 3, 5 > = (1)(3) + (2)(5) = 3 + 10 = 13
(def of the DOT) (by Calc) (by Calc)
(b) Compute the following DOT product: < 3, 5 > · < 3, 5 > Solution: < 3, 5 > · < 3, 5 > = (3)(3) + (5)(5) = 9 + 25 = 34
(def of the DOT) (by Calc) (by Calc)
(c) Compute the following DOT product: < −3, 2 > · < 1, 2 > Solution: < −3, 2 > · < 1, 2 > = (−3)(1) + (2)(2) = −3 + 4 =1
(def of the DOT) (by Calc) (by Calc)
(d) Compute the following DOT product: < 1, 4 > · < −3, −5 > Solution: < 1, 4 > · < −3, −5 > = (1)(−3) + (4)(−5) = −3 + −20 = −23
(def of the DOT) (by Calc) (by Calc)
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Trigonometry Sec. 8
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(e) Compute the following DOT product: < 2, 3 > · < 3, −3 > Solution: < 2, 3 > · < 3, −3 > = (2)(3) + (3)(−3)
(def of the DOT)
= 6 + −9
(by Calc)
= −3
(by Calc)
(f) Compute the following DOT product: < 1, −4 > · < 3, 7 > Solution: < 1, −4 > · < 3, 7 > = (1)(3) + (−4)(7)
(def of the DOT)
= 3 + −28 = −25
(by Calc) (by Calc)
(g) < 2, 1, 3 > · < 1, 2, 1 > Solution: 2 + 2 + 3 = 7 2. Vector Dot Product (a) Compute the following DOT product: (1i + 2j) · (0i + 5j) Solution: (1i + 2j) · (0i + 5j) = (1 < 1, 0 > +2 < 0, 1 >) · (0 < 1, 0 > +5 < 0, 1 >) =< 1, 2 > · < 0, 5 > = (1)(0) + (2)(5) = 0 + 10 = 10
(def of i and j) (by Inspection) (def of the DOT) (by Calc) (by Calc)
(b) Compute the following DOT product: (3i + 0j) · (3i + 5j)
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Trigonometry Sec. 8
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Solution: (3i + 0j) · (3i + 5j) = (3 < 1, 0 > +0 < 0, 1 >) · (3 < 1, 0 > +5 < 0, 1 >) =< 3, 0 > · < 3, 5 > = (3)(3) + (0)(5) =9+0 =9
(def of i and j) (by Inspection) (def of the DOT) (by Calc) (by Calc)
(c) Compute the following DOT product: (−3i + 2j) · (1i + 2j) Solution: (−3i + 2j) · (1i + 2j) = (−3 < 1, 0 > +2 < 0, 1 >) · (1 < 1, 0 > +2 < 0, 1 >) =< −3, 2 > · < 1, 2 > = (−3)(1) + (2)(2) = −3 + 4
(def of i and j) (by Inspection) (def of the DOT) (by Calc)
=1
(by Calc)
(d) Compute the following DOT product: (1i + 4j) · (−3i + −5j) Solution: (1i + 4j) · (−3i + −5j) = (1 < 1, 0 > +4 < 0, 1 >) · (−3 < 1, 0 > + − 5 < 0, 1 >) =< 1, 4 > · < −3, −5 > = (1)(−3) + (4)(−5)
= −3 + −20 = −23
(def of i and j) (by Inspection) (def of the DOT) (by Calc) (by Calc)
3. Vector Dot Product to find magnitude (a) Use the dot product to find the magnitude of the following vector: ~v =< 1, 2 >
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Trigonometry Sec. 8
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Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 1, 2 > · < 1, 2 > (given) = (1)(1) + (2)(2) =1+4
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 5 √ ||~v || = 5
(def of the DOT) (by Calc) (by Calc)
||~v || ≈ 2.24
(b) Use the dot product to find the magnitude of the following vector: ~v =< 3, 5 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 3, 5 > · < 3, 5 > (given) = (3)(3) + (5)(5) = 9 + 25
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 34 √ ||~v || = 34 ||~v || ≈ 5.83
(def of the DOT) (by Calc) (by Calc)
(c) Use the dot product to find the magnitude of the following vector: ~v =< −3, 2 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< −3, 2 > · < −3, 2 > (given) = (−3)(−3) + (2)(2)
=9+4 ....thus.... ....then.. (assuming norm is positive)....
2
||~v || = 13 √ ||~v || = 13 ||~v || ≈ 3.61
(def of the DOT)
(by Calc) (by Calc)
(d) Use the dot product to find the magnitude of the following vector: ~v =< 1, 4 >
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Trigonometry Sec. 8
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Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 1, 4 > · < 1, 4 > (given) = (1)(1) + (4)(4) = 1 + 16
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 17 √ ||~v || = 17
(def of the DOT) (by Calc) (by Calc)
||~v || ≈ 4.12
(e) Use the dot product to find the magnitude of the following vector: ~v =< 2, 3 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 2, 3 > · < 2, 3 > (given) = (2)(2) + (3)(3) =4+9
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 13 √ ||~v || = 13 ||~v || ≈ 3.61
(def of the DOT) (by Calc) (by Calc)
(f) Use the dot product to find the magnitude of the following vector: ~v =< 1, −4 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 1, −4 > · < 1, −4 > (given) = (1)(1) + (−4)(−4) = 1 + 16 ....thus.... ....then.. (assuming norm is positive)....
2
||~v || = 17 √ ||~v || = 17 ||~v || ≈ 4.12
(def of the DOT)
(by Calc) (by Calc)
(g) Use the dot product to find the magnitude of the following vector: ~v =< 3, −4 >
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Trigonometry Sec. 8
MathHands.com M´ arquez
Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 3, −4 > · < 3, −4 > (given) = (3)(3) + (−4)(−4) = 9 + 16
....thus.... ....then.. (assuming norm is positive)....
(def of the DOT) (by Calc)
||~v ||2 = 25 √ ||~v || = 25
(by Calc)
||~v || ≈ 5
4. Vector Dot Product to find ’distance’ The DOT product provides a way to define the ’distance’ between two vectors, ~v and w. ~ So long as we can define subtraction of the vectors we can define the distance between them as follows: p dist(~v , w) ~ = ||~v − w|| ~ = (~v − w) ~ · (~v − w) ~ (a) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 1, 2 >
~v =< 3, 5 >
Solution: First note that ~v − w ~ =< 2, 3 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 2, 3 > · < 2, 3 > = (2)(2) + (3)(3) =4+9 ....thus.... ....then.. (assuming the distance is positive)....
2
||~v − w|| ~ = 13 √ ||~v − w|| ~ = 13
(from given) (def of the DOT) (by Calc) (by Calc)
||~v − w|| ~ ≈ 3.61
(b) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< −3, 2 >
~v =< 1, 2 >
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Trigonometry Sec. 8
MathHands.com M´ arquez
Solution: First note that ~v − w ~ =< 4, 0 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 4, 0 > · < 4, 0 > = (4)(4) + (0)(0)
(from given) (def of the DOT)
= 16 + 0
(by Calc)
2
....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ = 16 √ ||~v − w|| ~ = 16
(by Calc)
||~v − w|| ~ ≈4
(c) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 1, 4 >
~v =< −3, −5 >
Solution: First note that ~v − w ~ =< −4, −9 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< −4, −9 > · < −4, −9 > = (−4)(−4) + (−9)(−9) = 16 + 81 ....thus.... ....then.. (assuming the distance is positive)....
(from given)
(def of the DOT) (by Calc)
2
||~v − w|| ~ = 97 √ ||~v − w|| ~ = 97 ||~v − w|| ~ ≈ 9.85
(by Calc)
(d) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 2, 3 >
~v =< 3, −3 >
Solution: First note that ~v − w ~ =< 1, −6 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 1, −6 > · < 1, −6 > = (1)(1) + (−6)(−6) = 1 + 36
....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ 2 = 37 √ ||~v − w|| ~ = 37 ||~v − w|| ~ ≈ 6.08
(from given) (def of the DOT) (by Calc) (by Calc)
(e) Use the dot product to find the ’distance’ between the indicated vectors.
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Trigonometry Sec. 8
MathHands.com M´ arquez w ~ =< 1, −4 >
~v =< 3, 7 >
Solution: First note that ~v − w ~ =< 2, 11 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 2, 11 > · < 2, 11 >
= (2)(2) + (11)(11) = 4 + 121 ....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ 2 = 125 √ ||~v − w|| ~ = 125 ||~v − w|| ~ ≈ 11.18
(from given) (def of the DOT) (by Calc) (by Calc)
(f) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 3, −4 >
~v =< 3, 7 >
Solution: First note that ~v − w ~ =< 0, 11 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 0, 11 > · < 0, 11 >
= (0)(0) + (11)(11) = 0 + 121 ....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ 2 = 121 √ ||~v − w|| ~ = 121
(from given) (def of the DOT) (by Calc) (by Calc)
||~v − w|| ~ ≈ 11
(g) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 1, −3, 5, 0, 6 >
~v =< 5, 2, −10, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you.. 5. Vector Dot Product to find ’angle’ between two vectors (a) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, 2 >
~v =< 3, −5 >
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Trigonometry Sec. 8
MathHands.com M´ arquez
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 1, 2 > · < 3, −5 > = (given) || < 1, 2 > |||| < 3, −5 > || −7 −7 ≈ ≈ (by Calc) (2.236)(5.831) 13.038 cos θ = −0.54 (by Calc)
cos θ =
....thus....
θ ≈ cos−1 (−0.54)
θ ≈ 2.14 radians ≈ 122.6◦
~v
....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
~ w
(b) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< −3, 2 >
~v =< 3, 5 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle
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Trigonometry Sec. 8
MathHands.com M´ arquez
’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < −3, 2 > · < 3, 5 > = (given) || < −3, 2 > |||| < 3, 5 > || 1 1 ≈ ≈ (by Calc) (3.606)(5.831) 21.027 cos θ = 0.05 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0.05)
θ ≈ 1.52 radians
w ~
≈ 87.1◦
~v
(c) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, 4 >
~v =< −3, −5 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 1, 4 > · < −3, −5 > = (given) || < 1, 4 > |||| < −3, −5 > || −23 −23 ≈ ≈ (by Calc) (4.123)(5.831) 24.041 cos θ = −0.96 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (−0.96)
θ ≈ 2.86 radians ≈ 163.9◦
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~v
Trigonometry Sec. 8
w ~
(d) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 2, 3 >
~v =< 3, −3 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 2, 3 > · < 3, −3 > = (given) || < 2, 3 > |||| < 3, −3 > || −3 −3 ≈ ≈ (by Calc) (3.606)(4.243) 15.3 cos θ = −0.2 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (−0.2) θ ≈ 1.77 radians ≈ 101.4◦
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~v
Trigonometry Sec. 8
w~
(e) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 3, 7 >
~v =< 1, −4 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 3, 7 > · < 1, −4 > = (given) || < 3, 7 > |||| < 1, −4 > || −25 −25 ≈ ≈ (by Calc) (7.616)(4.123) 31.401 cos θ = −0.8 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (−0.8) θ ≈ 2.5 radians ≈ 143.2◦
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~v
Trigonometry Sec. 8
~ w
(f) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 3, 9 >
~v =< 7, 1 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 3, 9 > · < 7, 1 > = (given) || < 3, 9 > |||| < 7, 1 > || 30 30 ≈ ≈ (by Calc) (9.487)(7.071) 67.083 cos θ = 0.45 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0.45) θ ≈ 1.1 radians ≈ 63◦
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~v
Trigonometry Sec. 8
~ w
(g) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< −5, 2 >
~v =< 2, 5 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < −5, 2 > · < 2, 5 > = (given) || < −5, 2 > |||| < 2, 5 > || 0 0 ≈ ≈ (by Calc) (5.385)(5.385) 28.998 cos θ = 0 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0) θ ≈ 1.57 radians ≈ 90◦
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w ~
Trigonometry Sec. 8
~v
(h) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, 7 >
~v =< 7, −1 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 1, 7 > · < 7, −1 > = (given) || < 1, 7 > |||| < 7, −1 > || 0 0 ≈ ≈ (by Calc) (7.071)(7.071) 49.999 cos θ = 0 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0) θ ≈ 1.57 radians ≈ 90◦
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~v
Trigonometry Sec. 8
w ~
(i) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 2, 3 >
~v =< 3, −2 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 2, 3 > · < 3, −2 > = (given) || < 2, 3 > |||| < 3, −2 > || 0 0 ≈ ≈ (by Calc) (3.606)(3.606) 13.003 cos θ = 0 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0) θ ≈ 1.57 radians ≈ 90◦
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~v
Trigonometry Sec. 8
w ~
(j) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, −3, 5, 0, 6 >
~v =< 5, 2, −10, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you.. 6. Perpendicular Test by the DOT Find the Dot product for each pair of vectors, then determine if they are perpendicular. (a) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 1, 2 >
~v =< 3, −5 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, −5 > · < 1, 2 > = −7
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (b) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 3, 2 >
~v =< 2, −3 >
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Trigonometry Sec. 8
MathHands.com M´ arquez
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 2, −3 > · < 3, 2 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (c) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 1, 4 >
~v =< −3, −5 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< −3, −5 > · < 1, 4 > = −23
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (d) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 2, 3 >
~v =< 3, −3 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, −3 > · < 2, 3 > = −3
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (e) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 3, 6 >
~v =< 2, −1 >
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Trigonometry Sec. 8
MathHands.com M´ arquez
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 2, −1 > · < 3, 6 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (f) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 3, 12 >
~v =< −4, 1 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< −4, 1 > · < 3, 12 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (g) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< −5, 2 >
~v =< 3, 5 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, 5 > · < −5, 2 > = −5
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (h) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 1, 7 >
~v =< 7, −2 >
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Trigonometry Sec. 8
MathHands.com M´ arquez
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 7, −2 > · < 1, 7 >
(given)
= −7
(by inspection)
....therefore....~v and w ~ are not perpendicular. (i) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 2, 3 >
~v =< 3, −2 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, −2 > · < 2, 3 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (j) test to see if perpendicular... w ~ =< 1, −3, 5, −2, 6 >
~v =< 5, 0, −1, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you.. 7. Projections by the DOT (a) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 1, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 1, 2 > · < 5, 0 > = < 5, 0 > || < 5, 0 > ||2 5 ≈ < 5, 0 > 25 ≈ 0.2< 5, 0 >
projw~ ~v =
≈ < 1, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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~v
Trigonometry Sec. 8
w ~
(b) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 5, 0 > < 5, 0 > = || < 5, 0 > ||2 15 ≈ < 5, 0 > 25 ≈ 0.6< 5, 0 >
projw~ ~v =
≈ < 3, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v w ~
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Trigonometry Sec. 8
MathHands.com M´ arquez
(c) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 4, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 4, 2 > · < 5, 0 > = < 5, 0 > || < 5, 0 > ||2 20 ≈ < 5, 0 > 25 ≈ 0.8< 5, 0 > ≈ < 4, 0 >
projw~ ~v =
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v w ~
(d) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 7, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 7, 2 > · < 5, 0 > = < 5, 0 > || < 5, 0 > ||2 35 ≈ < 5, 0 > 25 ≈ 1.4< 5, 0 >
projw~ ~v =
≈ < 7, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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~v w ~
(e) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< −3, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < −3, 2 > · < 5, 0 > < 5, 0 > = || < 5, 0 > ||2 −15 ≈ < 5, 0 > 25 ≈ −0.6< 5, 0 >
projw~ ~v =
≈ < −3, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v w ~
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(f) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 1, 2 >
w ~ =< 0, 6 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 1, 2 > · < 0, 6 > = < 0, 6 > || < 0, 6 > ||2 12 ≈ < 0, 6 > 36 ≈ 0.333< 0, 6 > ≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v
w ~
projw~ ~v =
(g) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 0, 7 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 0, 7 > = < 0, 7 > || < 0, 7 > ||2 14 ≈ < 0, 7 > 49 ≈ 0.286< 0, 7 >
projw~ ~v =
≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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w ~
Trigonometry Sec. 8
~v
(h) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 4, 2 >
w ~ =< 0, 2 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 4, 2 > · < 0, 2 > < 0, 2 > = || < 0, 2 > ||2 4 ≈ < 0, 2 > 4 ≈ 1< 0, 2 >
projw~ ~v =
w ~
≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v
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(i) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 7, 2 >
w ~ =< 0, 4 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 7, 2 > · < 0, 4 > = < 0, 4 > || < 0, 4 > ||2 8 ≈ < 0, 4 > 16 ≈ 0.5< 0, 4 > ≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
w ~
projw~ ~v =
~v
(j) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< −3, 2 >
w ~ =< 0, 3 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < −3, 2 > · < 0, 3 > = < 0, 3 > || < 0, 3 > ||2 6 ≈ < 0, 3 > 9 ≈ 0.667< 0, 3 >
projw~ ~v =
≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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w ~
Trigonometry Sec. 8
~v
(k) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 1, 2 >
w ~ =< 1, 6 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 1, 2 > · < 1, 6 > < 1, 6 > = || < 1, 6 > ||2 13 ≈ < 1, 6 > 37 ≈ 0.351< 1, 6 >
projw~ ~v =
(given) (by Calc) (by Calc) (by Calc)
~v
w ~
≈ < 0.35, 2.11 >
(famous DOT product property)
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(l) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 7, 1 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 7, 1 > = < 7, 1 > || < 7, 1 > ||2 23 ≈ < 7, 1 > 50 ≈ 0.46< 7, 1 > ≈ < 3.22, 0.46 >
projw~ ~v =
~v
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~ w
(m) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 4, 2 >
w ~ =< 4, 2 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 4, 2 > · < 4, 2 > = < 4, 2 > || < 4, 2 > ||2 20 ≈ < 4, 2 > 20 ≈ 1< 4, 2 >
projw~ ~v =
≈ < 4, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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~v w
(n) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 7, 2 >
w ~ =< 8, 1 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 7, 2 > · < 8, 1 > < 8, 1 > = || < 8, 1 > ||2 58 ≈ < 8, 1 > 65 ≈ 0.892< 8, 1 >
(famous DOT product property)
projw~ ~v =
(given) (by Calc) (by Calc)
≈ < 7.14, 0.89 >
(by Calc)
~v
~ w
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(o) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 9, 3 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 9, 3 > = < 9, 3 > || < 9, 3 > ||2 33 ≈ < 9, 3 > 90 ≈ 0.367< 9, 3 > ≈ < 3.3, 1.1 >
(famous DOT product property)
projw~ ~v =
(given) (by Calc) (by Calc) (by Calc)
~ w ~v
(p) find projection ~v =< 1, −3, 5, −2, 6 >
ontow ~ =< 5, 0, −1, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you..
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1. Vector Arithmetic (a) Vector ADDITION Suppose ~v = and w ~ = < − 3, 5>, compute ~v + w ~ Solution: ~v + w ~ = + < − 3, 5>
(definition of addition on vectors) (by inspection)
~v
~ w ~v +
~ w
= = < − 2, 7>
(given)
(b) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = +
= =
(given) (definition of addition on vectors) (by inspection)
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~v
Trigonometry Sec. 7
~ w
~v + ~ w
(c) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = + =
(given) (definition of addition on vectors)
=
~ w
~v ~v +
(by inspection)
w ~
(d) Vector ADDITION Suppose ~v = and w ~ = < − 3, 1>, compute ~v + w ~
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Solution: ~v + w ~ = + < − 3, 1> = =
(given) (definition of addition on vectors) (by inspection)
~v +
w ~
w ~
~v
(e) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = +
= =
(given) (definition of addition on vectors) (by inspection)
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~ w
~v ~v + w ~
(f) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = + =
(given) (definition of addition on vectors) (by inspection)
w ~
=
~v +
~ w
~v
(g) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~
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Solution: ~v + w ~ = + =
(given) (definition of addition on vectors)
=
~ w
~v ~v + w ~
(by inspection)
(h) Vector ADDITION Suppose ~v = < − 3, 2> and w ~ = < − 1, 5>, compute ~v + w ~ Solution: ~v + w ~ = < − 3, 2> + < − 1, 5> = < − 3 + −1, = < − 4, 7>
2 + 5>
(given) (definition of addition on vectors) (by inspection)
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~v +
~ w
~ w
Trigonometry Sec. 7
~v
2. Vector Arithmetic (a) Vector SUBTRACTION Suppose ~v = and w ~ = < − 3, 5>, compute ~v − w ~ Solution: ~v − w ~ = − < − 3, 5> 2 − 5>
(definition of subtraction on vectors) (by inspection)
~v
=
(given) (definition of subtraction on vectors) (by inspection)
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~ ~v − w ~ −w
Trigonometry Sec. 7
~v
(d) Vector SUBTRACTION Suppose ~v = and w ~ = < − 3, 1>, compute ~v − w ~ Solution: ~v − w ~ = − < − 3, 1> =
(given) (definition of subtraction on vectors)
=
(by inspection)
−w ~
~v
~ ~v − w
(e) Vector SUBTRACTION Suppose ~v = and w ~ = , compute ~v − w ~
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Solution: ~v − w ~ = − =
(given) (definition of subtraction on vectors)
w~
(by inspection)
~ −w
=
~v −
~v
(f) Vector SUBTRACTION Suppose ~v = and w ~ = , compute ~v − w ~ Solution: ~v − w ~ = −
= =
(given) (definition of subtraction on vectors) (by inspection)
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−w ~
~v
~v − w ~
(g) Vector SUBTRACTION Suppose ~v = and w ~ = , compute ~v − w ~ Solution: ~v − w ~ = − =
(given) (definition of subtraction on vectors)
w~
(by inspection)
~ −w
=
~v −
~v
(h) Vector SUBTRACTION Suppose ~v = < − 3, 2> and w ~ = < − 1, 5>, compute ~v − w ~
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Solution: ~v − w ~ = < − 3, 2> − < − 1, 5> = < − 3 − −1, 2 − 5>
(given) (definition of subtraction on vectors)
= < − 2, −3>
(by inspection)
~v ~v −
~ −w
w ~
3. Vector Arithmetic (a) Vector SCALARS Suppose ~v = compute 2~v Solution: 2~v = 2 = =
(given) (def of scalar multiplication) (by inspection)
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2~v ~v
(b) Vector SCALARS Suppose ~v = compute 3~v Solution: 3~v = 3 =
(given) (def of scalar multiplication)
=
(by inspection)
3~v ~v
(c) Vector SCALARS Suppose ~v = compute −2~v
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Solution: −2~v = −2 = < − 2(2), −2(0)>
(given) (def of scalar multiplication)
= < − 4, 0>
(by inspection)
−2~v ~v
(d) Vector SCALARS Suppose ~v = compute 2~v Solution: 2~v = 2 = =
(given) (def of scalar multiplication) (by inspection)
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2~v
~v
(e) Vector SCALARS Suppose ~v = compute 3~v Solution: 3~v = 3
(given)
= =
~v
(def of scalar multiplication) (by inspection)
3~v
(f) Vector SCALARS Suppose ~v = compute 2~v
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Solution: 2~v = 2 =
(given) (def of scalar multiplication) (by inspection)
~v
2~v
=
(g) Vector SCALARS Suppose ~v = compute 3~v Solution: 3~v = 3 = =
(given) (def of scalar multiplication) (by inspection)
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~v
3~v
Trigonometry Sec. 7
(h) Vector SCALARS Suppose ~v = compute 1.5~v Solution: 1.5~v = 1.5
(given) (def of scalar multiplication) (by inspection)
~v
1.5~v
= =
(i) Vector SCALARS Suppose ~v = compute .75~v
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Solution: .75~v = .75 =
(given) (def of scalar multiplication) (by inspection)
~v
.75~v
=
(j) Vector SCALARS Suppose ~v = compute −1~v Solution: −1~v = −1
= < − 1(4), −1(2)> = < − 4, −2>
(given) (def of scalar multiplication) (by inspection)
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~v −1
~v
4. Vector Arithmetic: Famous Vectors There are two very famous vectors, there are: i =< 1, 0 > and j =< 0, 1 >. (a) FAMOUS VECTORS i and j Compute and draw the following vectors 4i + 2j Solution: let ~v = 4i + 2j = 4 < 1, 0 > +2 < 0, 1 > =< 4, 0 > + < 0, 2 >
(given) (def of scalar multiplication)
4i +
2j
~ 2j
=< 4, 2 >
~ 4i
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(b) FAMOUS VECTORS i and j Compute and draw the following vectors 1i + −2j Solution: let ~v = 1i + −2j
= 1 < 1, 0 > + − 2 < 0, 1 >
=< 1, 0 > + < 0, −2 > =< 1, −2 >
(given) (def of scalar multiplication)
~ 1i ~ −2j j −2
1i +
(c) FAMOUS VECTORS i and j Compute and draw the following vectors 5i + −2j Solution: let ~v = 5i + −2j = 5 < 1, 0 > + − 2 < 0, 1 > =< 5, 0 > + < 0, −2 > =< 5, −2 >
(given) (def of scalar multiplication)
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~ 5i
−2j
~ −2j
5i +
(d) FAMOUS VECTORS i and j Compute and draw the following vectors 5i + 1j Solution: let ~v = 5i + 1j = 5 < 1, 0 > +1 < 0, 1 > =< 5, 0 > + < 0, 1 >
(given) (def of scalar multiplication)
j 5i + 1 ~ 5i
~ 1j
=< 5, 1 >
(e) FAMOUS VECTORS i and j Compute and draw the following vectors 6i + 3j
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Solution: let ~v = 6i + 3j = 6 < 1, 0 > +3 < 0, 1 >
(given) (def of scalar multiplication)
6i +
3j
~ 3j
=< 6, 0 > + < 0, 3 > =< 6, 3 >
~ 6i
(f) FAMOUS VECTORS i and j Compute and draw the following vectors 3i + 4j Solution: let ~v = 3i + 4j = 3 < 1, 0 > +4 < 0, 1 > =< 3, 0 > + < 0, 4 >
(given) (def of scalar multiplication)
=< 3, 4 >
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3i +
~ 4j
4j
Trigonometry Sec. 7
~ 3i
5. NORM of a VECTOR Find the norm of the indicated vector. (a) Compute: || < 4, 2 > || Solution: p (4)2 + (2)2 √ = 20 ≈ 4.47
.47 ≈4
(def of norm) (by Inspection) (by Calc)
2
|| < 4, 2 > || =
4
(b) Compute:
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Solution: p (1)2 + (−2)2 √ = 5 ≈ 2.24
|| < 1, −2 > || =
(def of norm) (by Inspection) (by Calc)
1 −2 .24 ≈2
(c) Compute: || < 5, −2 > || Solution: p (5)2 + (−2)2 √ = 29 ≈ 5.39
|| < 5, −2 > || =
(def of norm) (by Inspection) (by Calc)
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5 −2
≈5 .39
(d) Compute: || < 5, 1 > || Solution: p (5)2 + (1)2 √ = 26 ≈ 5.1
≈ 5.1 5
(def of norm) (by Inspection) (by Calc)
1
|| < 5, 1 > || =
(e) Compute: || < 6, 3 > || ©2007-2008 MathHands.com
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Solution: p (6)2 + (3)2 √ = 45
|| < 6, 3 > || =
(def of norm) (by Inspection)
.71 ≈6
(by Calc)
3
≈ 6.71
6
(f) Compute: || < 5, 0 > || Solution: p (5)2 + (0)2 √ = 25 ≈5
|| < 5, 0 > || =
(def of norm) (by Inspection) (by Calc)
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≈5 5
0
6. Which is Bigger? (a) Determine which number is larger: ||4i + 2j|| OR (||4i|| + ||2j||) Solution: ||4i + 2j|| = || < 4, 2 > || p = (4)2 + (2)2 √ = 20 ≈ 4.47
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||4i|| + ||2j|| = || < 4, 0 > || + || < 0, 2 > || =4+2 =6
(def of i and j) (def of norm) (by Inspection)
Therefore, ||4i|| + ||2j|| is larger. (b) Determine which number is larger: ||1i + −2j|| OR (||1i|| + || − 2j||) Solution: ||1i + −2j|| = || < 1, −2 > || p = (1)2 + (−2)2 √ = 5 ≈ 2.24
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
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||1i|| + || − 2j|| = || < 1, 0 > || + || < 0, −2 > || =1+2 =3
(def of i and j) (def of norm) (by Inspection)
Therefore, ||1i|| + || − 2j|| is larger. (c) Determine which number is larger: ||5i + −2j|| OR (||5i|| + || − 2j||) Solution: ||5i + −2j|| = || < 5, −2 > || p = (5)2 + (−2)2 √ = 29 ≈ 5.39
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||5i|| + || − 2j|| = || < 5, 0 > || + || < 0, −2 > || =5+2 =7
(def of i and j) (def of norm) (by Inspection)
Therefore, ||5i|| + || − 2j|| is larger. (d) Determine which number is larger: ||5i + 1j|| OR (||5i|| + ||1j||) Solution: ||5i + 1j|| = || < 5, 1 > || p = (5)2 + (1)2 √ = 26 ≈ 5.1
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||5i|| + ||1j|| = || < 5, 0 > || + || < 0, 1 > || =5+1 =6
(def of i and j) (def of norm) (by Inspection)
Therefore, ||5i|| + ||1j|| is larger. (e) Determine which number is larger: ||6i + 3j|| OR (||6i|| + ||3j||)
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Solution: ||6i + 3j|| = || < 6, 3 > || p = (6)2 + (3)2 √ = 45 ≈ 6.71
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||6i|| + ||3j|| = || < 6, 0 > || + || < 0, 3 > || =6+3 =9
(def of i and j) (def of norm) (by Inspection)
Therefore, ||6i|| + ||3j|| is larger. (f) Determine which number is larger: ||5i + 0j|| OR (||5i|| + ||0j||) Solution: ||5i + 0j|| = || < 5, 0 > || p = (5)2 + (0)2 √ = 25 ≈5
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||5i|| + ||0j|| = || < 5, 0 > || + || < 0, 0 > || =5+0 =5
(def of i and j) (def of norm) (by Inspection)
Therefore, ||5i|| + ||0j|| is larger. (g) ||u + v|| OR (||u|| + ||v||) Solution: the idea is to study the above pattern and understand that it will always be the case that the sum of the individual norms is larger or equal to the norm of the sum of the vectors. In some sense, this is equivalent to saying that the sum of the lengths of any two sides of a [Euclidean] triangle have to be larger than the size of the length of the the third side of the triangle. This is very famous, it is called the triangle inequality. 7. Normalize this.. Find the normalized vector for each: (a) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< 5, 2 >
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Solution: First we find the norm of the vector: p (5)2 + (2)2 √ = 29
||~v || = || < 5, 2 > || =
(def of norm) (by Inspection)
≈ 5.39 Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(by Calc)
Let us denote the normalized vector ~v as ”v~n ”
1 ||~v || 5 2 = , ||~v || ||~v || 5 2 ≈ , 5.39 5.39
v~n =
(given) (def of scalar multiplication) (approximate)
≈ h0.93, 0.37i
(by inspection)
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
~v v~n
(b) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< 5, −2 > Solution: First we find the norm of the vector: p (5)2 + (−2)2 √ = 29 ≈ 5.39
||~v || = || < 5, −2 > || =
(def of norm) (by Inspection) (by Calc)
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Then we scale the original vector ~v by multiplying by
1 ||~ v || .
Let us denote the normalized vector ~v as ”v~n ”
1 ||~v || 5 −2 = , ||~v || ||~v || −2 5 , ≈ 5.39 5.39
v~n =
(given) (def of scalar multiplication) (approximate) (by inspection)
≈ h0.93, −0.37i
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
v~n
~v
(c) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< −3, 4 > Solution: First we find the norm of the vector: p (−3)2 + (4)2 √ = 25 ≈5
||~v || = || < −3, 4 > || =
Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(def of norm) (by Inspection) (by Calc)
Let us denote the normalized vector ~v as ”v~n ”
©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez 1 < − 3, 4> ||~v || −3 4 = , ||~v || ||~v || −3 4 , ≈ 5 5
(given)
v~n =
(def of scalar multiplication) (approximate)
≈ h−0.6, 0.8i
(by inspection)
v~n
~v
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
(d) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< −3, −4 > Solution: First we find the norm of the vector: p (−3)2 + (−4)2 √ = 25
||~v || = || < −3, −4 > || =
≈5 Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(def of norm) (by Inspection) (by Calc)
Let us denote the normalized vector ~v as ”v~n ”
©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez 1 < − 3, −4> ||~v || −3 −4 = , ||~v || ||~v || −3 −4 , ≈ 5 5
(given)
v~n =
(def of scalar multiplication) (approximate) (by inspection)
≈ h−0.6, −0.8i
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
v~n
~v
(e) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< −3, −1 > Solution: First we find the norm of the vector: p (−3)2 + (−1)2 √ = 10
||~v || = || < −3, −1 > || =
≈ 3.16 Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(def of norm) (by Inspection) (by Calc)
Let us denote the normalized vector ~v as ”v~n ”
©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez 1 < − 3, −1> ||~v || −3 −1 = , ||~v || ||~v || −3 −1 , ≈ 3.16 3.16
v~n =
≈ h−0.95, −0.32i
(given) (def of scalar multiplication) (approximate) (by inspection)
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
v~n ~v
(f) ~v =< a, b > (not both zero..) Solution: 1 ||~v || a b = , ||~v || ||~v || b a ,√ = √ a2 + b 2 a2 + b 2
v~n =
(given) (def of scalar multiplication) (def of norm)
©2007-2008 MathHands.com
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
Euler’s Identity
◦
1. Convert Convert 3ei30 to standard form, a + bi Solution: According to Euler’s amazing identity..eiθ = cos θ + i sin θ..thus..
◦
3ei30
= 3[cos 30◦ + i sin 30◦ ] "√ # 3 1 = 3 +i 2 2 √ 3 3 3 +i = 2 2
Alternatively, you could convert using the picture... and convert in very much the same way we convert from polar to cartesian...
3
b
3
“√ ” 3 2
+3
`1´ 2
i
30◦
◦
2. Convert Convert 3e−i30 to standard form, a + bi Solution: According to Euler’s amazing identity..eiθ = cos θ + i sin θ..thus..
◦
3e−i30
= 3[cos −30◦ + i sin −30◦ ] # "√ −1 3 +i = 3 2 2 √ 3 3 −3 = + i 2 2
Or use the picture... Or use polar conversion ideas.. (r, θ) = (3, −30◦) 3. Convert Convert ei 2 to standard form, a + bi π
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math hands
pg. 3
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
Solution: According to Euler’s amazing identity..eiθ = cos θ + i sin θ..thus..
ei 2
π
= = =
4. Convert Convert e
3π 2 i
to standard form, a + bi
5. Convert Convert e
7π 6 i
to standard form, a + bi
π π + i sin 2 2 0 + i(1) i cos
6. Convert Convert 3e2i to standard form, a + bi 7. Convert Convert 5e3.14i to standard form, a + bi 8. Convert Convert e6.28i to standard form, a + bi 9. Convert Convert 2 + i to Euler form, reiθ
Solution: 2 + i =
√ i26.57◦ 5e many other correct possibilities for theta..
10. Convert Convert 2 − 2i to Euler form, reiθ √ √ √ ◦ ◦ ◦ Solution: 2 − 2i = (2 2)e−i45 OR (2 2)ei315 or (2 2)ei765 or infinite many more possibilities for θ 11. Convert Convert −2 − 2i to Euler form, reiθ √ ◦ Solution: −2 − 2i = (2 2)ei225 many other correct possibilities for theta.. 12. Convert Convert −3 + 2i to Euler form, reiθ 13. Convert Convert i to Euler form, reiθ ◦
Solution: i = ei90 many other correct possibilities for theta.. note r = 1 14. Convert Convert 3i to Euler form, reiθ ◦
Solution: i = 3ei90 many other correct possibilities for theta.. note r = 1 15. Convert Convert .5 + 3i to Euler form, reiθ 16. Convert Convert −1 to Euler form, reiθ ◦
Solution: i = ei180 many other correct possibilities for theta.. note r = 1
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math hands
pg. 4
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
17. Convert Convert −i to Euler form, reiθ ◦
Solution: i = ei270 many other correct possibilities for theta.. note r = 1
18. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Multiply the old fashion way... ! √ 3 1 + i · 2 2
◦
◦
e30 i · e60
√ ! 1 3 + i 2 2
i
◦
(b) Multiply the Euler way, and simplify answer.. (note
Solution: ei90 = i
19. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Multiply the old fashion way... ! √ 3 3 3 + i · 2 2
◦
◦
3e30 i · 2e60
√ ! 2 2 3 + i 2 2
i
◦
(b) Multiply the Euler way, and simplify answer.. (note
Solution: 6ei90 = 6i
20. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Multiply the old fashion way... ! √ 3 1 + i · 2 2
◦
◦
e30 i · e30
! √ 3 1 + i 2 2
i
◦
(b) Multiply the Euler way, and simplify answer.. (note
Solution: ei60
21. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Divide the old fashion way...
◦
e30 i e60◦ i
√ 3 1 2 +√2 i 1 3 2 + 2 i ◦
(b) Divide the Euler way, and simplify answer.. (note
Solution: e−i30
22. Euler’s World Multiply or divide as indicated (b) Calculate, the Euler way, and simplify answer.. (note these are the same numbers..) ◦ 3 e30 i
(a) Expand the binomial, the old fashion way... !3 √ 3 1 + i 2 2
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pg. 5
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez ◦
Solution: ei90 = i compare this problem with #57 from two sections ago.
23. Euler’s World Multiply or divide as indicated (a) Expand the binomial, the old fashion way (or not..) ... !7 √ 3 1 + i 2 2
these are the same numbers..)
◦
e30
i
7
◦
Solution: ei210
(b) Calculate, the Euler way, and simplify answer.. (note
24. Euler’s World Convert the numbers to Euler form, then compute. (a)
(b) 3 + 4i −2 − 5i
(3 + 4i)(−2 − 5i) 25. Euler’s World to find Roots (a) Try to find a square root of i, i.e. a number x such that x2 = i √ i
(note these are the same numbers..)
◦
e90
i
21
◦
(b) Find a root, the Euler way, and simplify answer..
Solution: ei45
26. Euler’s World to find Roots (a) Try to find a square root: (after a good honest attempt, go to part b.) s
(note these are the same numbers..)
√ 1 3 + i 2 2
◦
e60
i
12
◦
(b) Find a root, the Euler way, and simplify answer..
Solution: ei30
27. Euler’s World to find Roots (a) Try to find a fourth root: (after a good honest attempt, go to part b.) s 4
swer.. (note these are the same numbers..)
√ 16 16 3 + i 2 2
◦
16e60
i
41
◦
(b) Find a fourth root, the Euler way, and simplify an-
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math hands
Solution: 2ei15
pg. 6
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
28. FTA and Unity Roots The two solutions to x2 = 1 are x = 1, −1. Show how one can derive these solutions. 29. FTA and Unity Roots The three solutions to x3 = 1 are x = 1, these solutions.
−1 2
+
√ 3 −1 2 i, 2
−
√ 3 2 i.
Show how one can derive
30. FTA and Unity Roots The four solutions to x4 = 1 are x = 1, i, −1, −i. Show how one can derive these solutions. ◦
31. FTA and Unity Roots To find the five solutions to x5 = 1, start with writing 1 = ek360 i , then start finding 5th roots, every value of k should produce one... ◦
Solution: for integers k, eik72
32. Euler’s World Another Proof of MOTA (a) Multiply the old fashion way
(b) multiply, the Euler way, and simplify answer.. (note these are the same numbers..)
(cos x + i sin x)(cos y + i sin y)
eix · eiy (c) Compare the real part of the answer for a, with the real part of the answer for part b.
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math hands
pg. 7
Trigonometry Sec. 05 notes
MathHands.com M´ arquez
Converting Complex numbers From and To EULER FORM Main Idea Perhaps the most celebrated identity of mathematics is Euler’s World known timeless identity: eiθ = cos θ + i sin θ Among other things, it’s power lies on the idea that a binomial (on the right side) can be converted to a monomial (left side). Indeed, every complex number, a + bi can be written in Euler Form as reiθ . It will serve us well to think about Euler’s identity as a sort of bridge between the binomials numbers which make up the entire complex plane and the monomial world expressed in Euler Form. Moreover, we have already crossed this bridge. You will see the ideas are identical to the ideas used in converting from cartesian coordinated to polar coordinates in first section of this chapter. Converting: EULER TO & FORM Standard Form The translating ’dictionary’ to go from Euler-to-Standard is as follows: From the defining features of the polar and cartesian cordinates, we conclude that generally:
reiθ = r[eiθ ]
(getting ready to use Euler’s ID, eiθ = cos (θ) + i sin (θ))
= r[cos (θ) + i sin (θ)]
(used eiθ = cos (θ) + i sin (θ))
= r cos (θ) + ir sin (θ) = x + iy
(algebra) (used polar dictionary)
and similarly going backwards. Therefore, we have the translating ’dictionary’ to translate from EULER TO& FROM STANDARD Form
MONOMIALS
BINOMIALS
90◦
6 60◦ 120◦
5 4 30◦
reiθ = x + yi 3
150◦
reb iθ r
x+ yi b
2
r
y
y
1
θ
x
θ
0◦
-6
180◦
-5
-4
-3
-2
-1
1
x 2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
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pg. 1
Trigonometry Sec. 05 notes
MathHands.com M´ arquez
EXAMPLEs: Polar Coordinates < −− > Cartesian Coordinates ◦
Convert the Euler Form, 5ei250 to Standard form of the complex number.
6 90◦
5 60◦
5ei250 = 5 cos (i250◦) + i5 sin (i250◦) (Euler, Baby!!) ◦
120◦
4
≈ −1.71 + −4.7i
30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦ 210◦
-3 b
5ei250
-4
◦
300◦
b
240◦
−1.71 + −4.7i-5
270◦
-6
Convert the cartesian coordinates, (-4, 3), to polar coordinates.
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pg. 2
Trigonometry Sec. 05 notes
MathHands.com M´ arquez r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4
(r, θ) 150◦
(−4, 3)
30◦
b
b
3 2 1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 3/ − 4 (note: eq has infinite solutions) ◦ θ ≈ −36.87 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−4)2 + (3)2 r2 =25 r =±
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√ 25 ≈ ±5
note: if we use the angle −36.87◦, then we must choose the negative value of r, r ≈ −5, thus the polar cordinates (−5, −36.87◦) is one way to represent the point (−4, 3)
math hands
pg. 3
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
Converting Complex numbers From and To EULER FORM ◦
1. Convert the Euler Form, 5ei240 to Standard form of the complex number. Solution: 5ei(240
◦
)
= 5 cos (240◦) + i5 sin (240◦ ) √ −5 −5 3 + i = 2 2 ≈ −2.5 + −4.33i
(Euler, Baby!!)
◦
2. Convert the Euler Form, 7ei210 to Standard form of the complex number. Solution: 7ei(210
◦
)
= 7 cos (210◦) + i7 sin (210◦ ) √ −7 3 −7 = + i 2 2 ≈ −6.06 + −3.5i
(Euler, Baby!!)
◦
3. Convert the Euler Form, 7ei330 to Standard form of the complex number. Solution: 7ei(330
◦
)
= 7 cos (330◦) + i7 sin (330◦ ) √ 7 3 −7 + i = 2 2 ≈ 6.06 + −3.5i
(Euler, Baby!!)
◦
4. Convert the Euler Form, 5ei150 to Standard form of the complex number. Solution: 5ei(150
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2007-2009 MathHands.com
◦
)
= 5 cos (150◦) + i5 sin (150◦ ) √ −5 3 5 + i = 2 2 ≈ −4.33 + 2.5i
math hands
(Euler, Baby!!)
pg. 4
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez ◦
5. Convert the Euler Form, 2ei240 to Standard form of the complex number. Solution: 2ei(240
◦
)
= 2 cos (240◦) + i2 sin (240◦ ) √ −2 −2 3 = + i 2 2 ≈ −1 + −1.73i
(Euler, Baby!!)
◦
6. Convert the Euler Form, 7ei135 to Standard form of the complex number. Solution: 7ei(135
◦
)
= 7 cos (135◦) + i7 sin (135◦ ) ≈ −4.95 + 4.95i
(Euler, Baby!!)
◦
7. Convert the Euler Form, 3ei−30 to Standard form of the complex number. Solution: ◦
3ei(−30
)
= 3 cos (−30◦ ) + i3 sin (−30◦ ) √ 3 3 −3 + i = 2 2 ≈ 2.6 + −1.5i
(Euler, Baby!!)
◦
8. Convert the Euler Form, 2ei−90 to Standard form of the complex number. Solution: ◦
2ei(−90
)
= 2 cos (−90◦ ) + i2 sin (−90◦ ) √ 0 −2.31 3 = + i 2 2 ≈ 0 + −2i
(Euler, Baby!!)
◦
9. Convert the Euler Form, 4ei180 to Standard form of the complex number.
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pg. 5
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
Solution: 4ei(180
◦
)
= 4 cos (180◦) + i4 sin (180◦ ) √ −8 0 3 + i = 2 2 ≈ −4 + 0i
(Euler, Baby!!)
◦
10. Convert the Euler Form, 7ei225 to Standard form of the complex number. Solution: 7ei(225
◦
)
= 7 cos (225◦) + i7 sin (225◦ )
(Euler, Baby!!)
≈ −4.95 + −4.95i
◦
11. Convert the Euler Form, 5ei−150 to Standard form of the complex number. Solution: ◦
5ei(−150
)
= 5 cos (−150◦) + i5 sin (−150◦) √ −5 3 −5 = + i 2 2 ≈ −4.33 + −2.5i
(Euler, Baby!!)
◦
12. Convert the Euler Form, 5ei−300 to Standard form of the complex number. Solution: ◦
5ei(−300
)
= 5 cos (−300◦) + i5 sin (−300◦) √ 5 5 3 = + i 2 2 ≈ 2.5 + 4.33i
(Euler, Baby!!)
◦
13. Convert the Euler Form, 2ei400 to Standard form of the complex number. Solution: 2ei(400
◦
)
= 2 cos (400◦) + i2 sin (400◦ )
(Euler, Baby!!)
≈ 1.53 + 1.29i
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pg. 6
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez ◦
14. Convert the Euler Form, 3ei360 to Standard form of the complex number. Solution: 3ei(360
◦
)
= 3 cos (360◦) + i3 sin (360◦ ) √ 6 −0 3 = + i 2 2 ≈ 3 + −0i
(Euler, Baby!!)
◦
15. Convert the Euler Form, 2ei3600 to Standard form of the complex number. Solution: 2ei(3600
◦
)
= 2 cos (3600◦) + i2 sin (3600◦) √ 4 −0 3 i = + 2 2 ≈ 2 + −0i
(Euler, Baby!!)
◦
16. Convert the Euler Form, 4ei−180 to Standard form of the complex number. Solution: ◦
4ei(−180
)
= 4 cos (−180◦) + i4 sin (−180◦) √ −8 −0 3 + i = 2 2 ≈ −4 + −0i
(Euler, Baby!!)
◦
17. Convert the Euler Form, 5ei45 to Standard form of the complex number. Solution: ◦
5ei(45
)
= 5 cos (45◦ ) + i5 sin (45◦ ) ≈ 3.54 + 3.54i
(Euler, Baby!!)
◦
18. Convert the Euler Form, 5ei225 to Standard form of the complex number.
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pg. 7
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
Solution: 5ei(225
◦
)
= 5 cos (225◦) + i5 sin (225◦ ) ≈ −3.54 + −3.54i
(Euler, Baby!!)
19. Convert the complex number 4 + 5i to EULER form. Solution: r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
reb iθ
120◦
6
4+ 5i b
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 5/4
(note: eq has infinite solutions)
θ ≈ 51.34◦ (note: this is just one [acceptable] value for θ) r2 =x2 + y 2 note: if we use the angle 51.34◦, then we must choose the positive value of r, r ≈ 6.4, thus the Euler Form
r2 =(4)2 + (5)2 r2 =41 r =±
√
◦
6.4ei(51.34
41 ≈ ±6.4
)
is one way to represent the complex number 4 + 5i
20. Convert the complex number −4 + 1i to EULER form. Solution:
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pg. 8
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2
reb iθ
−4b + 1i
1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
tan θ = 1/ − 4
270◦
(note:-6 eq has infinite solutions)
θ ≈ −14.04◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2
note: if we use the angle −14.04◦, then we must choose the negative value of r, r ≈ −4.12, thus the polar cordinates
r2 =(−4)2 + (1)2 r2 =17 r =±
√ 17 ≈ ±4.12
◦
−4.12ei(−14.04
)
is one way to represent the complex number −4 + 1i
21. Convert the complex number −3 + −4i to EULER form. Solution:
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pg. 9
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦ 210◦
-3
reb iθ
−3 b + −4i
-4 -5
300◦ 240◦
-6 270◦
tan θ = −4/ − 3
(note: eq has infinite solutions)
θ ≈ 53.13◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 note: if we use the angle 53.13◦, then we must choose the negative value of r, r ≈ −5, thus the polar cordinates ◦ −5ei(53.13 )
r2 =(−3)2 + (−4)2 r2 =25 r =±
√ 25 ≈ ±5
is one way to represent the complex number −3+−4i
22. Convert the complex number −1 + 5i to EULER form. Solution:
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pg. 10
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦
120◦
60◦
reb iθ
6
−1b + 5i 5 4
30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
tan θ = 5/ − 1
270◦
(note:-6 eq has infinite solutions)
θ ≈ −78.69◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2
note: if we use the angle −78.69◦, then we must choose the negative value of r, r ≈ −5.1, thus the polar cordinates
r2 =(−1)2 + (5)2 r2 =26 r =±
√
◦
−5.1ei(−78.69
26 ≈ ±5.1
)
is one way to represent the complex number −1 + 5i
23. Convert the complex number 1 + −5i to EULER form. Solution:
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pg. 11
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4
reb iθ
-5 300◦
1 +b −5i
240◦
-6 270◦
tan θ = −5/1
(note: eq has infinite solutions)
θ ≈ −78.69◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(1)2 + (−5)2 r2 =26 r =±
√
26 ≈ ±5.1
note: if we use the angle −78.69◦, then we must choose the positive value of r, r ≈ 5.1, thus the Euler Form ◦ 5.1ei(−78.69 ) is one way to represent the complex number 1 + −5i
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pg. 12
Trigonometry Sec. 5
MathHands.com M´ arquez
More onComplex Numbers The Idea Last section was ’out of this world’. We introduced the imaginary number, their standard form ’a + bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide. The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3
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Trigonometry Sec. 5
MathHands.com M´ arquez 5i − 3 5i − 3 = ·1 i+3 i+3 = =
5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi = 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 − 11i −3 11 = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so.. 2i + 3 − 5i − 2 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
Instructions 1. Divide
7i + 1 7i + 3
Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
2. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
3. Divide
4i + 2 5i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
4. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
5. Divide
3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
6. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
7. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
8. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
9. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 10. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 11. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 12. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2 3
= (1) · i2 =i
(just add exponents) (setting up to use fact i4 = 1) (yipei-kae-yeh...)
2
= −1
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
13. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
14. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
15. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
16. Compute w/ Complex Numbers Calculate and write in standard form. i−5
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Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
17. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
18. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
19. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
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Trigonometry Sec. 5
MathHands.com M´ arquez
20. Compute w/ Complex Numbers Calculate and write in standard form. i2
Solution: i2 = i0 · i1 0 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
0
(used i4 = 1) (ez as sundays)
= (1) · i =i
21. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
22. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 = i4 ·i = (1)
−1
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
23. Compute w/ Complex Numbers Calculate and write in standard form. i11
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1) (ez as sundays)
= (1) · i =i
24. Compute w/ Complex Numbers Calculate and write in standard form. i−6
Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
25. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
8
(used i4 = 1) (ez as sundays)
= (1) · i =i
26. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
27. Compute w/ Complex Numbers Calculate and write in standard form.
1 1 √ +√ i 2 2
2
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
=
(FOIL) (used i2 = −1)
28. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 1 3 + i 2 2
(FOIL) (used i2 = −1)
29. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
30. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
31. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
10
= (i)
= −1
(getting ready to use A3 = i) (used A3 = i) (ez as Sundays...)
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
32. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 33. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
34. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 35. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 36. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 37. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
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Trigonometry Sec. 04 notes
MathHands.com M´ arquez
Intro to Complex Numbers The Question In the last section we wondered what might be far to the right and to the left of the real number line. In doing so, we entertained the existence of the very special unreal number ∞. In this section, we continue on this path and now consider what lies above or below the real number line?
? ∞
−∞ -5
-4
Out of Audacity, a number is born
-3
-2
-1
0
1
2
3
4
5
?
Here is the account, directly from the world famous, the almost immortal, mathematician Euler himself.... ”I sat in that soft comfortable chair leaned back and enjoyed the million thoughts dancing inside my head. A blank paper, a pencil and my awesome coffee was all there was on the desk. On the paper the equation x2 = −1 The equation was screaming, enticing, talking trash, challenging me, saying ”you can’t solve me!” Hours went by faster than I would have liked. Days past by, weeks and months...There was no real number that would solve the equation. But the forces were greater. The inspiration divine. I would not be stopped.. and one day it happened. There was no real number solution, I had looked on the positive side on the negative side and all numbers between. Resolved to avoid defeat at all costs, I invented a number. From my own imagination, I gathered all my might, my courage, and my audacity, √ and I thought...I will create a number. I will call it i, and I will solve my problem by declaring i = −1. It’s my number so I can make it behave however I please, just as the artists paints the clouds at his whim... This solves the equation x2 = −1 and marks the birth of a grand elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.” Needless, to say, I have taken some artistic liberties with this account of events. In fact, traces of complex numbers or ’imaginary’ numbers can be found in 9th century’s Al-Khwarizmi’s Algebra text. During the next couple centuries these ideas made their way to Italy and France, as people were learning to solve degree 3 equations.√ By the turn of the 17th century Descartes coined the phrase ”imaginary” numbers, referring √ to numbers such as −1. At last, it was Euler, in the 18th century who named such number i, declaring i = −1. Thus... by definition of i;
i2 = −1 and, i is a solution to x2 = −1 In addition to solving the equation x2 =− 1, and this marks the birth of a grand and elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.
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math hands
pg. 1
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
Where do all the i’s Live With the invent of i came other numbers such as 2i, 3i, −5i, or 2 + 3i. Generally, complex numbers are numbers that can be written in the form a + bi where a and b are real numbers. Now, in the previous sections we noted a visual representation of the real numbers using the real number line. Considering this, the natural question is where or how do we represent the complex numbers visually? Over the centuries, the most powerful and common way to represent the complex numbers is to place them as an extension of the real number line, extending it above and below to make what we commonly call the complex plane. In essence, this is done by placing an ’imaginary’ axis perpendicular to the real number line. Then we position every number a+ bi on the plane similarly to placing the ordered point (a, b) on the cartesian [xy]plane. came numbers such as 2i, 3i, 4i, . . . as well as numbers such as 2 + 3i. Here are some visual representations of a few complex numbers along with their position relative to the known real number line. 3i
The Complex Numbers, C
2i b
−4 + i
b
3 + 2i
3
4
1i 2 + 0i
R
b
-5
-4
-3
-2
-1
0
1
2
5
−1i −2i −3i b
4 − 3i
Negative Radicals With the invention of i we can now make sense of radicals (i.e. square roots) of √ negative real numbers. Consider √ the radical −1, the number whose square is −1. Recall when we first defined 4 we did so as ’the number whose number square is 4.’ But there are two such numbers 2 and −2. By default, we declared to radical to mean the positive number √ whose square is 4. We follow a similar logic here, as we are confronted with the same dilemma. If we define −1 as the number whose square is −1, we will find there are two possible choices, i and −i (see examples). By convention, we will define negative radical to be, the positive i rather than −i. Examples 1. Simplify
√ −4
√
√ −4 = i 4 = i2 = 2i
2. Simplify
√ −10
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√ √ −10 = i 10
math hands
(neg rad) (def rad) (CoLM)
(neg rad)
pg. 2
Trigonometry Sec. 04 notes 3. Simplify
4. Simplify
√ −15
MathHands.com M´ arquez √ √ −15 = i 15
(neg rad)
√ −x
Solution: Notice, we do not know the value of x. We don’t know if x is positive √ or negative. This means we don’t know if −x is positive or negative therefore we don’t know if the radical −x is positive or negative.
5. Adding in C
3 + 5i + 2 + 3i
Solution:
3 + 5i + 2 + 3i = = (3 + 2) + (5i + 3i)
(given) (ALA)
= 3 + 2 + (5 + 3)i
(DL)
= 5 + 8i
(AT)
6. Multiplying in C (3 + 5i)(2 + 3i)
Solution:
(3 + 5i)(2 + 3i) = = 3 · 2 + 5i · 2 + 3 · 3i + 5i · 3i = 6 + 10i + 9i + 15i2
= 6 + (10 + 9)i + 15i
2
(given) (FOIL) (BI) (DL)
2
(AT) (Def of i)
= 6 + 19i + −15 = −9 + 19i
(BI) (BI)
= 6 + 19i + 15i = 6 + 19i + 15 · −1
7. Multiplying in C (4 +− 5i)(2 + 3i)
Solution:
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math hands
pg. 3
Trigonometry Sec. 04 notes
MathHands.com M´ arquez (4 +− 5i)(2 + 3i) =
(given)
= 4 · 2 + 5i · 2 + 4 · 3i + 5i · 3i −
−
= 8 + 10i + 12i + 15i −
−
2
= 8 + ( 10 + 12)i + 15i −
−
= 6 + 2i + 15i −
2
2
(FOIL) (BI) (DL) (BI)
= 6 + 2i + 15 · −1 = 6 + 2i + 15
(Def of i) (NNT)
−
= 21 + 2i
(BI)
8. Multiplying in C (4 + 3i)(2 + 3i)
Solution:
(4 + 3i)(2 + 3i) =
(given)
= 8 + 12i + 6i + 9i
2
= 8 + 12i + 6i + 9 · 1 −
(FOIL) (def i)
= 8 + 12i + 6i + 9
(BI)
= 1 + 18i
(BI)
−
−
9. Multiplying in C i7
Solution:
i7 = iiiiiii 2 2 2
(+Expo)
=i i i i
(+Expo)
= −1 · −1 · −1 · i = −1 · i
(Def of i) (BI)
= −i
(MT)
And Now, Divide We have now introduced the imaginary number, their standard form ’a+bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide.
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math hands
pg. 4
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3 5i − 3 5i − 3 = ·1 i+3 i+3 = =
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2007-2009 MathHands.com
5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
math hands
pg. 5
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
=
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 11 −3 − 11i = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
2i + 1 −i+1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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2007-2009 MathHands.com
math hands
pg. 6
Trigonometry Sec. 04 notes
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 5i − 2 2i + 3 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
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2007-2009 MathHands.com
2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
math hands
pg. 7
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Intro to Complex Numbers 1. Plot the following points. (a) 2 + 3i (b) 5 + 1i (c) −4 + −3i
(d) −4i
Solution:
8i 7i 6i 5i 4i b
3i
2 + 3i
2i b
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5 + 1i
5
6
7
8
9 10 11 12
-2i
−4 + −3i b
-3i
0 + −4i
-4i b -5i -6i -7i -8i
2. Plot the following points. (a) −1 + 4i
(b) 2 + 2i
(c) 5 + −5i
(d) −7i
Solution:
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2007-2009 MathHands.com
math hands
pg. 8
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
8i 7i 6i 5i
−1 + 4i b 4i 3i b
2i
2 + 2i
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5
6
7
8
9 10 11 12
-2i -3i -4i -5i
5b + −5i
-6i
0 + −7i
-7i b -8i
Simplify
√ −90
Solution: √ −90
4. 3. Simplify
√ = i 90 √ = i 9 · 10 √ √ = i 9 10 √ = i · 3 10 √ = 3i 10
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −150
Solution:
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2007-2009 MathHands.com
math hands
pg. 9
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −150
5. Simplify
√ = i 150 √ = i 25 · 6 √ √ = i 25 6 √ = i·5 6 √ = 5i 6
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −375
Solution: √ −375
6. Simplify
√ = i 375 √ = i 25 · 15 √ √ = i 25 15 √ = i · 5 15 √ = 5i 15
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −18
Solution: √ −18
7. Simplify
√ = i 18 √ =i 9·2 √ √ =i 9 2 √ = i·3 2 √ = 3i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −50
Solution:
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2007-2009 MathHands.com
math hands
pg. 10
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −50
8. Simplify
√ = i 50 √ = i 25 · 2 √ √ = i 25 2 √ = i·5 2 √ = 5i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −8
Solution: √ −8
9. Simplify
√ =i 8 √ = i 4·2 √ √ =i 4 2 √ = i·2 2 √ = 2i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −125
Solution: √ −125
10. Simplify
√ = i 125 √ = i 25 · 5 √ √ = i 25 5 √ = i·5 5 √ = 5i 5
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ 3 −8
Solution: −2 since (−2)3 = −8 11. Simplify
√ 3 −125
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math hands
pg. 11
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: −5 since (−5)3 = −125 12. Simplify
√ 3 −1
Solution: −1 since (−1)3 = −1 13. Add 3i + 2 + 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 5i) = 2 + (3i + 4) + 5i = 2 + (4 + 3i) + 5i = (2 + 4) + (3i + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i
(ALA) (CoLA) (ALA) (DL) (BI)
14. Add 2i + 7 + 9i + 3 Solution:
by convention, we write in ’standard form’, ”a + bi” (7 + 2i) + (3 + 9i) = 7 + (2i + 3) + 9i = 7 + (3 + 2i) + 9i = (7 + 3) + (2i + 9i) = (7 + 3) + (2 + 9)i = 10 + 11i
(ALA) (CoLA) (ALA) (DL) (BI)
15. Add 3i − 2 + − 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (−2 + 3i) + (4 + −5i) = −2 + (3i + 4) + −5i
= −2 + (4 + 3i) + −5i = (−2 + 4) + (3i + −5i)
= (−2 + 4) + (3 + −5)i = 2 + −2i
(ALA) (CoLA) (ALA) (DL) (BI)
16. Add 30i + 11 + 5i + 5
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math hands
pg. 12
Trigonometry Sec. 04 exercises Solution:
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” (11 + 30i) + (5 + 5i) = 11 + (30i + 5) + 5i = 11 + (5 + 30i) + 5i = (11 + 5) + (30i + 5i) = (11 + 5) + (30 + 5)i = 16 + 35i
(ALA) (CoLA) (ALA) (DL) (BI)
17. Add i + 1 + i + 2 Solution:
by convention, we write in ’standard form’, ”a + bi” (1 + 1i) + (2 + 1i) = 1 + (1i + 2) + 1i = 1 + (2 + 1i) + 1i = (1 + 2) + (1i + 1i) = (1 + 2) + (1 + 1)i = 3 + 2i
(ALA) (CoLA) (ALA) (DL) (BI)
18. Add 3i + 2 + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 0i) = 2 + (3i + 4) + 0i = 2 + (4 + 3i) + 0i = (2 + 4) + (3i + 0i) = (2 + 4) + (3 + 0)i = 6 + 3i
(ALA) (CoLA) (ALA) (DL) (BI)
19. Add 3i + i Solution:
by convention, we write in ’standard form’, ”a + bi” (0 + 3i) + (0 + 1i) = 0 + (3i + 0) + 1i = 0 + (0 + 3i) + 1i = (0 + 0) + (3i + 1i) = (0 + 0) + (3 + 1)i = 0 + 4i
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math hands
(ALA) (CoLA) (ALA) (DL) (BI)
pg. 13
Trigonometry Sec. 04 exercises 20. Multiply 3i + 2
MathHands.com M´ arquez 5i + 4
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 5i + 4 = (2)(4) + (2)(5i) + (3i)(4) + (3i)(5i) = 8 + 10i + 12i + 15i
2
9i + 3
(BI)
= 8 + 10i + 12i + 15(−1)
(def i)
= (8 + −15) + (10i + 12i) = (8 + −15) + (10 + 12)i
(CoLA, ALA,BI) (DL)
= −7 + 22i
21. Multiply 2i + 7
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 2i + 7 9i + 3 = (7)(3) + (7)(9i) + (2i)(3) + (2i)(9i) 2
= 21 + 63i + 6i + 18i = 21 + 63i + 6i + 18(−1)
(BI) (def i)
= (21 + −18) + (63i + 6i) = (21 + −18) + (63 + 6)i
(CoLA, ALA,BI) (DL)
= 3 + 69i
22. Multiply 3i − 2
− 5i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i − 2 − 5i + 4 = (−2)(4) + (−2)(−5i) + (3i)(4) + (3i)(−5i) = −8 + 10i + 12i + −15i
2
= −8 + 10i + 12i + −15(−1) = (−8 + 15) + (10i + 12i)
= (−8 + 15) + (10 + 12)i = 7 + 22i
23. Multiply 30i + 11
5i + 5
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(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
math hands
pg. 14
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: by convention, we write in ’standard form’, ”a + bi” 30i + 11 5i + 5 = (11)(5) + (11)(5i) + (30i)(5) + (30i)(5i) = 55 + 55i + 150i + 150i
2
= 55 + 55i + 150i + 150(−1) = (55 + −150) + (55i + 150i)
= (55 + −150) + (55 + 150)i = −95 + 205i
24. Multiply i + 1
i+2
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
25. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i)
(BI) (def i)
= (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i
(CoLA, ALA,BI) (DL)
= 8 + −8i + 12i + −12i = 8 + −8i + 12i + −12(−1)
= 20 + 4i
26. Multiply 3i − 1
i−1
(FOIL)
2
(BI)
Solution:
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math hands
pg. 15
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i)
(DL)
= −2 + −4i
i+2
(BI) (def i) (CoLA, ALA,BI)
= (1 + −3) + (−1 + −3)i
27. Multiply i + 1
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
28. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i) = 8 + −8i + 12i + −12i
2
= 8 + −8i + 12i + −12(−1) = (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i = 20 + 4i
29. Multiply 3i − 1
i−1
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution:
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math hands
pg. 16
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i) = (1 + −3) + (−1 + −3)i = −2 + −4i
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
30. multiply i3 Solution: i3 = i2 · i
= −1 · i = −i
(JAE) (def of i) (MT)
31. multiply i4 Solution: i4 = i2 · i2 = −1 · −1 =1
(JAE) (def of i) (NotNot)
32. multiply i6 Solution: i6 = i4 · i2
= 1 · −1 = −1
(JAE) (see previous problem) (MiD)
3
33. (3i + 2)
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math hands
pg. 17
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution:
(3i + 2)3
(given) 3
= (3i)
2
+
3
2
3 · (3i) (2)
2
+
3(3i)(2)
+
3
(2)
(famous, PP3)
= 27i + 54i + 36i + 8
(BI)
= 27(−i) + 54(−1) + 36(i) + 8 = −46 + 9i
(BI) (BI)
3
34. (2i + 1)
Solution:
(2i + 1)3
(given) 3
= (2i) 3
2
+
3 · (2i) (1)
2
2
+
3(2i)(1)
+
3
(1)
(famous, PP3)
= 8i + 12i + 6i + 1
(BI)
= 8(−i) + 12(−1) + 6(i) + 1 = −11 + −2i
(BI) (BI)
3
35. (−2i + 1)
Solution:
3
(−2i + 1)
(given) 3
= (−2i)
3
2
+ 2
3 · (−2i) (1)
+
= − 8i + 12i − 6i + 1
= −8(−i) + 12(−1) + −6(i) + 1 = −11 + 2i
2
3(−2i)(1)
+
3
(1)
(famous, PP3) (BI) (BI) (BI)
3
36. (−2i + 3)
Solution:
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math hands
pg. 18
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 3
(−2i + 3)
(given) 3
= (−2i)
2
+
3
2
3 · (−2i) (3)
+
2
3(−2i)(3)
3
+
(3)
= − 8i + 36i − 54i + 27 = −8(−i) + 36(−1) + −54(i) + 27
(famous, PP3) (BI) (BI)
= −9 + −46i
(BI)
37. (−1i + 1)3 Solution:
3
(−1i + 1)
(given) 3
= (−1i)
+
3
2
2
3 · (−1i) (1)
+
2
3(−1i)(1)
3
+
(1)
= − i + 3i − 3i + 1 = −1(−i) + 3(−1) + −3(i) + 1
(famous, PP3) (BI) (BI)
= −2 + −2i
(BI)
38. (1i + 1)3 Solution:
3
(1i + 1)
(given) 3
= (1i) 3
39. Divide
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+ 2
2
3 · (1i) (1)
+
2
3(1i)(1)
+
3
(1)
(famous, PP3)
= i + 3i + 3i + 1 = 1(−i) + 3(−1) + 3(i) + 1
(BI) (BI)
= −2 + 2i
(BI)
7i + 1 7i + 3
math hands
pg. 19
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
40. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
41. Divide
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2007-2009 MathHands.com
5i + 2 4i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
math hands
pg. 20
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
42. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
43. Divide
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3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
math hands
pg. 21
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
44. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
45. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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math hands
pg. 22
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
46. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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math hands
pg. 23
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
47. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 48. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 49. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 50. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2
(just add exponents) (setting up to use fact i4 = 1)
3
= (1) · i2 =i
(yipei-kae-yeh...)
2
= −1
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math hands
pg. 24
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
51. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
52. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
53. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
54. Compute w/ Complex Numbers Calculate and write in standard form. i−5
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math hands
pg. 25
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
55. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
56. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
57. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
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·i
math hands
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
pg. 26
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
58. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
59. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
60. Compute w/ Complex Numbers Calculate and write in standard form. i11
Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1)
= (1) · i
=i
(ez as sundays)
61. Compute w/ Complex Numbers Calculate and write in standard form. i−6
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pg. 27
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
62. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
= (1)8 · i
(used i4 = 1)
=i
(ez as sundays)
63. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
64. Compute w/ Complex Numbers Calculate and write in standard form.
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1 1 √ +√ i 2 2
math hands
2
pg. 28
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i = √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
(FOIL) (used i2 = −1)
65. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 3 1 + i 2 2
(FOIL) (used i2 = −1)
66. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
67. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
68. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
(getting ready to use A3 = i)
10
= (i)
(used A3 = i)
= −1
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(ez as Sundays...)
math hands
pg. 30
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
69. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 70. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
71. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 72. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 73. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 74. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
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pg. 31
Trigonometry Sec. 03 notes
MathHands.com M´ arquez
Converting Equations Polar & Cartesian The IDEA In the last couple sections we’ve established, among other things, a ’dictionary’ to convert coordinate points from cartesian to polar and vise-varsa. In this section, we would like to exploit that same dictionary to convert equations. As a reminder, the dictionary has two components, ’from’ polar, and ’to’ polar. The goal in converting, in most cases, is to eliminate all x’s and y’s and replace them with appropriate expressions involving r’s and θ’s if we are translating to polar. When converting from polar, the idea is to eliminate the r’s and θ’s. We summarize the ’dictionary’ below. Translating
Polar
Cartesian
90◦ 120◦
y = r sin θ x = r cos θ
60◦ 30◦
150◦ b
b
(r, θ)
(x, y)
0◦
180◦
210◦
330◦
θ = tan−1 300◦
240◦
r2 = x2 + y
270◦
y x 2
Example: Converting Equation of a line to Polar Convert the following equation to polar coordinates: y=x Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
y y x tan θ θ
=x
(given)
=1
(div by x, setting up to use translating dictionary, tan θ = y/x)
=1 = 45◦
(used tan θ = y/x) (notice: this there are infinite possibilities for θ (ie −135 , 225 ...), this is just one of them.) ◦
◦
Thus, one way to translate the equation y = x to polar is to convert to θ = 45◦ Example: Converting Equation of a circle to Polar Convert the following equation to polar coordinates: x2 + y 2 = 25 Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
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Trigonometry Sec. 03 notes
MathHands.com M´ arquez x2 + y 2 = 25
(given)
2
2
(used the dictionary r = x2 + y 2 )
r = 25
at this point we have eliminated all x’s and y’s from the equation, thus the converting is complete. It is sometimes desirable to solve for r explicitly, thus we could also say r = 5 or r = −5, in fact all three of these equations, r2 = 25, r = 5, AND r = −5 all yield the same graph as x2 + y 2 = 25. Example: Converting the equation of a circle to Polar Convert the following equation to polar coordinates: x2 + 3x + 5 + y 2 − 4y + 2 = 25 Solution:
x2 + 3x + 5 + y 2 − 4y + 2 = 25 2
2
(given)
2
(x + y ) + 3(x) − 4(y) = 18
(just cleaning up, getting ready to use the dictionary)
(r ) + 3(r cos θ) − 4(r sin θ) = 18
(used the dictionary)
2
r + 3r cos θ − 4r sin θ = 18
(algebra)
Example: Converting the equation of a circle from Polar Convert the following equation of a circle to polar coordinates: r = 12 cos θ Solution: r = 12 cos θ r · r = r(12 cos θ)
r2 = 12(r cos θ) 2
2
x + y = 12x 2
2
2
y + x + − 12x = 0 2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
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math hands
(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 2
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Converting Equations Polar & Cartesian 1. Convert the following equation to polar coordinates: y = 2x
Solution: Note: the ’translating’ of this equation is not unique.
y = 2x y =2 x tan θ = 2 θ ≈ 63.435
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) ◦
(notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 63.435◦ 2. Convert the following equation to polar coordinates: 2 y= − x 3
Solution: Note: the ’translating’ of this equation is not unique. 2 y= − x 3 2 y = − x 3 2 tan θ = − 3 θ ≈ −33.69◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ −33.69◦ 3. Convert the following equation to polar coordinates: y=
2 x 5
Solution: Note: the ’translating’ of this equation is not unique.
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez 2 x 5 2 = 5 2 = 5 ≈ 21.801◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 21.801◦ 4. Convert the following equation to polar coordinates: y=
4 x 3
Solution: Note: the ’translating’ of this equation is not unique. 4 x 3 4 = 3 4 = 3 ≈ 53.13◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 4x to polar is to convert to θ ≈ 53.13◦ 5. Convert the following equation to polar coordinates: 4 y= − x 3 Solution: Note: the ’translating’ of this equation is not unique. 4 y= − x 3 4 y = − x 3 4 tan θ = − 3 θ ≈ −53.13◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = − 4x to polar is to convert to θ ≈ −53.13◦ 6. Convert the following equation of a circle to polar coordinates: 2x2 + 3x + 2y 2 + − 5y = 7
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution:
2x2 + 3x + 2y 2 + − 5y = 7 2 x2 + y 2 + 3(x) + − 5(y) = 7 2 r2 + 3(r cos θ) + − 5(r sin θ) = 7
(given) (ready to use the dictionary....) (used the dictionary)
7. Convert the following equation of a circle to polar coordinates: − 5x2 + 2x + − 5y 2 + 7y = 25 Solution:
− 5x2 + 2x + − 5y 2 + 7y = 25 − 5 x2 + y 2 + 2(x) + 7(y) = 25 − 5 r2 + 2(r cos θ) + 7(r sin θ) = 25
(given) (ready to use the dictionary....) (used the dictionary)
8. Convert the following equation of a circle to polar coordinates: 3 4x2 + x + 4y 2 + 1y = 9 2
Solution: 3 4x2 + x + 4y 2 + 1y = 9 2 3 4 x2 + y 2 + (x) + 1(y) = 9 2 3 2 4 r + (r cos θ) + 1(r sin θ) = 9 2
(given) (ready to use the dictionary....) (used the dictionary)
9. Convert the following equation of a circle to polar coordinates: 3x2 + 3x + 3y 2 + 5y = 4
Solution:
3x2 + 3x + 3y 2 + 5y = 4 3 x2 + y 2 + 3(x) + 5(y) = 4 3 r2 + 3(r cos θ) + 5(r sin θ) = 4
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(given) (ready to use the dictionary....) (used the dictionary)
pg. 5
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
10. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 9 25
Solution: y2 x2 + =1 9 25 (r cos θ)2 (r sin θ)2 + =1 9 25 ! cos2 θ sin2 θ 2 + =1 r 9 25
(given) (ready to use the dictionary....) (alebra)
11. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 16
Solution: y2 x2 + =1 4 16 (r cos θ)2 (r sin θ)2 + =1 4 16 ! cos2 θ sin2 θ + =1 r2 4 16
(given) (ready to use the dictionary....) (alebra)
12. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 49
Solution: y2 x2 + =1 4 49 (r cos θ)2 (r sin θ)2 + =1 4 49 ! cos2 θ sin2 θ =1 + r2 4 49
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(given) (ready to use the dictionary....) (alebra)
pg. 6
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
13. Convert the following equation of a circle to polar coordinates: r = 2 sin θ
Solution: r = 2 sin θ
(given)
r · r = r(2 sin θ)
(clever little idea, mult both sides by r)
r2 = 2(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 2y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 2y = 0
(optional step: complete the square)
2
x + y + − 2y + 1 = 1 2 x2 + y + − 1 = 1
(optional to step: complete the square) (Shows its a circle, shows center/radius)
14. Convert the following equation of a circle to polar coordinates: r = 5 sin θ
Solution: r = 5 sin θ
(given)
r · r = r(5 sin θ)
(clever little idea, mult both sides by r)
2
r = 5(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 5y = 0 25 25 x2 + y 2 + − 5y + = 4 4 2 25 5 = x2 + y + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
15. Convert the following equation of a circle to polar coordinates: r = 6 sin θ
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 6 sin θ r · r = r(6 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 6(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 6y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 6y = 0
(optional step: complete the square)
2
x + y + − 6y + 9 = 9 2 x2 + y + − 3 = 9
(optional to step: complete the square) (Shows its a circle, shows center/radius)
16. Convert the following equation of a circle to polar coordinates: r = 11 sin θ
Solution: r = 11 sin θ r · r = r(11 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 11y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 11y = 0 121 121 = x2 + y 2 + − 11y + 4 4 2 11 121 x2 + y + − = 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
17. Convert the following equation of a circle to polar coordinates: r = 8 cos θ
Solution: r = 8 cos θ
(given)
r · r = r(8 cos θ)
(clever little idea, mult both sides by r)
2
r = 8(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 8x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 8x = 0
(optional step: complete the square)
2
y + x + − 8x + 16 = 16 2 y 2 + x + − 4 = 16
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(optional to step: complete the square) (Shows its a circle, shows center/radius)
math hands
pg. 8
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
18. Convert the following equation of a circle to polar coordinates: r = 12 cos θ
Solution: r = 12 cos θ
(given)
r · r = r(12 cos θ)
(clever little idea, mult both sides by r)
r2 = 12(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 12x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 12x = 0
(optional step: complete the square)
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
(optional to step: complete the square) (Shows its a circle, shows center/radius)
19. Convert the following equation of a circle to polar coordinates: r = 5 cos θ
Solution: r = 5 cos θ
(given)
r · r = r(5 cos θ)
(clever little idea, mult both sides by r)
2
r = 5(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 5x = 0 25 25 y 2 + x2 + − 5x + = 4 4 2 25 5 = y2 + x + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
20. Convert the following equation of a circle to polar coordinates: r = 11 cos θ
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 11 cos θ r · r = r(11 cos θ) r2 = 11(r cos θ) 2
2
x + y = 11x 2
2
y + x + − 11x = 0 121 121 = y 2 + x2 + − 11x + 4 4 2 121 11 = y2 + x + − 2 4
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(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 10
Trigonometry Sec. 02 notes
MathHands.com M´ arquez
Introduction to Polar Graphs Main Idea Plot Points You should be very conformable plotting points in radians as well as degrees. Simply make a table of values with r’s and corresponding θ’s, then start plotting until you can make sense of the the graph. If connecting the points is uneasy, choose smaller angle intervals, i.e. instead of plotting and calculating the r every 30 degrees, plot every 5 degrees. Get to Know Your Calculator I am usually not a big calculator fan, but this may be a good time when we can use it appropriately to graph some of these functions. After all, there is limited beauty and creativity that occurs when plotting points. Get to know Famous Questions 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ? Get to know Famous Graphs Get to know Famous Graphs 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ?
r = sin θ
r = cos θ
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r = cos(2θ)
r = cos(3θ)
r = 5 + 3 sin θ
r = sin(2θ)
r = sin(4θ)
r = 3 + 5 sin θ
r = sin(3θ)
r = cos(5θ)
math hands
pg. 1
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
Introduction to Polar Graphs 1. Graph and Understand the graph of r = 5 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 2. Graph and Understand the graph of r = 5 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 3. Graph and Understand the graph of r = −3 cos θ
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pg. 2
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 4. Graph and Understand the graph of r = 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 5. Graph and Understand the graph of r = −6 sin θ
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pg. 3
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 6. Graph and Understand the graph of r = 5 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 7. Graph and Understand the graph of r = 5 sin θ. Limit the study of this graph to the values of θ ranging from −90◦ to 0◦ .
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pg. 4
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 8. Graph and Understand the graph of r = −3 cos θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 9. Graph and Understand the graph of r = 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 180◦.
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pg. 5
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 10. Graph and Understand the graph of r = −6 sin θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 11. Graph and Understand the graph of r = 5 + cos θ
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pg. 6
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 12. Graph and Understand the graph of r = 4 − 2 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 13. Graph and Understand the graph of r = −3 + 2 cos θ
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pg. 7
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 14. Graph and Understand the graph of r = 3 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 15. Graph and Understand the graph of r = 4 − 2 cos θ
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pg. 8
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 16. Graph and Understand the graph of r = −2 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 17. Graph and Understand the graph of r = 3 + 2 sin θ
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pg. 9
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 18. Graph and Understand the graph of r = 3 − 2 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 19. Graph and Understand the graph of r = −3 + 2 sin θ
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pg. 10
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 20. Graph and Understand the graph of r = 3 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 21. Graph and Understand the graph of r = 2 − 3 sin θ
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pg. 11
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 22. Graph and Understand the graph of r = −2 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 23. Graph and Understand the graph of r = 2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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pg. 12
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 24. Graph and Understand the graph of r = 2 − 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 60◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 25. Graph and Understand the graph of r = −2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
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pg. 13
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 26. Graph and Understand the graph of r = 5 + 5 sin θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 27. Graph and Understand the graph of r = 3 − 2 sin θ. Limit the study of this graph to the values of θ ranging from 135◦ to 225◦ .
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2007-2009 MathHands.com
math hands
pg. 14
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 28. Graph and Understand the graph of r = −3 + 2 sin θ. Limit the study of this graph to the values of θ ranging from 360◦ to 540◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 29. Graph and Understand the graph of r = 5 cos(2θ)
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2007-2009 MathHands.com
math hands
pg. 15
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 30. Graph and Understand the graph of r = 6 cos(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 31. Graph and Understand the graph of r = 5 sin(2θ)
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2007-2009 MathHands.com
math hands
pg. 16
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 32. Graph and Understand the graph of r = 6 sin(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 33. Graph and Understand the graph of r = 5 cos(3θ)
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2007-2009 MathHands.com
math hands
pg. 17
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 34. Graph and Understand the graph of r = 6 cos(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 35. Graph and Understand the graph of r = 5 sin(3θ)
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2007-2009 MathHands.com
math hands
pg. 18
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 36. Graph and Understand the graph of r = 4 sin(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 37. Graph and Understand the graph of r = 5 sin(2θ). Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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2007-2009 MathHands.com
math hands
pg. 19
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 38. Graph and Understand the graph of r = 6 sin(4θ). Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 39. Graph and Understand the graph of r = 5 cos(3θ). Limit the study of this graph to the values of θ ranging from 0◦ to 180◦ .
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2007-2009 MathHands.com
math hands
pg. 20
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 40. Graph and Understand the graph of r = 6 cos(5θ). Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 41. Graph and Understand the graph of r2 = 25 sin2 θ √ Solution: Note, we can solve for r as r√= ± 25 sin2 θ... then we can plot each r separately, that is we graph √ r = 25 sin2 θ and then we graph r = − 25 sin2 θ
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2007-2009 MathHands.com
math hands
pg. 21
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
42. Graph and Understand the graph of r2 = 25 sin2 3θ √ 25 sin2 3θ... then we can plot each r separately, that is we graph Solution: Note, we can solve for r as r = ± √ √ 2 r = 25 sin 3θ and then we graph r = − 25 sin2 3θ 90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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2007-2009 MathHands.com
math hands
pg. 22
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
43. Understanding The graph of r = 2 − 3 cos θ has an inner loop. What values of θ trace precisely this inner loop? i.e. ?◦ ≤ θ ≤?◦ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center
44. Understanding On the graph of r = 5 sin(3θ) (a) ’sin 3θ’ will take values between 1 and −1; never larger than one, never smaller than −1. Find the angles θ, at which sin 3θ take on its largest value 1, or is smallest value -1. (i.e solve sin 3θ = 1 and solve sin 3θ = −1 )
(b) Graph r = 5 sin(3θ) and mark the points obtained from part a). What can you say about these points? A) nothing B) they are the ’tip of the leaves’ C) they are not on the leaves. (c) Find the angles at which the graph goes through (0, 0) (d) Find the interval/s for theta that trace the second ’leaf’ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center, such leaf starts and ends at center.. 45. Understanding On the graph of r = −5 + 2 cos θ
(a) ’cos θ’ will take values between 1 and −1; never larger than one, never smaller than −1. What is the largest or the smallest that the quantity ”−5 + 2 cos θ” can become?
(b) Find the angles at which ”−5 + 2 cos θ” takes on its largest value -3, or is smallest value -7. (c) Graph r = −5 + 2 cos θ and mark the points obtained from part b). What can you say about these points? A) nothing B) its the furthest point from the origin (d) Find the angles at which the graph goes through (0, 0) (e) Find the interval/s for theta that trace ONLY the outer leaf.
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2007-2009 MathHands.com
math hands
pg. 23
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Intro to Polar Coordinates 1. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (3, 120◦), p2 = (−6, 120◦), p3 = (−3, −120◦), p4 = (3, −300◦), p5 = (−3, 135◦) Solution: 90◦ 60◦ 120◦
30◦
p1
150◦
b
p4
p3 b
0◦ 180◦
p5
b
330◦ 210◦
p2 b
300◦
240◦ 270◦
2. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (2, −120◦), p2 = (−4, −120◦), p3 = (−4, 120◦), p4 = (4, −60◦ ), p5 = (−4, 135◦) Solution:
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2007-2009 MathHands.com
math hands
pg. 6
MathHands.com M´ arquez
Intro to Polar Coordinates 1. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (3, 120◦), p2 = (−6, 120◦), p3 = (−3, −120◦), p4 = (3, −300◦), p5 = (−3, 135◦) Solution: 90◦ 60◦ 120◦
30◦
p1
150◦
b
p4
p3 b
0◦ 180◦
p5
b
330◦ 210◦
p2 b
300◦
240◦ 270◦
2. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (2, −120◦), p2 = (−4, −120◦), p3 = (−4, 120◦), p4 = (4, −60◦ ), p5 = (−4, 135◦) Solution:
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2007-2009 MathHands.com
math hands
pg. 6
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez 90◦ 60◦ 120◦
p2 b
30◦
150◦
0◦ 180◦
p1 b
210◦
p4
pp35 b
b
330◦
300◦ 240◦ 270◦
3. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (1, 45◦ ), p2 = (−2, 45◦), p3 = (−5, −45◦), p4 = (5, −225◦), p5 = (−5, 135◦) Solution:
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2007-2009 MathHands.com
math hands
pg. 7
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez 90◦ 60◦ 120◦
p4
b
p3 30◦
150◦
p1 b
0◦ 180◦
p2 b
330◦ 210◦
p5
b
300◦ 240◦ 270◦
4. Convert the polar coordinates, (−4, 210◦) to cartisian coordinates Solution:
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2007-2009 MathHands.com
math hands
pg. 8
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦
6 60◦
120◦
5
30◦ 150◦
(−4, 210◦) b
x = r cos θ y = r sin θ so...
4 3
(3.46, 2)
x = r cos θ = −4 cos(210◦) ≈ 3.46 0◦ -6
180◦
b
2 1
-5
-4
-3
-2
-1
1
2
3
4
5
6
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = −4 sin(210◦) ≈2
-2 -3 -4 -5
300◦ 240◦
-6 270◦
5. Convert the polar coordinates, (4, 210◦) to cartisian coordinates Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 4 cos(210◦) ≈ −3.46 0◦
2 1
-6
180◦
-5
-4
-3
-2
-1 -1
(4, 210◦) b
330◦ 210◦
y = r sin θ = 4 sin(210◦) ≈ −2
(−3.46, −2) b
-2 -3 -4 -5
300◦ 240◦
-6 270◦
6. Convert the polar coordinates, (5, 150◦) to cartisian coordinates
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2007-2009 MathHands.com
math hands
pg. 9
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solution:
90◦
6 60◦
120◦
150◦
5
30◦
(5, 150◦) b
x = r cos θ y = r sin θ so...
4
(−4.33, 2.5)
3
b
x = r cos θ = 5 cos(150◦) ≈ −4.33 0◦
2 1
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = 5 sin(150◦) ≈ 2.5
-2 -3 -4 -5
300◦ 240◦
-6 270◦
7. Convert the polar coordinates, (6, −240◦) to cartisian coordinates Solution:
90◦
6
(6, −240◦)
(−3, 5.2)
60◦
120◦ b
b
30◦ 150◦
5
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 6 cos(−240◦) ≈ −3 0◦ -6
180◦
2 1
-5
-4
-3
-2
-1 -1
330◦ 210◦
y = r sin θ = 6 sin(−240◦) ≈ 5.2
-2 -3 -4 -5
300◦ 240◦
-6 270◦
8. Convert the polar coordinates, (−5, 0◦ ) to cartisian coordinates
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2007-2009 MathHands.com
math hands
pg. 10
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = −5 cos(0◦) ≈ −5 0◦
(−5, 0◦) b
2 1
(−5, 0) b
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = −5 sin(0◦) ≈0
-2 -3 -4 -5
300◦ 240◦
-6 270◦
9. Convert the polar coordinates, (6, −180◦) to cartisian coordinates Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 6 cos(−180◦) ≈ −6 (−6, −0) 0◦
(6, −180◦) b
2 1
b
-6
180◦
-5
-4
-3
-2
-1 -1
330◦ 210◦
y = r sin θ = 6 sin(−180◦) ≈ −0
-2 -3 -4 -5
300◦ 240◦
-6 270◦
10. Convert the polar coordinates, (6, 180◦) to cartisian coordinates
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2007-2009 MathHands.com
math hands
pg. 11
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Solution:
90◦
6 60◦
120◦
5
30◦ 150◦
x = r cos θ y = r sin θ so...
4 3
x = r cos θ = 6 cos(180◦) ≈ −6 (−6, 0) 0◦
(6, 180◦) b
2 1
b
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1
330◦ 210◦
y = r sin θ = 6 sin(180◦) ≈0
-2 -3 -4 -5
300◦ 240◦
-6 270◦
11. Convert the cartesian coordinates, (-2, 3), to polar coordinates. Solution:
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2007-2009 MathHands.com
math hands
pg. 12
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4
(r, θ) b
150◦
(−2, 3)
30◦
b
3 2 1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 3/ − 2 (note: eq has infinite solutions) ◦ θ ≈ −56.31 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−2)2 + (3)2 r2 =13
√ r = ± 13 ≈ ±3.61
note: if we use the angle −56.31◦, then we must choose the negative value of r, r ≈ −3.61, thus the polar cordinates (−3.61, −56.31◦) is one way to represent the point (−2, 3)
12. Convert the cartesian coordinates, (2, -3), to polar coordinates. Solution:
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2007-2009 MathHands.com
math hands
pg. 13
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2
(r, θ)
330◦
b
-3
(2, −3) b
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = −3/2 (note: eq has infinite solutions) ◦ θ ≈ −56.31 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(2)2 + (−3)2 r2 =13
√ r = ± 13 ≈ ±3.61
note: if we use the angle −56.31◦, then we must choose the positive value of r, r ≈ 3.61, thus the polar cordinates (3.61, −56.31◦) is one way to represent the point (2, −3)
13. Convert the cartesian coordinates, (-5, -3), to polar coordinates. Solution:
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2007-2009 MathHands.com
math hands
pg. 14
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2
(r, θ)
(−5, −3)
330◦
b
b
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = −3/ − 5 (note: eq has infinite solutions) θ ≈ 30.96◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−5)2 + (−3)2 r2 =34
√ r = ± 34 ≈ ±5.83
note: if we use the angle 30.96◦, then we must choose the negative value of r, r ≈ −5.83, thus the polar cordinates (−5.83, 30.96◦) is one way to represent the point (−5, −3)
14. Convert the cartesian coordinates, (-4, -1), to polar coordinates. Solution:
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2007-2009 MathHands.com
math hands
pg. 15
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦ 180◦
-6
(r, θ) b
-5
-4
-3
(−4, −1) b
-2
-1
1
2
3
4
5
6
-1 -2
330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = −1/ − 4 (note: eq has infinite solutions) θ ≈ 14.04◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−4)2 + (−1)2 r2 =17
√ r = ± 17 ≈ ±4.12
note: if we use the angle 14.04◦, then we must choose the negative value of r, r ≈ −4.12, thus the polar cordinates (−4.12, 14.04◦) is one way to represent the point (−4, −1)
15. Convert the cartesian coordinates, (-6, 3), to polar coordinates. Solution:
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2007-2009 MathHands.com
math hands
pg. 16
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4
(r, θ) b
(−6, 3)
30◦
b
150◦
3 2 1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 3/ − 6 (note: eq has infinite solutions) ◦ θ ≈ −26.57 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−6)2 + (3)2 r2 =45
√ r = ± 45 ≈ ±6.71
note: if we use the angle −26.57◦, then we must choose the negative value of r, r ≈ −6.71, thus the polar cordinates (−6.71, −26.57◦) is one way to represent the point (−6, 3)
16. Convert the cartesian coordinates, (-8, 4), to polar coordinates. Solution:
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2007-2009 MathHands.com
math hands
pg. 17
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5
(r, θ)
(−8, 4)
b
b
4
30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 4/ − 8 (note: eq has infinite solutions) ◦ θ ≈ −26.57 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−8)2 + (4)2 r2 =80
√ r = ± 80 ≈ ±8.94
note: if we use the angle −26.57◦, then we must choose the negative value of r, r ≈ −8.94, thus the polar cordinates (−8.94, −26.57◦) is one way to represent the point (−8, 4)
17. Plot Points, for a Famous Graph Plot the following points and connect the points.
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2007-2009 MathHands.com
math hands
pg. 18
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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2007-2009 MathHands.com
math hands
angle theta r 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 345 360
5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00
pg. 19
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
angle theta r 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 345 360
Solution: 18. Plot Points, for a Famous Graph Plot the following points and connect the points.
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2007-2009 MathHands.com
math hands
pg. 20
5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00 4.33 2.50 0.00 -2.50 -4.33 -5.00 -4.33 -2.50 -0.00 2.50 4.33 5.00
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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2007-2009 MathHands.com
math hands
angle theta r 0 0.00 15 1.29 30 2.50 45 3.54 60 4.33 75 4.83 90 5.00 105 4.83 120 4.33 135 3.54 150 2.50 165 1.29 180 0.00 195 -1.29 210 -2.50 225 -3.54 240 -4.33 255 -4.83 270 -5.00 285 -4.83 300 -4.33 315 -3.54 330 -2.50 345 -1.29 360 -0.00 5sin th
pg. 21
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
angle theta r 0 0.00 15 1.29 30 2.50 45 3.54 60 4.33 75 4.83 90 5.00 105 4.83 120 4.33 135 3.54 150 2.50 165 1.29 180 0.00 195 -1.29 210 -2.50 225 -3.54 240 -4.33 255 -4.83 270 -5.00 285 -4.83 300 -4.33 315 -3.54 330 -2.50 345 -1.29 360 -0.00 5sin th
Solution:
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2007-2009 MathHands.com
math hands
pg. 22
Trigonometry Sec. 02 notes
MathHands.com M´ arquez
Introduction to Polar Graphs Main Idea Plot Points You should be very conformable plotting points in radians as well as degrees. Simply make a table of values with r’s and corresponding θ’s, then start plotting until you can make sense of the the graph. If connecting the points is uneasy, choose smaller angle intervals, i.e. instead of plotting and calculating the r every 30 degrees, plot every 5 degrees. Get to Know Your Calculator I am usually not a big calculator fan, but this may be a good time when we can use it appropriately to graph some of these functions. After all, there is limited beauty and creativity that occurs when plotting points. Get to know Famous Questions 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ? Get to know Famous Graphs Get to know Famous Graphs 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ?
r = sin θ
r = cos θ
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r = cos(2θ)
r = cos(3θ)
r = 5 + 3 sin θ
r = sin(2θ)
r = sin(4θ)
r = 3 + 5 sin θ
r = sin(3θ)
r = cos(5θ)
math hands
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
Introduction to Polar Graphs 1. Graph and Understand the graph of r = 5 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 2. Graph and Understand the graph of r = 5 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 3. Graph and Understand the graph of r = −3 cos θ
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 4. Graph and Understand the graph of r = 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 5. Graph and Understand the graph of r = −6 sin θ
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pg. 3
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 6. Graph and Understand the graph of r = 5 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 7. Graph and Understand the graph of r = 5 sin θ. Limit the study of this graph to the values of θ ranging from −90◦ to 0◦ .
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pg. 4
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 8. Graph and Understand the graph of r = −3 cos θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 9. Graph and Understand the graph of r = 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 180◦.
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pg. 5
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 10. Graph and Understand the graph of r = −6 sin θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 11. Graph and Understand the graph of r = 5 + cos θ
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pg. 6
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 12. Graph and Understand the graph of r = 4 − 2 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 13. Graph and Understand the graph of r = −3 + 2 cos θ
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pg. 7
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 14. Graph and Understand the graph of r = 3 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 15. Graph and Understand the graph of r = 4 − 2 cos θ
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pg. 8
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 16. Graph and Understand the graph of r = −2 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 17. Graph and Understand the graph of r = 3 + 2 sin θ
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pg. 9
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 18. Graph and Understand the graph of r = 3 − 2 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 19. Graph and Understand the graph of r = −3 + 2 sin θ
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pg. 10
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 20. Graph and Understand the graph of r = 3 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 21. Graph and Understand the graph of r = 2 − 3 sin θ
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pg. 11
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 22. Graph and Understand the graph of r = −2 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 23. Graph and Understand the graph of r = 2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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pg. 12
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 24. Graph and Understand the graph of r = 2 − 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 60◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 25. Graph and Understand the graph of r = −2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
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pg. 13
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 26. Graph and Understand the graph of r = 5 + 5 sin θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 27. Graph and Understand the graph of r = 3 − 2 sin θ. Limit the study of this graph to the values of θ ranging from 135◦ to 225◦ .
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pg. 14
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 28. Graph and Understand the graph of r = −3 + 2 sin θ. Limit the study of this graph to the values of θ ranging from 360◦ to 540◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 29. Graph and Understand the graph of r = 5 cos(2θ)
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pg. 15
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 30. Graph and Understand the graph of r = 6 cos(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 31. Graph and Understand the graph of r = 5 sin(2θ)
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pg. 16
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 32. Graph and Understand the graph of r = 6 sin(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 33. Graph and Understand the graph of r = 5 cos(3θ)
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pg. 17
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 34. Graph and Understand the graph of r = 6 cos(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 35. Graph and Understand the graph of r = 5 sin(3θ)
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pg. 18
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 36. Graph and Understand the graph of r = 4 sin(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 37. Graph and Understand the graph of r = 5 sin(2θ). Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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pg. 19
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 38. Graph and Understand the graph of r = 6 sin(4θ). Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 39. Graph and Understand the graph of r = 5 cos(3θ). Limit the study of this graph to the values of θ ranging from 0◦ to 180◦ .
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pg. 20
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 40. Graph and Understand the graph of r = 6 cos(5θ). Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 41. Graph and Understand the graph of r2 = 25 sin2 θ √ Solution: Note, we can solve for r as r√= ± 25 sin2 θ... then we can plot each r separately, that is we graph √ r = 25 sin2 θ and then we graph r = − 25 sin2 θ
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
42. Graph and Understand the graph of r2 = 25 sin2 3θ √ 25 sin2 3θ... then we can plot each r separately, that is we graph Solution: Note, we can solve for r as r = ± √ √ 2 r = 25 sin 3θ and then we graph r = − 25 sin2 3θ 90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
43. Understanding The graph of r = 2 − 3 cos θ has an inner loop. What values of θ trace precisely this inner loop? i.e. ?◦ ≤ θ ≤?◦ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center
44. Understanding On the graph of r = 5 sin(3θ) (a) ’sin 3θ’ will take values between 1 and −1; never larger than one, never smaller than −1. Find the angles θ, at which sin 3θ take on its largest value 1, or is smallest value -1. (i.e solve sin 3θ = 1 and solve sin 3θ = −1 )
(b) Graph r = 5 sin(3θ) and mark the points obtained from part a). What can you say about these points? A) nothing B) they are the ’tip of the leaves’ C) they are not on the leaves. (c) Find the angles at which the graph goes through (0, 0) (d) Find the interval/s for theta that trace the second ’leaf’ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center, such leaf starts and ends at center.. 45. Understanding On the graph of r = −5 + 2 cos θ
(a) ’cos θ’ will take values between 1 and −1; never larger than one, never smaller than −1. What is the largest or the smallest that the quantity ”−5 + 2 cos θ” can become?
(b) Find the angles at which ”−5 + 2 cos θ” takes on its largest value -3, or is smallest value -7. (c) Graph r = −5 + 2 cos θ and mark the points obtained from part b). What can you say about these points? A) nothing B) its the furthest point from the origin (d) Find the angles at which the graph goes through (0, 0) (e) Find the interval/s for theta that trace ONLY the outer leaf.
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Trigonometry Sec. 03 notes
MathHands.com M´ arquez
Converting Equations Polar & Cartesian The IDEA In the last couple sections we’ve established, among other things, a ’dictionary’ to convert coordinate points from cartesian to polar and vise-varsa. In this section, we would like to exploit that same dictionary to convert equations. As a reminder, the dictionary has two components, ’from’ polar, and ’to’ polar. The goal in converting, in most cases, is to eliminate all x’s and y’s and replace them with appropriate expressions involving r’s and θ’s if we are translating to polar. When converting from polar, the idea is to eliminate the r’s and θ’s. We summarize the ’dictionary’ below. Translating
Polar
Cartesian
90◦ 120◦
y = r sin θ x = r cos θ
60◦ 30◦
150◦ b
b
(r, θ)
(x, y)
0◦
180◦
210◦
330◦
θ = tan−1 300◦
240◦
r2 = x2 + y
270◦
y x 2
Example: Converting Equation of a line to Polar Convert the following equation to polar coordinates: y=x Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
y y x tan θ θ
=x
(given)
=1
(div by x, setting up to use translating dictionary, tan θ = y/x)
=1 = 45◦
(used tan θ = y/x) (notice: this there are infinite possibilities for θ (ie −135 , 225 ...), this is just one of them.) ◦
◦
Thus, one way to translate the equation y = x to polar is to convert to θ = 45◦ Example: Converting Equation of a circle to Polar Convert the following equation to polar coordinates: x2 + y 2 = 25 Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
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Trigonometry Sec. 03 notes
MathHands.com M´ arquez x2 + y 2 = 25
(given)
2
2
(used the dictionary r = x2 + y 2 )
r = 25
at this point we have eliminated all x’s and y’s from the equation, thus the converting is complete. It is sometimes desirable to solve for r explicitly, thus we could also say r = 5 or r = −5, in fact all three of these equations, r2 = 25, r = 5, AND r = −5 all yield the same graph as x2 + y 2 = 25. Example: Converting the equation of a circle to Polar Convert the following equation to polar coordinates: x2 + 3x + 5 + y 2 − 4y + 2 = 25 Solution:
x2 + 3x + 5 + y 2 − 4y + 2 = 25 2
2
(given)
2
(x + y ) + 3(x) − 4(y) = 18
(just cleaning up, getting ready to use the dictionary)
(r ) + 3(r cos θ) − 4(r sin θ) = 18
(used the dictionary)
2
r + 3r cos θ − 4r sin θ = 18
(algebra)
Example: Converting the equation of a circle from Polar Convert the following equation of a circle to polar coordinates: r = 12 cos θ Solution: r = 12 cos θ r · r = r(12 cos θ)
r2 = 12(r cos θ) 2
2
x + y = 12x 2
2
2
y + x + − 12x = 0 2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
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math hands
(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Converting Equations Polar & Cartesian 1. Convert the following equation to polar coordinates: y = 2x
Solution: Note: the ’translating’ of this equation is not unique.
y = 2x y =2 x tan θ = 2 θ ≈ 63.435
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) ◦
(notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 63.435◦ 2. Convert the following equation to polar coordinates: 2 y= − x 3
Solution: Note: the ’translating’ of this equation is not unique. 2 y= − x 3 2 y = − x 3 2 tan θ = − 3 θ ≈ −33.69◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ −33.69◦ 3. Convert the following equation to polar coordinates: y=
2 x 5
Solution: Note: the ’translating’ of this equation is not unique.
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pg. 3
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez 2 x 5 2 = 5 2 = 5 ≈ 21.801◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 21.801◦ 4. Convert the following equation to polar coordinates: y=
4 x 3
Solution: Note: the ’translating’ of this equation is not unique. 4 x 3 4 = 3 4 = 3 ≈ 53.13◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 4x to polar is to convert to θ ≈ 53.13◦ 5. Convert the following equation to polar coordinates: 4 y= − x 3 Solution: Note: the ’translating’ of this equation is not unique. 4 y= − x 3 4 y = − x 3 4 tan θ = − 3 θ ≈ −53.13◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = − 4x to polar is to convert to θ ≈ −53.13◦ 6. Convert the following equation of a circle to polar coordinates: 2x2 + 3x + 2y 2 + − 5y = 7
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution:
2x2 + 3x + 2y 2 + − 5y = 7 2 x2 + y 2 + 3(x) + − 5(y) = 7 2 r2 + 3(r cos θ) + − 5(r sin θ) = 7
(given) (ready to use the dictionary....) (used the dictionary)
7. Convert the following equation of a circle to polar coordinates: − 5x2 + 2x + − 5y 2 + 7y = 25 Solution:
− 5x2 + 2x + − 5y 2 + 7y = 25 − 5 x2 + y 2 + 2(x) + 7(y) = 25 − 5 r2 + 2(r cos θ) + 7(r sin θ) = 25
(given) (ready to use the dictionary....) (used the dictionary)
8. Convert the following equation of a circle to polar coordinates: 3 4x2 + x + 4y 2 + 1y = 9 2
Solution: 3 4x2 + x + 4y 2 + 1y = 9 2 3 4 x2 + y 2 + (x) + 1(y) = 9 2 3 2 4 r + (r cos θ) + 1(r sin θ) = 9 2
(given) (ready to use the dictionary....) (used the dictionary)
9. Convert the following equation of a circle to polar coordinates: 3x2 + 3x + 3y 2 + 5y = 4
Solution:
3x2 + 3x + 3y 2 + 5y = 4 3 x2 + y 2 + 3(x) + 5(y) = 4 3 r2 + 3(r cos θ) + 5(r sin θ) = 4
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(given) (ready to use the dictionary....) (used the dictionary)
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
10. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 9 25
Solution: y2 x2 + =1 9 25 (r cos θ)2 (r sin θ)2 + =1 9 25 ! cos2 θ sin2 θ 2 + =1 r 9 25
(given) (ready to use the dictionary....) (alebra)
11. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 16
Solution: y2 x2 + =1 4 16 (r cos θ)2 (r sin θ)2 + =1 4 16 ! cos2 θ sin2 θ + =1 r2 4 16
(given) (ready to use the dictionary....) (alebra)
12. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 49
Solution: y2 x2 + =1 4 49 (r cos θ)2 (r sin θ)2 + =1 4 49 ! cos2 θ sin2 θ =1 + r2 4 49
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(given) (ready to use the dictionary....) (alebra)
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Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
13. Convert the following equation of a circle to polar coordinates: r = 2 sin θ
Solution: r = 2 sin θ
(given)
r · r = r(2 sin θ)
(clever little idea, mult both sides by r)
r2 = 2(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 2y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 2y = 0
(optional step: complete the square)
2
x + y + − 2y + 1 = 1 2 x2 + y + − 1 = 1
(optional to step: complete the square) (Shows its a circle, shows center/radius)
14. Convert the following equation of a circle to polar coordinates: r = 5 sin θ
Solution: r = 5 sin θ
(given)
r · r = r(5 sin θ)
(clever little idea, mult both sides by r)
2
r = 5(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 5y = 0 25 25 x2 + y 2 + − 5y + = 4 4 2 25 5 = x2 + y + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
15. Convert the following equation of a circle to polar coordinates: r = 6 sin θ
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pg. 7
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 6 sin θ r · r = r(6 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 6(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 6y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 6y = 0
(optional step: complete the square)
2
x + y + − 6y + 9 = 9 2 x2 + y + − 3 = 9
(optional to step: complete the square) (Shows its a circle, shows center/radius)
16. Convert the following equation of a circle to polar coordinates: r = 11 sin θ
Solution: r = 11 sin θ r · r = r(11 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 11y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 11y = 0 121 121 = x2 + y 2 + − 11y + 4 4 2 11 121 x2 + y + − = 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
17. Convert the following equation of a circle to polar coordinates: r = 8 cos θ
Solution: r = 8 cos θ
(given)
r · r = r(8 cos θ)
(clever little idea, mult both sides by r)
2
r = 8(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 8x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 8x = 0
(optional step: complete the square)
2
y + x + − 8x + 16 = 16 2 y 2 + x + − 4 = 16
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(optional to step: complete the square) (Shows its a circle, shows center/radius)
math hands
pg. 8
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
18. Convert the following equation of a circle to polar coordinates: r = 12 cos θ
Solution: r = 12 cos θ
(given)
r · r = r(12 cos θ)
(clever little idea, mult both sides by r)
r2 = 12(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 12x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 12x = 0
(optional step: complete the square)
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
(optional to step: complete the square) (Shows its a circle, shows center/radius)
19. Convert the following equation of a circle to polar coordinates: r = 5 cos θ
Solution: r = 5 cos θ
(given)
r · r = r(5 cos θ)
(clever little idea, mult both sides by r)
2
r = 5(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 5x = 0 25 25 y 2 + x2 + − 5x + = 4 4 2 25 5 = y2 + x + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
20. Convert the following equation of a circle to polar coordinates: r = 11 cos θ
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math hands
pg. 9
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 11 cos θ r · r = r(11 cos θ) r2 = 11(r cos θ) 2
2
x + y = 11x 2
2
y + x + − 11x = 0 121 121 = y 2 + x2 + − 11x + 4 4 2 121 11 = y2 + x + − 2 4
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math hands
(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 10
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
Intro to Complex Numbers The Question In the last section we wondered what might be far to the right and to the left of the real number line. In doing so, we entertained the existence of the very special unreal number ∞. In this section, we continue on this path and now consider what lies above or below the real number line?
? ∞
−∞ -5
-4
Out of Audacity, a number is born
-3
-2
-1
0
1
2
3
4
5
?
Here is the account, directly from the world famous, the almost immortal, mathematician Euler himself.... ”I sat in that soft comfortable chair leaned back and enjoyed the million thoughts dancing inside my head. A blank paper, a pencil and my awesome coffee was all there was on the desk. On the paper the equation x2 = −1 The equation was screaming, enticing, talking trash, challenging me, saying ”you can’t solve me!” Hours went by faster than I would have liked. Days past by, weeks and months...There was no real number that would solve the equation. But the forces were greater. The inspiration divine. I would not be stopped.. and one day it happened. There was no real number solution, I had looked on the positive side on the negative side and all numbers between. Resolved to avoid defeat at all costs, I invented a number. From my own imagination, I gathered all my might, my courage, and my audacity, √ and I thought...I will create a number. I will call it i, and I will solve my problem by declaring i = −1. It’s my number so I can make it behave however I please, just as the artists paints the clouds at his whim... This solves the equation x2 = −1 and marks the birth of a grand elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.” Needless, to say, I have taken some artistic liberties with this account of events. In fact, traces of complex numbers or ’imaginary’ numbers can be found in 9th century’s Al-Khwarizmi’s Algebra text. During the next couple centuries these ideas made their way to Italy and France, as people were learning to solve degree 3 equations.√ By the turn of the 17th century Descartes coined the phrase ”imaginary” numbers, referring √ to numbers such as −1. At last, it was Euler, in the 18th century who named such number i, declaring i = −1. Thus... by definition of i;
i2 = −1 and, i is a solution to x2 = −1 In addition to solving the equation x2 =− 1, and this marks the birth of a grand and elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.
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pg. 1
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
Where do all the i’s Live With the invent of i came other numbers such as 2i, 3i, −5i, or 2 + 3i. Generally, complex numbers are numbers that can be written in the form a + bi where a and b are real numbers. Now, in the previous sections we noted a visual representation of the real numbers using the real number line. Considering this, the natural question is where or how do we represent the complex numbers visually? Over the centuries, the most powerful and common way to represent the complex numbers is to place them as an extension of the real number line, extending it above and below to make what we commonly call the complex plane. In essence, this is done by placing an ’imaginary’ axis perpendicular to the real number line. Then we position every number a+ bi on the plane similarly to placing the ordered point (a, b) on the cartesian [xy]plane. came numbers such as 2i, 3i, 4i, . . . as well as numbers such as 2 + 3i. Here are some visual representations of a few complex numbers along with their position relative to the known real number line. 3i
The Complex Numbers, C
2i b
−4 + i
b
3 + 2i
3
4
1i 2 + 0i
R
b
-5
-4
-3
-2
-1
0
1
2
5
−1i −2i −3i b
4 − 3i
Negative Radicals With the invention of i we can now make sense of radicals (i.e. square roots) of √ negative real numbers. Consider √ the radical −1, the number whose square is −1. Recall when we first defined 4 we did so as ’the number whose number square is 4.’ But there are two such numbers 2 and −2. By default, we declared to radical to mean the positive number √ whose square is 4. We follow a similar logic here, as we are confronted with the same dilemma. If we define −1 as the number whose square is −1, we will find there are two possible choices, i and −i (see examples). By convention, we will define negative radical to be, the positive i rather than −i. Examples 1. Simplify
√ −4
√
√ −4 = i 4 = i2 = 2i
2. Simplify
√ −10
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√ √ −10 = i 10
math hands
(neg rad) (def rad) (CoLM)
(neg rad)
pg. 2
Trigonometry Sec. 04 notes 3. Simplify
4. Simplify
√ −15
MathHands.com M´ arquez √ √ −15 = i 15
(neg rad)
√ −x
Solution: Notice, we do not know the value of x. We don’t know if x is positive √ or negative. This means we don’t know if −x is positive or negative therefore we don’t know if the radical −x is positive or negative.
5. Adding in C
3 + 5i + 2 + 3i
Solution:
3 + 5i + 2 + 3i = = (3 + 2) + (5i + 3i)
(given) (ALA)
= 3 + 2 + (5 + 3)i
(DL)
= 5 + 8i
(AT)
6. Multiplying in C (3 + 5i)(2 + 3i)
Solution:
(3 + 5i)(2 + 3i) = = 3 · 2 + 5i · 2 + 3 · 3i + 5i · 3i = 6 + 10i + 9i + 15i2
= 6 + (10 + 9)i + 15i
2
(given) (FOIL) (BI) (DL)
2
(AT) (Def of i)
= 6 + 19i + −15 = −9 + 19i
(BI) (BI)
= 6 + 19i + 15i = 6 + 19i + 15 · −1
7. Multiplying in C (4 +− 5i)(2 + 3i)
Solution:
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pg. 3
Trigonometry Sec. 04 notes
MathHands.com M´ arquez (4 +− 5i)(2 + 3i) =
(given)
= 4 · 2 + 5i · 2 + 4 · 3i + 5i · 3i −
−
= 8 + 10i + 12i + 15i −
−
2
= 8 + ( 10 + 12)i + 15i −
−
= 6 + 2i + 15i −
2
2
(FOIL) (BI) (DL) (BI)
= 6 + 2i + 15 · −1 = 6 + 2i + 15
(Def of i) (NNT)
−
= 21 + 2i
(BI)
8. Multiplying in C (4 + 3i)(2 + 3i)
Solution:
(4 + 3i)(2 + 3i) =
(given)
= 8 + 12i + 6i + 9i
2
= 8 + 12i + 6i + 9 · 1 −
(FOIL) (def i)
= 8 + 12i + 6i + 9
(BI)
= 1 + 18i
(BI)
−
−
9. Multiplying in C i7
Solution:
i7 = iiiiiii 2 2 2
(+Expo)
=i i i i
(+Expo)
= −1 · −1 · −1 · i = −1 · i
(Def of i) (BI)
= −i
(MT)
And Now, Divide We have now introduced the imaginary number, their standard form ’a+bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide.
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math hands
pg. 4
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3 5i − 3 5i − 3 = ·1 i+3 i+3 = =
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5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
math hands
pg. 5
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
=
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 11 −3 − 11i = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
2i + 1 −i+1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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math hands
pg. 6
Trigonometry Sec. 04 notes
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 5i − 2 2i + 3 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
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2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
math hands
pg. 7
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Intro to Complex Numbers 1. Plot the following points. (a) 2 + 3i (b) 5 + 1i (c) −4 + −3i
(d) −4i
Solution:
8i 7i 6i 5i 4i b
3i
2 + 3i
2i b
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5 + 1i
5
6
7
8
9 10 11 12
-2i
−4 + −3i b
-3i
0 + −4i
-4i b -5i -6i -7i -8i
2. Plot the following points. (a) −1 + 4i
(b) 2 + 2i
(c) 5 + −5i
(d) −7i
Solution:
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math hands
pg. 8
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
8i 7i 6i 5i
−1 + 4i b 4i 3i b
2i
2 + 2i
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5
6
7
8
9 10 11 12
-2i -3i -4i -5i
5b + −5i
-6i
0 + −7i
-7i b -8i
Simplify
√ −90
Solution: √ −90
4. 3. Simplify
√ = i 90 √ = i 9 · 10 √ √ = i 9 10 √ = i · 3 10 √ = 3i 10
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −150
Solution:
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math hands
pg. 9
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −150
5. Simplify
√ = i 150 √ = i 25 · 6 √ √ = i 25 6 √ = i·5 6 √ = 5i 6
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −375
Solution: √ −375
6. Simplify
√ = i 375 √ = i 25 · 15 √ √ = i 25 15 √ = i · 5 15 √ = 5i 15
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −18
Solution: √ −18
7. Simplify
√ = i 18 √ =i 9·2 √ √ =i 9 2 √ = i·3 2 √ = 3i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −50
Solution:
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math hands
pg. 10
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −50
8. Simplify
√ = i 50 √ = i 25 · 2 √ √ = i 25 2 √ = i·5 2 √ = 5i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −8
Solution: √ −8
9. Simplify
√ =i 8 √ = i 4·2 √ √ =i 4 2 √ = i·2 2 √ = 2i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −125
Solution: √ −125
10. Simplify
√ = i 125 √ = i 25 · 5 √ √ = i 25 5 √ = i·5 5 √ = 5i 5
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ 3 −8
Solution: −2 since (−2)3 = −8 11. Simplify
√ 3 −125
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pg. 11
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: −5 since (−5)3 = −125 12. Simplify
√ 3 −1
Solution: −1 since (−1)3 = −1 13. Add 3i + 2 + 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 5i) = 2 + (3i + 4) + 5i = 2 + (4 + 3i) + 5i = (2 + 4) + (3i + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i
(ALA) (CoLA) (ALA) (DL) (BI)
14. Add 2i + 7 + 9i + 3 Solution:
by convention, we write in ’standard form’, ”a + bi” (7 + 2i) + (3 + 9i) = 7 + (2i + 3) + 9i = 7 + (3 + 2i) + 9i = (7 + 3) + (2i + 9i) = (7 + 3) + (2 + 9)i = 10 + 11i
(ALA) (CoLA) (ALA) (DL) (BI)
15. Add 3i − 2 + − 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (−2 + 3i) + (4 + −5i) = −2 + (3i + 4) + −5i
= −2 + (4 + 3i) + −5i = (−2 + 4) + (3i + −5i)
= (−2 + 4) + (3 + −5)i = 2 + −2i
(ALA) (CoLA) (ALA) (DL) (BI)
16. Add 30i + 11 + 5i + 5
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pg. 12
Trigonometry Sec. 04 exercises Solution:
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” (11 + 30i) + (5 + 5i) = 11 + (30i + 5) + 5i = 11 + (5 + 30i) + 5i = (11 + 5) + (30i + 5i) = (11 + 5) + (30 + 5)i = 16 + 35i
(ALA) (CoLA) (ALA) (DL) (BI)
17. Add i + 1 + i + 2 Solution:
by convention, we write in ’standard form’, ”a + bi” (1 + 1i) + (2 + 1i) = 1 + (1i + 2) + 1i = 1 + (2 + 1i) + 1i = (1 + 2) + (1i + 1i) = (1 + 2) + (1 + 1)i = 3 + 2i
(ALA) (CoLA) (ALA) (DL) (BI)
18. Add 3i + 2 + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 0i) = 2 + (3i + 4) + 0i = 2 + (4 + 3i) + 0i = (2 + 4) + (3i + 0i) = (2 + 4) + (3 + 0)i = 6 + 3i
(ALA) (CoLA) (ALA) (DL) (BI)
19. Add 3i + i Solution:
by convention, we write in ’standard form’, ”a + bi” (0 + 3i) + (0 + 1i) = 0 + (3i + 0) + 1i = 0 + (0 + 3i) + 1i = (0 + 0) + (3i + 1i) = (0 + 0) + (3 + 1)i = 0 + 4i
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(ALA) (CoLA) (ALA) (DL) (BI)
pg. 13
Trigonometry Sec. 04 exercises 20. Multiply 3i + 2
MathHands.com M´ arquez 5i + 4
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 5i + 4 = (2)(4) + (2)(5i) + (3i)(4) + (3i)(5i) = 8 + 10i + 12i + 15i
2
9i + 3
(BI)
= 8 + 10i + 12i + 15(−1)
(def i)
= (8 + −15) + (10i + 12i) = (8 + −15) + (10 + 12)i
(CoLA, ALA,BI) (DL)
= −7 + 22i
21. Multiply 2i + 7
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 2i + 7 9i + 3 = (7)(3) + (7)(9i) + (2i)(3) + (2i)(9i) 2
= 21 + 63i + 6i + 18i = 21 + 63i + 6i + 18(−1)
(BI) (def i)
= (21 + −18) + (63i + 6i) = (21 + −18) + (63 + 6)i
(CoLA, ALA,BI) (DL)
= 3 + 69i
22. Multiply 3i − 2
− 5i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i − 2 − 5i + 4 = (−2)(4) + (−2)(−5i) + (3i)(4) + (3i)(−5i) = −8 + 10i + 12i + −15i
2
= −8 + 10i + 12i + −15(−1) = (−8 + 15) + (10i + 12i)
= (−8 + 15) + (10 + 12)i = 7 + 22i
23. Multiply 30i + 11
5i + 5
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(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
math hands
pg. 14
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: by convention, we write in ’standard form’, ”a + bi” 30i + 11 5i + 5 = (11)(5) + (11)(5i) + (30i)(5) + (30i)(5i) = 55 + 55i + 150i + 150i
2
= 55 + 55i + 150i + 150(−1) = (55 + −150) + (55i + 150i)
= (55 + −150) + (55 + 150)i = −95 + 205i
24. Multiply i + 1
i+2
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
25. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i)
(BI) (def i)
= (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i
(CoLA, ALA,BI) (DL)
= 8 + −8i + 12i + −12i = 8 + −8i + 12i + −12(−1)
= 20 + 4i
26. Multiply 3i − 1
i−1
(FOIL)
2
(BI)
Solution:
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pg. 15
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i)
(DL)
= −2 + −4i
i+2
(BI) (def i) (CoLA, ALA,BI)
= (1 + −3) + (−1 + −3)i
27. Multiply i + 1
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
28. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i) = 8 + −8i + 12i + −12i
2
= 8 + −8i + 12i + −12(−1) = (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i = 20 + 4i
29. Multiply 3i − 1
i−1
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution:
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Trigonometry Sec. 04 exercises
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by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i) = (1 + −3) + (−1 + −3)i = −2 + −4i
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
30. multiply i3 Solution: i3 = i2 · i
= −1 · i = −i
(JAE) (def of i) (MT)
31. multiply i4 Solution: i4 = i2 · i2 = −1 · −1 =1
(JAE) (def of i) (NotNot)
32. multiply i6 Solution: i6 = i4 · i2
= 1 · −1 = −1
(JAE) (see previous problem) (MiD)
3
33. (3i + 2)
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Solution:
(3i + 2)3
(given) 3
= (3i)
2
+
3
2
3 · (3i) (2)
2
+
3(3i)(2)
+
3
(2)
(famous, PP3)
= 27i + 54i + 36i + 8
(BI)
= 27(−i) + 54(−1) + 36(i) + 8 = −46 + 9i
(BI) (BI)
3
34. (2i + 1)
Solution:
(2i + 1)3
(given) 3
= (2i) 3
2
+
3 · (2i) (1)
2
2
+
3(2i)(1)
+
3
(1)
(famous, PP3)
= 8i + 12i + 6i + 1
(BI)
= 8(−i) + 12(−1) + 6(i) + 1 = −11 + −2i
(BI) (BI)
3
35. (−2i + 1)
Solution:
3
(−2i + 1)
(given) 3
= (−2i)
3
2
+ 2
3 · (−2i) (1)
+
= − 8i + 12i − 6i + 1
= −8(−i) + 12(−1) + −6(i) + 1 = −11 + 2i
2
3(−2i)(1)
+
3
(1)
(famous, PP3) (BI) (BI) (BI)
3
36. (−2i + 3)
Solution:
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Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 3
(−2i + 3)
(given) 3
= (−2i)
2
+
3
2
3 · (−2i) (3)
+
2
3(−2i)(3)
3
+
(3)
= − 8i + 36i − 54i + 27 = −8(−i) + 36(−1) + −54(i) + 27
(famous, PP3) (BI) (BI)
= −9 + −46i
(BI)
37. (−1i + 1)3 Solution:
3
(−1i + 1)
(given) 3
= (−1i)
+
3
2
2
3 · (−1i) (1)
+
2
3(−1i)(1)
3
+
(1)
= − i + 3i − 3i + 1 = −1(−i) + 3(−1) + −3(i) + 1
(famous, PP3) (BI) (BI)
= −2 + −2i
(BI)
38. (1i + 1)3 Solution:
3
(1i + 1)
(given) 3
= (1i) 3
39. Divide
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+ 2
2
3 · (1i) (1)
+
2
3(1i)(1)
+
3
(1)
(famous, PP3)
= i + 3i + 3i + 1 = 1(−i) + 3(−1) + 3(i) + 1
(BI) (BI)
= −2 + 2i
(BI)
7i + 1 7i + 3
math hands
pg. 19
Trigonometry Sec. 04 exercises
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Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
40. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
41. Divide
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5i + 2 4i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
math hands
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Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
42. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
43. Divide
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3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
math hands
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Trigonometry Sec. 04 exercises
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Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
44. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
45. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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pg. 22
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
46. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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pg. 23
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
47. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 48. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 49. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 50. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2
(just add exponents) (setting up to use fact i4 = 1)
3
= (1) · i2 =i
(yipei-kae-yeh...)
2
= −1
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51. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
52. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
53. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
54. Compute w/ Complex Numbers Calculate and write in standard form. i−5
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Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
55. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
56. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
57. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
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·i
math hands
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
pg. 26
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
58. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
59. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
60. Compute w/ Complex Numbers Calculate and write in standard form. i11
Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1)
= (1) · i
=i
(ez as sundays)
61. Compute w/ Complex Numbers Calculate and write in standard form. i−6
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Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
62. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
= (1)8 · i
(used i4 = 1)
=i
(ez as sundays)
63. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
64. Compute w/ Complex Numbers Calculate and write in standard form.
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1 1 √ +√ i 2 2
math hands
2
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Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i = √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
(FOIL) (used i2 = −1)
65. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 3 1 + i 2 2
(FOIL) (used i2 = −1)
66. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
67. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
68. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
(getting ready to use A3 = i)
10
= (i)
(used A3 = i)
= −1
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(ez as Sundays...)
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pg. 30
Trigonometry Sec. 04 exercises
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69. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 70. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
71. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 72. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 73. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 74. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
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Trigonometry Sec. 5
MathHands.com M´ arquez
More onComplex Numbers The Idea Last section was ’out of this world’. We introduced the imaginary number, their standard form ’a + bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide. The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3
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Trigonometry Sec. 5
MathHands.com M´ arquez 5i − 3 5i − 3 = ·1 i+3 i+3 = =
5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi = 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 − 11i −3 11 = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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Trigonometry Sec. 5
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so.. 2i + 3 − 5i − 2 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
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Trigonometry Sec. 5
MathHands.com M´ arquez
Instructions 1. Divide
7i + 1 7i + 3
Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
2. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
3. Divide
4i + 2 5i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
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Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
4. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
5. Divide
3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
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Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
6. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
7. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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Trigonometry Sec. 5
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
8. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
c
2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
9. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 10. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 11. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 12. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2 3
= (1) · i2 =i
(just add exponents) (setting up to use fact i4 = 1) (yipei-kae-yeh...)
2
= −1
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
13. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
14. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
15. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
16. Compute w/ Complex Numbers Calculate and write in standard form. i−5
c
2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
17. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
18. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
19. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
c
2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
20. Compute w/ Complex Numbers Calculate and write in standard form. i2
Solution: i2 = i0 · i1 0 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
0
(used i4 = 1) (ez as sundays)
= (1) · i =i
21. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
22. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 = i4 ·i = (1)
−1
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
23. Compute w/ Complex Numbers Calculate and write in standard form. i11
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1) (ez as sundays)
= (1) · i =i
24. Compute w/ Complex Numbers Calculate and write in standard form. i−6
Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
25. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
8
(used i4 = 1) (ez as sundays)
= (1) · i =i
26. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
c
2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez
27. Compute w/ Complex Numbers Calculate and write in standard form.
1 1 √ +√ i 2 2
2
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
=
(FOIL) (used i2 = −1)
28. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 1 3 + i 2 2
(FOIL) (used i2 = −1)
29. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
30. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
31. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
10
= (i)
= −1
(getting ready to use A3 = i) (used A3 = i) (ez as Sundays...)
c
2007-2009 MathHands.com
Trigonometry Sec. 5
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32. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 33. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
34. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 35. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 36. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 37. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
c
2007-2009 MathHands.com
Trigonometry Sec. 8
MathHands.com M´ arquez
1. Vector Dot Product (a) Compute the following DOT product: < 1, 2 > · < 3, 5 > Solution: < 1, 2 > · < 3, 5 > = (1)(3) + (2)(5) = 3 + 10 = 13
(def of the DOT) (by Calc) (by Calc)
(b) Compute the following DOT product: < 3, 5 > · < 3, 5 > Solution: < 3, 5 > · < 3, 5 > = (3)(3) + (5)(5) = 9 + 25 = 34
(def of the DOT) (by Calc) (by Calc)
(c) Compute the following DOT product: < −3, 2 > · < 1, 2 > Solution: < −3, 2 > · < 1, 2 > = (−3)(1) + (2)(2) = −3 + 4 =1
(def of the DOT) (by Calc) (by Calc)
(d) Compute the following DOT product: < 1, 4 > · < −3, −5 > Solution: < 1, 4 > · < −3, −5 > = (1)(−3) + (4)(−5) = −3 + −20 = −23
(def of the DOT) (by Calc) (by Calc)
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(e) Compute the following DOT product: < 2, 3 > · < 3, −3 > Solution: < 2, 3 > · < 3, −3 > = (2)(3) + (3)(−3)
(def of the DOT)
= 6 + −9
(by Calc)
= −3
(by Calc)
(f) Compute the following DOT product: < 1, −4 > · < 3, 7 > Solution: < 1, −4 > · < 3, 7 > = (1)(3) + (−4)(7)
(def of the DOT)
= 3 + −28 = −25
(by Calc) (by Calc)
(g) < 2, 1, 3 > · < 1, 2, 1 > Solution: 2 + 2 + 3 = 7 2. Vector Dot Product (a) Compute the following DOT product: (1i + 2j) · (0i + 5j) Solution: (1i + 2j) · (0i + 5j) = (1 < 1, 0 > +2 < 0, 1 >) · (0 < 1, 0 > +5 < 0, 1 >) =< 1, 2 > · < 0, 5 > = (1)(0) + (2)(5) = 0 + 10 = 10
(def of i and j) (by Inspection) (def of the DOT) (by Calc) (by Calc)
(b) Compute the following DOT product: (3i + 0j) · (3i + 5j)
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Solution: (3i + 0j) · (3i + 5j) = (3 < 1, 0 > +0 < 0, 1 >) · (3 < 1, 0 > +5 < 0, 1 >) =< 3, 0 > · < 3, 5 > = (3)(3) + (0)(5) =9+0 =9
(def of i and j) (by Inspection) (def of the DOT) (by Calc) (by Calc)
(c) Compute the following DOT product: (−3i + 2j) · (1i + 2j) Solution: (−3i + 2j) · (1i + 2j) = (−3 < 1, 0 > +2 < 0, 1 >) · (1 < 1, 0 > +2 < 0, 1 >) =< −3, 2 > · < 1, 2 > = (−3)(1) + (2)(2) = −3 + 4
(def of i and j) (by Inspection) (def of the DOT) (by Calc)
=1
(by Calc)
(d) Compute the following DOT product: (1i + 4j) · (−3i + −5j) Solution: (1i + 4j) · (−3i + −5j) = (1 < 1, 0 > +4 < 0, 1 >) · (−3 < 1, 0 > + − 5 < 0, 1 >) =< 1, 4 > · < −3, −5 > = (1)(−3) + (4)(−5)
= −3 + −20 = −23
(def of i and j) (by Inspection) (def of the DOT) (by Calc) (by Calc)
3. Vector Dot Product to find magnitude (a) Use the dot product to find the magnitude of the following vector: ~v =< 1, 2 >
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Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 1, 2 > · < 1, 2 > (given) = (1)(1) + (2)(2) =1+4
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 5 √ ||~v || = 5
(def of the DOT) (by Calc) (by Calc)
||~v || ≈ 2.24
(b) Use the dot product to find the magnitude of the following vector: ~v =< 3, 5 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 3, 5 > · < 3, 5 > (given) = (3)(3) + (5)(5) = 9 + 25
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 34 √ ||~v || = 34 ||~v || ≈ 5.83
(def of the DOT) (by Calc) (by Calc)
(c) Use the dot product to find the magnitude of the following vector: ~v =< −3, 2 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< −3, 2 > · < −3, 2 > (given) = (−3)(−3) + (2)(2)
=9+4 ....thus.... ....then.. (assuming norm is positive)....
2
||~v || = 13 √ ||~v || = 13 ||~v || ≈ 3.61
(def of the DOT)
(by Calc) (by Calc)
(d) Use the dot product to find the magnitude of the following vector: ~v =< 1, 4 >
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Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 1, 4 > · < 1, 4 > (given) = (1)(1) + (4)(4) = 1 + 16
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 17 √ ||~v || = 17
(def of the DOT) (by Calc) (by Calc)
||~v || ≈ 4.12
(e) Use the dot product to find the magnitude of the following vector: ~v =< 2, 3 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 2, 3 > · < 2, 3 > (given) = (2)(2) + (3)(3) =4+9
....thus.... ....then.. (assuming norm is positive)....
||~v ||2 = 13 √ ||~v || = 13 ||~v || ≈ 3.61
(def of the DOT) (by Calc) (by Calc)
(f) Use the dot product to find the magnitude of the following vector: ~v =< 1, −4 > Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 1, −4 > · < 1, −4 > (given) = (1)(1) + (−4)(−4) = 1 + 16 ....thus.... ....then.. (assuming norm is positive)....
2
||~v || = 17 √ ||~v || = 17 ||~v || ≈ 4.12
(def of the DOT)
(by Calc) (by Calc)
(g) Use the dot product to find the magnitude of the following vector: ~v =< 3, −4 >
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Trigonometry Sec. 8
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Solution: ||~v ||2 = ~v · ~v (famous DOT product property) =< 3, −4 > · < 3, −4 > (given) = (3)(3) + (−4)(−4) = 9 + 16
....thus.... ....then.. (assuming norm is positive)....
(def of the DOT) (by Calc)
||~v ||2 = 25 √ ||~v || = 25
(by Calc)
||~v || ≈ 5
4. Vector Dot Product to find ’distance’ The DOT product provides a way to define the ’distance’ between two vectors, ~v and w. ~ So long as we can define subtraction of the vectors we can define the distance between them as follows: p dist(~v , w) ~ = ||~v − w|| ~ = (~v − w) ~ · (~v − w) ~ (a) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 1, 2 >
~v =< 3, 5 >
Solution: First note that ~v − w ~ =< 2, 3 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 2, 3 > · < 2, 3 > = (2)(2) + (3)(3) =4+9 ....thus.... ....then.. (assuming the distance is positive)....
2
||~v − w|| ~ = 13 √ ||~v − w|| ~ = 13
(from given) (def of the DOT) (by Calc) (by Calc)
||~v − w|| ~ ≈ 3.61
(b) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< −3, 2 >
~v =< 1, 2 >
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Trigonometry Sec. 8
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Solution: First note that ~v − w ~ =< 4, 0 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 4, 0 > · < 4, 0 > = (4)(4) + (0)(0)
(from given) (def of the DOT)
= 16 + 0
(by Calc)
2
....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ = 16 √ ||~v − w|| ~ = 16
(by Calc)
||~v − w|| ~ ≈4
(c) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 1, 4 >
~v =< −3, −5 >
Solution: First note that ~v − w ~ =< −4, −9 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< −4, −9 > · < −4, −9 > = (−4)(−4) + (−9)(−9) = 16 + 81 ....thus.... ....then.. (assuming the distance is positive)....
(from given)
(def of the DOT) (by Calc)
2
||~v − w|| ~ = 97 √ ||~v − w|| ~ = 97 ||~v − w|| ~ ≈ 9.85
(by Calc)
(d) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 2, 3 >
~v =< 3, −3 >
Solution: First note that ~v − w ~ =< 1, −6 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 1, −6 > · < 1, −6 > = (1)(1) + (−6)(−6) = 1 + 36
....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ 2 = 37 √ ||~v − w|| ~ = 37 ||~v − w|| ~ ≈ 6.08
(from given) (def of the DOT) (by Calc) (by Calc)
(e) Use the dot product to find the ’distance’ between the indicated vectors.
©2007-2008 MathHands.com
Trigonometry Sec. 8
MathHands.com M´ arquez w ~ =< 1, −4 >
~v =< 3, 7 >
Solution: First note that ~v − w ~ =< 2, 11 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 2, 11 > · < 2, 11 >
= (2)(2) + (11)(11) = 4 + 121 ....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ 2 = 125 √ ||~v − w|| ~ = 125 ||~v − w|| ~ ≈ 11.18
(from given) (def of the DOT) (by Calc) (by Calc)
(f) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 3, −4 >
~v =< 3, 7 >
Solution: First note that ~v − w ~ =< 0, 11 > ||~v − w|| ~ 2 = (~v − w) ~ · (~v − w) ~ (famous DOT product property) =< 0, 11 > · < 0, 11 >
= (0)(0) + (11)(11) = 0 + 121 ....thus.... ....then.. (assuming the distance is positive)....
||~v − w|| ~ 2 = 121 √ ||~v − w|| ~ = 121
(from given) (def of the DOT) (by Calc) (by Calc)
||~v − w|| ~ ≈ 11
(g) Use the dot product to find the ’distance’ between the indicated vectors. w ~ =< 1, −3, 5, 0, 6 >
~v =< 5, 2, −10, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you.. 5. Vector Dot Product to find ’angle’ between two vectors (a) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, 2 >
~v =< 3, −5 >
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Trigonometry Sec. 8
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Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 1, 2 > · < 3, −5 > = (given) || < 1, 2 > |||| < 3, −5 > || −7 −7 ≈ ≈ (by Calc) (2.236)(5.831) 13.038 cos θ = −0.54 (by Calc)
cos θ =
....thus....
θ ≈ cos−1 (−0.54)
θ ≈ 2.14 radians ≈ 122.6◦
~v
....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
~ w
(b) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< −3, 2 >
~v =< 3, 5 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle
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Trigonometry Sec. 8
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’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < −3, 2 > · < 3, 5 > = (given) || < −3, 2 > |||| < 3, 5 > || 1 1 ≈ ≈ (by Calc) (3.606)(5.831) 21.027 cos θ = 0.05 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0.05)
θ ≈ 1.52 radians
w ~
≈ 87.1◦
~v
(c) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, 4 >
~v =< −3, −5 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 1, 4 > · < −3, −5 > = (given) || < 1, 4 > |||| < −3, −5 > || −23 −23 ≈ ≈ (by Calc) (4.123)(5.831) 24.041 cos θ = −0.96 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (−0.96)
θ ≈ 2.86 radians ≈ 163.9◦
©2007-2008 MathHands.com
MathHands.com M´ arquez
~v
Trigonometry Sec. 8
w ~
(d) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 2, 3 >
~v =< 3, −3 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 2, 3 > · < 3, −3 > = (given) || < 2, 3 > |||| < 3, −3 > || −3 −3 ≈ ≈ (by Calc) (3.606)(4.243) 15.3 cos θ = −0.2 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (−0.2) θ ≈ 1.77 radians ≈ 101.4◦
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~v
Trigonometry Sec. 8
w~
(e) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 3, 7 >
~v =< 1, −4 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 3, 7 > · < 1, −4 > = (given) || < 3, 7 > |||| < 1, −4 > || −25 −25 ≈ ≈ (by Calc) (7.616)(4.123) 31.401 cos θ = −0.8 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (−0.8) θ ≈ 2.5 radians ≈ 143.2◦
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~v
Trigonometry Sec. 8
~ w
(f) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 3, 9 >
~v =< 7, 1 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 3, 9 > · < 7, 1 > = (given) || < 3, 9 > |||| < 7, 1 > || 30 30 ≈ ≈ (by Calc) (9.487)(7.071) 67.083 cos θ = 0.45 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0.45) θ ≈ 1.1 radians ≈ 63◦
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~v
Trigonometry Sec. 8
~ w
(g) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< −5, 2 >
~v =< 2, 5 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < −5, 2 > · < 2, 5 > = (given) || < −5, 2 > |||| < 2, 5 > || 0 0 ≈ ≈ (by Calc) (5.385)(5.385) 28.998 cos θ = 0 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0) θ ≈ 1.57 radians ≈ 90◦
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w ~
Trigonometry Sec. 8
~v
(h) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, 7 >
~v =< 7, −1 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 1, 7 > · < 7, −1 > = (given) || < 1, 7 > |||| < 7, −1 > || 0 0 ≈ ≈ (by Calc) (7.071)(7.071) 49.999 cos θ = 0 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0) θ ≈ 1.57 radians ≈ 90◦
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~v
Trigonometry Sec. 8
w ~
(i) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 2, 3 >
~v =< 3, −2 >
Solution: Let us assume the angle between the vectors is between 0 and 180◦ Suppose we call such angle ’θ’, then.... ~v · w ~ (famous DOT product property) ||~v ||||w|| ~ < 2, 3 > · < 3, −2 > = (given) || < 2, 3 > |||| < 3, −2 > || 0 0 ≈ ≈ (by Calc) (3.606)(3.606) 13.003 cos θ = 0 (by Calc)
cos θ =
....thus.... ....then.. (assume the sought angle is 0 ≤ θ ≤ 180◦ )....
θ ≈ cos−1 (0) θ ≈ 1.57 radians ≈ 90◦
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~v
Trigonometry Sec. 8
w ~
(j) Use the dot product to find the ’angle’ between the indicated vectors. w ~ =< 1, −3, 5, 0, 6 >
~v =< 5, 2, −10, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you.. 6. Perpendicular Test by the DOT Find the Dot product for each pair of vectors, then determine if they are perpendicular. (a) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 1, 2 >
~v =< 3, −5 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, −5 > · < 1, 2 > = −7
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (b) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 3, 2 >
~v =< 2, −3 >
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Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 2, −3 > · < 3, 2 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (c) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 1, 4 >
~v =< −3, −5 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< −3, −5 > · < 1, 4 > = −23
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (d) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 2, 3 >
~v =< 3, −3 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, −3 > · < 2, 3 > = −3
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (e) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 3, 6 >
~v =< 2, −1 >
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Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 2, −1 > · < 3, 6 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (f) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 3, 12 >
~v =< −4, 1 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< −4, 1 > · < 3, 12 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (g) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< −5, 2 >
~v =< 3, 5 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, 5 > · < −5, 2 > = −5
(given) (by inspection)
....therefore....~v and w ~ are not perpendicular. (h) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 1, 7 >
~v =< 7, −2 >
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Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 7, −2 > · < 1, 7 >
(given)
= −7
(by inspection)
....therefore....~v and w ~ are not perpendicular. (i) Use the dot product to find test if the indicated vectors are ’perpendicular’. w ~ =< 2, 3 >
~v =< 3, −2 >
Solution: The perpendicular test is quite simple and elegant. Vector ~v is perpendicular to w ~ if and only if ~v · w ~ = 0, or more concisely, ~v ⊥w ~ ⇐⇒ ~v · w ~ =0 Thus we check..
~v · w ~ =< 3, −2 > · < 2, 3 > =0
(given) (by inspection)
....therefore....~v and w ~ are perpendicular. (j) test to see if perpendicular... w ~ =< 1, −3, 5, −2, 6 >
~v =< 5, 0, −1, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you.. 7. Projections by the DOT (a) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 1, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 1, 2 > · < 5, 0 > = < 5, 0 > || < 5, 0 > ||2 5 ≈ < 5, 0 > 25 ≈ 0.2< 5, 0 >
projw~ ~v =
≈ < 1, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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~v
Trigonometry Sec. 8
w ~
(b) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 5, 0 > < 5, 0 > = || < 5, 0 > ||2 15 ≈ < 5, 0 > 25 ≈ 0.6< 5, 0 >
projw~ ~v =
≈ < 3, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v w ~
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(c) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 4, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 4, 2 > · < 5, 0 > = < 5, 0 > || < 5, 0 > ||2 20 ≈ < 5, 0 > 25 ≈ 0.8< 5, 0 > ≈ < 4, 0 >
projw~ ~v =
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v w ~
(d) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 7, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 7, 2 > · < 5, 0 > = < 5, 0 > || < 5, 0 > ||2 35 ≈ < 5, 0 > 25 ≈ 1.4< 5, 0 >
projw~ ~v =
≈ < 7, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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~v w ~
(e) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< −3, 2 >
w ~ =< 5, 0 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < −3, 2 > · < 5, 0 > < 5, 0 > = || < 5, 0 > ||2 −15 ≈ < 5, 0 > 25 ≈ −0.6< 5, 0 >
projw~ ~v =
≈ < −3, 0 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v w ~
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(f) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 1, 2 >
w ~ =< 0, 6 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 1, 2 > · < 0, 6 > = < 0, 6 > || < 0, 6 > ||2 12 ≈ < 0, 6 > 36 ≈ 0.333< 0, 6 > ≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v
w ~
projw~ ~v =
(g) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 0, 7 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 0, 7 > = < 0, 7 > || < 0, 7 > ||2 14 ≈ < 0, 7 > 49 ≈ 0.286< 0, 7 >
projw~ ~v =
≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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w ~
Trigonometry Sec. 8
~v
(h) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 4, 2 >
w ~ =< 0, 2 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 4, 2 > · < 0, 2 > < 0, 2 > = || < 0, 2 > ||2 4 ≈ < 0, 2 > 4 ≈ 1< 0, 2 >
projw~ ~v =
w ~
≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~v
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(i) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 7, 2 >
w ~ =< 0, 4 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 7, 2 > · < 0, 4 > = < 0, 4 > || < 0, 4 > ||2 8 ≈ < 0, 4 > 16 ≈ 0.5< 0, 4 > ≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
w ~
projw~ ~v =
~v
(j) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< −3, 2 >
w ~ =< 0, 3 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < −3, 2 > · < 0, 3 > = < 0, 3 > || < 0, 3 > ||2 6 ≈ < 0, 3 > 9 ≈ 0.667< 0, 3 >
projw~ ~v =
≈ < 0, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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w ~
Trigonometry Sec. 8
~v
(k) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 1, 2 >
w ~ =< 1, 6 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 1, 2 > · < 1, 6 > < 1, 6 > = || < 1, 6 > ||2 13 ≈ < 1, 6 > 37 ≈ 0.351< 1, 6 >
projw~ ~v =
(given) (by Calc) (by Calc) (by Calc)
~v
w ~
≈ < 0.35, 2.11 >
(famous DOT product property)
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(l) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 7, 1 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 7, 1 > = < 7, 1 > || < 7, 1 > ||2 23 ≈ < 7, 1 > 50 ≈ 0.46< 7, 1 > ≈ < 3.22, 0.46 >
projw~ ~v =
~v
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
~ w
(m) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 4, 2 >
w ~ =< 4, 2 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 4, 2 > · < 4, 2 > = < 4, 2 > || < 4, 2 > ||2 20 ≈ < 4, 2 > 20 ≈ 1< 4, 2 >
projw~ ~v =
≈ < 4, 2 >
(famous DOT product property) (given) (by Calc) (by Calc) (by Calc)
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~v w
(n) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 7, 2 >
w ~ =< 8, 1 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 7, 2 > · < 8, 1 > < 8, 1 > = || < 8, 1 > ||2 58 ≈ < 8, 1 > 65 ≈ 0.892< 8, 1 >
(famous DOT product property)
projw~ ~v =
(given) (by Calc) (by Calc)
≈ < 7.14, 0.89 >
(by Calc)
~v
~ w
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(o) Use the dot product to find the ’projection’ of ~v onto w ~ then draw such projection. ~v =< 3, 2 >
w ~ =< 9, 3 >
Solution: ~v · w ~ w ~ ||w|| ~ 2 < 3, 2 > · < 9, 3 > = < 9, 3 > || < 9, 3 > ||2 33 ≈ < 9, 3 > 90 ≈ 0.367< 9, 3 > ≈ < 3.3, 1.1 >
(famous DOT product property)
projw~ ~v =
(given) (by Calc) (by Calc) (by Calc)
~ w ~v
(p) find projection ~v =< 1, −3, 5, −2, 6 >
ontow ~ =< 5, 0, −1, 3, 1 >
Solution: don’t be afraid, don’t google it, don’t ask anyone.. just you and the problem.. its on..., if it knocks you down, just get up... don’t let it beat you..
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Trigonometry Sec. 7
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1. Vector Arithmetic (a) Vector ADDITION Suppose ~v = and w ~ = < − 3, 5>, compute ~v + w ~ Solution: ~v + w ~ = + < − 3, 5>
(definition of addition on vectors) (by inspection)
~v
~ w ~v +
~ w
= = < − 2, 7>
(given)
(b) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = +
= =
(given) (definition of addition on vectors) (by inspection)
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~v
Trigonometry Sec. 7
~ w
~v + ~ w
(c) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = + =
(given) (definition of addition on vectors)
=
~ w
~v ~v +
(by inspection)
w ~
(d) Vector ADDITION Suppose ~v = and w ~ = < − 3, 1>, compute ~v + w ~
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Solution: ~v + w ~ = + < − 3, 1> = =
(given) (definition of addition on vectors) (by inspection)
~v +
w ~
w ~
~v
(e) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = +
= =
(given) (definition of addition on vectors) (by inspection)
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~ w
~v ~v + w ~
(f) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~ Solution: ~v + w ~ = + =
(given) (definition of addition on vectors) (by inspection)
w ~
=
~v +
~ w
~v
(g) Vector ADDITION Suppose ~v = and w ~ = , compute ~v + w ~
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Solution: ~v + w ~ = + =
(given) (definition of addition on vectors)
=
~ w
~v ~v + w ~
(by inspection)
(h) Vector ADDITION Suppose ~v = < − 3, 2> and w ~ = < − 1, 5>, compute ~v + w ~ Solution: ~v + w ~ = < − 3, 2> + < − 1, 5> = < − 3 + −1, = < − 4, 7>
2 + 5>
(given) (definition of addition on vectors) (by inspection)
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~v +
~ w
~ w
Trigonometry Sec. 7
~v
2. Vector Arithmetic (a) Vector SUBTRACTION Suppose ~v = and w ~ = < − 3, 5>, compute ~v − w ~ Solution: ~v − w ~ = − < − 3, 5> 2 − 5>
(definition of subtraction on vectors) (by inspection)
~v
=
(given) (definition of subtraction on vectors) (by inspection)
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~ ~v − w ~ −w
Trigonometry Sec. 7
~v
(d) Vector SUBTRACTION Suppose ~v = and w ~ = < − 3, 1>, compute ~v − w ~ Solution: ~v − w ~ = − < − 3, 1> =
(given) (definition of subtraction on vectors)
=
(by inspection)
−w ~
~v
~ ~v − w
(e) Vector SUBTRACTION Suppose ~v = and w ~ = , compute ~v − w ~
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Solution: ~v − w ~ = − =
(given) (definition of subtraction on vectors)
w~
(by inspection)
~ −w
=
~v −
~v
(f) Vector SUBTRACTION Suppose ~v = and w ~ = , compute ~v − w ~ Solution: ~v − w ~ = −
= =
(given) (definition of subtraction on vectors) (by inspection)
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−w ~
~v
~v − w ~
(g) Vector SUBTRACTION Suppose ~v = and w ~ = , compute ~v − w ~ Solution: ~v − w ~ = − =
(given) (definition of subtraction on vectors)
w~
(by inspection)
~ −w
=
~v −
~v
(h) Vector SUBTRACTION Suppose ~v = < − 3, 2> and w ~ = < − 1, 5>, compute ~v − w ~
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Solution: ~v − w ~ = < − 3, 2> − < − 1, 5> = < − 3 − −1, 2 − 5>
(given) (definition of subtraction on vectors)
= < − 2, −3>
(by inspection)
~v ~v −
~ −w
w ~
3. Vector Arithmetic (a) Vector SCALARS Suppose ~v = compute 2~v Solution: 2~v = 2 = =
(given) (def of scalar multiplication) (by inspection)
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2~v ~v
(b) Vector SCALARS Suppose ~v = compute 3~v Solution: 3~v = 3 =
(given) (def of scalar multiplication)
=
(by inspection)
3~v ~v
(c) Vector SCALARS Suppose ~v = compute −2~v
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Solution: −2~v = −2 = < − 2(2), −2(0)>
(given) (def of scalar multiplication)
= < − 4, 0>
(by inspection)
−2~v ~v
(d) Vector SCALARS Suppose ~v = compute 2~v Solution: 2~v = 2 = =
(given) (def of scalar multiplication) (by inspection)
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2~v
~v
(e) Vector SCALARS Suppose ~v = compute 3~v Solution: 3~v = 3
(given)
= =
~v
(def of scalar multiplication) (by inspection)
3~v
(f) Vector SCALARS Suppose ~v = compute 2~v
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Solution: 2~v = 2 =
(given) (def of scalar multiplication) (by inspection)
~v
2~v
=
(g) Vector SCALARS Suppose ~v = compute 3~v Solution: 3~v = 3 = =
(given) (def of scalar multiplication) (by inspection)
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~v
3~v
Trigonometry Sec. 7
(h) Vector SCALARS Suppose ~v = compute 1.5~v Solution: 1.5~v = 1.5
(given) (def of scalar multiplication) (by inspection)
~v
1.5~v
= =
(i) Vector SCALARS Suppose ~v = compute .75~v
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Solution: .75~v = .75 =
(given) (def of scalar multiplication) (by inspection)
~v
.75~v
=
(j) Vector SCALARS Suppose ~v = compute −1~v Solution: −1~v = −1
= < − 1(4), −1(2)> = < − 4, −2>
(given) (def of scalar multiplication) (by inspection)
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~v −1
~v
4. Vector Arithmetic: Famous Vectors There are two very famous vectors, there are: i =< 1, 0 > and j =< 0, 1 >. (a) FAMOUS VECTORS i and j Compute and draw the following vectors 4i + 2j Solution: let ~v = 4i + 2j = 4 < 1, 0 > +2 < 0, 1 > =< 4, 0 > + < 0, 2 >
(given) (def of scalar multiplication)
4i +
2j
~ 2j
=< 4, 2 >
~ 4i
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(b) FAMOUS VECTORS i and j Compute and draw the following vectors 1i + −2j Solution: let ~v = 1i + −2j
= 1 < 1, 0 > + − 2 < 0, 1 >
=< 1, 0 > + < 0, −2 > =< 1, −2 >
(given) (def of scalar multiplication)
~ 1i ~ −2j j −2
1i +
(c) FAMOUS VECTORS i and j Compute and draw the following vectors 5i + −2j Solution: let ~v = 5i + −2j = 5 < 1, 0 > + − 2 < 0, 1 > =< 5, 0 > + < 0, −2 > =< 5, −2 >
(given) (def of scalar multiplication)
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~ 5i
−2j
~ −2j
5i +
(d) FAMOUS VECTORS i and j Compute and draw the following vectors 5i + 1j Solution: let ~v = 5i + 1j = 5 < 1, 0 > +1 < 0, 1 > =< 5, 0 > + < 0, 1 >
(given) (def of scalar multiplication)
j 5i + 1 ~ 5i
~ 1j
=< 5, 1 >
(e) FAMOUS VECTORS i and j Compute and draw the following vectors 6i + 3j
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Solution: let ~v = 6i + 3j = 6 < 1, 0 > +3 < 0, 1 >
(given) (def of scalar multiplication)
6i +
3j
~ 3j
=< 6, 0 > + < 0, 3 > =< 6, 3 >
~ 6i
(f) FAMOUS VECTORS i and j Compute and draw the following vectors 3i + 4j Solution: let ~v = 3i + 4j = 3 < 1, 0 > +4 < 0, 1 > =< 3, 0 > + < 0, 4 >
(given) (def of scalar multiplication)
=< 3, 4 >
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3i +
~ 4j
4j
Trigonometry Sec. 7
~ 3i
5. NORM of a VECTOR Find the norm of the indicated vector. (a) Compute: || < 4, 2 > || Solution: p (4)2 + (2)2 √ = 20 ≈ 4.47
.47 ≈4
(def of norm) (by Inspection) (by Calc)
2
|| < 4, 2 > || =
4
(b) Compute:
©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez || < 1, −2 > ||
Solution: p (1)2 + (−2)2 √ = 5 ≈ 2.24
|| < 1, −2 > || =
(def of norm) (by Inspection) (by Calc)
1 −2 .24 ≈2
(c) Compute: || < 5, −2 > || Solution: p (5)2 + (−2)2 √ = 29 ≈ 5.39
|| < 5, −2 > || =
(def of norm) (by Inspection) (by Calc)
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Trigonometry Sec. 7
MathHands.com M´ arquez
5 −2
≈5 .39
(d) Compute: || < 5, 1 > || Solution: p (5)2 + (1)2 √ = 26 ≈ 5.1
≈ 5.1 5
(def of norm) (by Inspection) (by Calc)
1
|| < 5, 1 > || =
(e) Compute: || < 6, 3 > || ©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez
Solution: p (6)2 + (3)2 √ = 45
|| < 6, 3 > || =
(def of norm) (by Inspection)
.71 ≈6
(by Calc)
3
≈ 6.71
6
(f) Compute: || < 5, 0 > || Solution: p (5)2 + (0)2 √ = 25 ≈5
|| < 5, 0 > || =
(def of norm) (by Inspection) (by Calc)
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Trigonometry Sec. 7
MathHands.com M´ arquez
≈5 5
0
6. Which is Bigger? (a) Determine which number is larger: ||4i + 2j|| OR (||4i|| + ||2j||) Solution: ||4i + 2j|| = || < 4, 2 > || p = (4)2 + (2)2 √ = 20 ≈ 4.47
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||4i|| + ||2j|| = || < 4, 0 > || + || < 0, 2 > || =4+2 =6
(def of i and j) (def of norm) (by Inspection)
Therefore, ||4i|| + ||2j|| is larger. (b) Determine which number is larger: ||1i + −2j|| OR (||1i|| + || − 2j||) Solution: ||1i + −2j|| = || < 1, −2 > || p = (1)2 + (−2)2 √ = 5 ≈ 2.24
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
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Trigonometry Sec. 7
MathHands.com M´ arquez
||1i|| + || − 2j|| = || < 1, 0 > || + || < 0, −2 > || =1+2 =3
(def of i and j) (def of norm) (by Inspection)
Therefore, ||1i|| + || − 2j|| is larger. (c) Determine which number is larger: ||5i + −2j|| OR (||5i|| + || − 2j||) Solution: ||5i + −2j|| = || < 5, −2 > || p = (5)2 + (−2)2 √ = 29 ≈ 5.39
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||5i|| + || − 2j|| = || < 5, 0 > || + || < 0, −2 > || =5+2 =7
(def of i and j) (def of norm) (by Inspection)
Therefore, ||5i|| + || − 2j|| is larger. (d) Determine which number is larger: ||5i + 1j|| OR (||5i|| + ||1j||) Solution: ||5i + 1j|| = || < 5, 1 > || p = (5)2 + (1)2 √ = 26 ≈ 5.1
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||5i|| + ||1j|| = || < 5, 0 > || + || < 0, 1 > || =5+1 =6
(def of i and j) (def of norm) (by Inspection)
Therefore, ||5i|| + ||1j|| is larger. (e) Determine which number is larger: ||6i + 3j|| OR (||6i|| + ||3j||)
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Trigonometry Sec. 7
MathHands.com M´ arquez
Solution: ||6i + 3j|| = || < 6, 3 > || p = (6)2 + (3)2 √ = 45 ≈ 6.71
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||6i|| + ||3j|| = || < 6, 0 > || + || < 0, 3 > || =6+3 =9
(def of i and j) (def of norm) (by Inspection)
Therefore, ||6i|| + ||3j|| is larger. (f) Determine which number is larger: ||5i + 0j|| OR (||5i|| + ||0j||) Solution: ||5i + 0j|| = || < 5, 0 > || p = (5)2 + (0)2 √ = 25 ≈5
(def of i and j) (def of norm) (by Inspection) (by Calc)
Meanwhile
||5i|| + ||0j|| = || < 5, 0 > || + || < 0, 0 > || =5+0 =5
(def of i and j) (def of norm) (by Inspection)
Therefore, ||5i|| + ||0j|| is larger. (g) ||u + v|| OR (||u|| + ||v||) Solution: the idea is to study the above pattern and understand that it will always be the case that the sum of the individual norms is larger or equal to the norm of the sum of the vectors. In some sense, this is equivalent to saying that the sum of the lengths of any two sides of a [Euclidean] triangle have to be larger than the size of the length of the the third side of the triangle. This is very famous, it is called the triangle inequality. 7. Normalize this.. Find the normalized vector for each: (a) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< 5, 2 >
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Trigonometry Sec. 7
MathHands.com M´ arquez
Solution: First we find the norm of the vector: p (5)2 + (2)2 √ = 29
||~v || = || < 5, 2 > || =
(def of norm) (by Inspection)
≈ 5.39 Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(by Calc)
Let us denote the normalized vector ~v as ”v~n ”
1 ||~v || 5 2 = , ||~v || ||~v || 5 2 ≈ , 5.39 5.39
v~n =
(given) (def of scalar multiplication) (approximate)
≈ h0.93, 0.37i
(by inspection)
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
~v v~n
(b) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< 5, −2 > Solution: First we find the norm of the vector: p (5)2 + (−2)2 √ = 29 ≈ 5.39
||~v || = || < 5, −2 > || =
(def of norm) (by Inspection) (by Calc)
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Trigonometry Sec. 7
MathHands.com M´ arquez
Then we scale the original vector ~v by multiplying by
1 ||~ v || .
Let us denote the normalized vector ~v as ”v~n ”
1 ||~v || 5 −2 = , ||~v || ||~v || −2 5 , ≈ 5.39 5.39
v~n =
(given) (def of scalar multiplication) (approximate) (by inspection)
≈ h0.93, −0.37i
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
v~n
~v
(c) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< −3, 4 > Solution: First we find the norm of the vector: p (−3)2 + (4)2 √ = 25 ≈5
||~v || = || < −3, 4 > || =
Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(def of norm) (by Inspection) (by Calc)
Let us denote the normalized vector ~v as ”v~n ”
©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez 1 < − 3, 4> ||~v || −3 4 = , ||~v || ||~v || −3 4 , ≈ 5 5
(given)
v~n =
(def of scalar multiplication) (approximate)
≈ h−0.6, 0.8i
(by inspection)
v~n
~v
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
(d) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< −3, −4 > Solution: First we find the norm of the vector: p (−3)2 + (−4)2 √ = 25
||~v || = || < −3, −4 > || =
≈5 Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(def of norm) (by Inspection) (by Calc)
Let us denote the normalized vector ~v as ”v~n ”
©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez 1 < − 3, −4> ||~v || −3 −4 = , ||~v || ||~v || −3 −4 , ≈ 5 5
(given)
v~n =
(def of scalar multiplication) (approximate) (by inspection)
≈ h−0.6, −0.8i
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
v~n
~v
(e) NORMALIZE them Compute and draw the corresponding normalized vector: ~v =< −3, −1 > Solution: First we find the norm of the vector: p (−3)2 + (−1)2 √ = 10
||~v || = || < −3, −1 > || =
≈ 3.16 Then we scale the original vector ~v by multiplying by
1 ||~ v || .
(def of norm) (by Inspection) (by Calc)
Let us denote the normalized vector ~v as ”v~n ”
©2007-2008 MathHands.com
Trigonometry Sec. 7
MathHands.com M´ arquez 1 < − 3, −1> ||~v || −3 −1 = , ||~v || ||~v || −3 −1 , ≈ 3.16 3.16
v~n =
≈ h−0.95, −0.32i
(given) (def of scalar multiplication) (approximate) (by inspection)
Now, NOTE: ~v and v~n have the same direction, BUT, v~n has norm ”1” as intended. (also keep in mind position does not matter for vectors.. only direction and size, thus nothing should be interpreted from their position, only their size and direction)
v~n ~v
(f) ~v =< a, b > (not both zero..) Solution: 1 ||~v || a b = , ||~v || ||~v || b a ,√ = √ a2 + b 2 a2 + b 2
v~n =
(given) (def of scalar multiplication) (def of norm)
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Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
Euler’s Identity
◦
1. Convert Convert 3ei30 to standard form, a + bi Solution: According to Euler’s amazing identity..eiθ = cos θ + i sin θ..thus..
◦
3ei30
= 3[cos 30◦ + i sin 30◦ ] "√ # 3 1 = 3 +i 2 2 √ 3 3 3 +i = 2 2
Alternatively, you could convert using the picture... and convert in very much the same way we convert from polar to cartesian...
3
b
3
“√ ” 3 2
+3
`1´ 2
i
30◦
◦
2. Convert Convert 3e−i30 to standard form, a + bi Solution: According to Euler’s amazing identity..eiθ = cos θ + i sin θ..thus..
◦
3e−i30
= 3[cos −30◦ + i sin −30◦ ] # "√ −1 3 +i = 3 2 2 √ 3 3 −3 = + i 2 2
Or use the picture... Or use polar conversion ideas.. (r, θ) = (3, −30◦) 3. Convert Convert ei 2 to standard form, a + bi π
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math hands
pg. 3
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
Solution: According to Euler’s amazing identity..eiθ = cos θ + i sin θ..thus..
ei 2
π
= = =
4. Convert Convert e
3π 2 i
to standard form, a + bi
5. Convert Convert e
7π 6 i
to standard form, a + bi
π π + i sin 2 2 0 + i(1) i cos
6. Convert Convert 3e2i to standard form, a + bi 7. Convert Convert 5e3.14i to standard form, a + bi 8. Convert Convert e6.28i to standard form, a + bi 9. Convert Convert 2 + i to Euler form, reiθ
Solution: 2 + i =
√ i26.57◦ 5e many other correct possibilities for theta..
10. Convert Convert 2 − 2i to Euler form, reiθ √ √ √ ◦ ◦ ◦ Solution: 2 − 2i = (2 2)e−i45 OR (2 2)ei315 or (2 2)ei765 or infinite many more possibilities for θ 11. Convert Convert −2 − 2i to Euler form, reiθ √ ◦ Solution: −2 − 2i = (2 2)ei225 many other correct possibilities for theta.. 12. Convert Convert −3 + 2i to Euler form, reiθ 13. Convert Convert i to Euler form, reiθ ◦
Solution: i = ei90 many other correct possibilities for theta.. note r = 1 14. Convert Convert 3i to Euler form, reiθ ◦
Solution: i = 3ei90 many other correct possibilities for theta.. note r = 1 15. Convert Convert .5 + 3i to Euler form, reiθ 16. Convert Convert −1 to Euler form, reiθ ◦
Solution: i = ei180 many other correct possibilities for theta.. note r = 1
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math hands
pg. 4
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
17. Convert Convert −i to Euler form, reiθ ◦
Solution: i = ei270 many other correct possibilities for theta.. note r = 1
18. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Multiply the old fashion way... ! √ 3 1 + i · 2 2
◦
◦
e30 i · e60
√ ! 1 3 + i 2 2
i
◦
(b) Multiply the Euler way, and simplify answer.. (note
Solution: ei90 = i
19. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Multiply the old fashion way... ! √ 3 3 3 + i · 2 2
◦
◦
3e30 i · 2e60
√ ! 2 2 3 + i 2 2
i
◦
(b) Multiply the Euler way, and simplify answer.. (note
Solution: 6ei90 = 6i
20. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Multiply the old fashion way... ! √ 3 1 + i · 2 2
◦
◦
e30 i · e30
! √ 3 1 + i 2 2
i
◦
(b) Multiply the Euler way, and simplify answer.. (note
Solution: ei60
21. Euler’s World Multiply or divide as indicated these are the same numbers..)
(a) Divide the old fashion way...
◦
e30 i e60◦ i
√ 3 1 2 +√2 i 1 3 2 + 2 i ◦
(b) Divide the Euler way, and simplify answer.. (note
Solution: e−i30
22. Euler’s World Multiply or divide as indicated (b) Calculate, the Euler way, and simplify answer.. (note these are the same numbers..) ◦ 3 e30 i
(a) Expand the binomial, the old fashion way... !3 √ 3 1 + i 2 2
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math hands
pg. 5
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez ◦
Solution: ei90 = i compare this problem with #57 from two sections ago.
23. Euler’s World Multiply or divide as indicated (a) Expand the binomial, the old fashion way (or not..) ... !7 √ 3 1 + i 2 2
these are the same numbers..)
◦
e30
i
7
◦
Solution: ei210
(b) Calculate, the Euler way, and simplify answer.. (note
24. Euler’s World Convert the numbers to Euler form, then compute. (a)
(b) 3 + 4i −2 − 5i
(3 + 4i)(−2 − 5i) 25. Euler’s World to find Roots (a) Try to find a square root of i, i.e. a number x such that x2 = i √ i
(note these are the same numbers..)
◦
e90
i
21
◦
(b) Find a root, the Euler way, and simplify answer..
Solution: ei45
26. Euler’s World to find Roots (a) Try to find a square root: (after a good honest attempt, go to part b.) s
(note these are the same numbers..)
√ 1 3 + i 2 2
◦
e60
i
12
◦
(b) Find a root, the Euler way, and simplify answer..
Solution: ei30
27. Euler’s World to find Roots (a) Try to find a fourth root: (after a good honest attempt, go to part b.) s 4
swer.. (note these are the same numbers..)
√ 16 16 3 + i 2 2
◦
16e60
i
41
◦
(b) Find a fourth root, the Euler way, and simplify an-
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math hands
Solution: 2ei15
pg. 6
Trigonometry Sec. 06 exercise
MathHands.com M´ arquez
28. FTA and Unity Roots The two solutions to x2 = 1 are x = 1, −1. Show how one can derive these solutions. 29. FTA and Unity Roots The three solutions to x3 = 1 are x = 1, these solutions.
−1 2
+
√ 3 −1 2 i, 2
−
√ 3 2 i.
Show how one can derive
30. FTA and Unity Roots The four solutions to x4 = 1 are x = 1, i, −1, −i. Show how one can derive these solutions. ◦
31. FTA and Unity Roots To find the five solutions to x5 = 1, start with writing 1 = ek360 i , then start finding 5th roots, every value of k should produce one... ◦
Solution: for integers k, eik72
32. Euler’s World Another Proof of MOTA (a) Multiply the old fashion way
(b) multiply, the Euler way, and simplify answer.. (note these are the same numbers..)
(cos x + i sin x)(cos y + i sin y)
eix · eiy (c) Compare the real part of the answer for a, with the real part of the answer for part b.
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math hands
pg. 7
Trigonometry Sec. 05 notes
MathHands.com M´ arquez
Converting Complex numbers From and To EULER FORM Main Idea Perhaps the most celebrated identity of mathematics is Euler’s World known timeless identity: eiθ = cos θ + i sin θ Among other things, it’s power lies on the idea that a binomial (on the right side) can be converted to a monomial (left side). Indeed, every complex number, a + bi can be written in Euler Form as reiθ . It will serve us well to think about Euler’s identity as a sort of bridge between the binomials numbers which make up the entire complex plane and the monomial world expressed in Euler Form. Moreover, we have already crossed this bridge. You will see the ideas are identical to the ideas used in converting from cartesian coordinated to polar coordinates in first section of this chapter. Converting: EULER TO & FORM Standard Form The translating ’dictionary’ to go from Euler-to-Standard is as follows: From the defining features of the polar and cartesian cordinates, we conclude that generally:
reiθ = r[eiθ ]
(getting ready to use Euler’s ID, eiθ = cos (θ) + i sin (θ))
= r[cos (θ) + i sin (θ)]
(used eiθ = cos (θ) + i sin (θ))
= r cos (θ) + ir sin (θ) = x + iy
(algebra) (used polar dictionary)
and similarly going backwards. Therefore, we have the translating ’dictionary’ to translate from EULER TO& FROM STANDARD Form
MONOMIALS
BINOMIALS
90◦
6 60◦ 120◦
5 4 30◦
reiθ = x + yi 3
150◦
reb iθ r
x+ yi b
2
r
y
y
1
θ
x
θ
0◦
-6
180◦
-5
-4
-3
-2
-1
1
x 2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
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pg. 1
Trigonometry Sec. 05 notes
MathHands.com M´ arquez
EXAMPLEs: Polar Coordinates < −− > Cartesian Coordinates ◦
Convert the Euler Form, 5ei250 to Standard form of the complex number.
6 90◦
5 60◦
5ei250 = 5 cos (i250◦) + i5 sin (i250◦) (Euler, Baby!!) ◦
120◦
4
≈ −1.71 + −4.7i
30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦ 210◦
-3 b
5ei250
-4
◦
300◦
b
240◦
−1.71 + −4.7i-5
270◦
-6
Convert the cartesian coordinates, (-4, 3), to polar coordinates.
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pg. 2
Trigonometry Sec. 05 notes
MathHands.com M´ arquez r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4
(r, θ) 150◦
(−4, 3)
30◦
b
b
3 2 1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 3/ − 4 (note: eq has infinite solutions) ◦ θ ≈ −36.87 (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(−4)2 + (3)2 r2 =25 r =±
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2007-2009 MathHands.com
√ 25 ≈ ±5
note: if we use the angle −36.87◦, then we must choose the negative value of r, r ≈ −5, thus the polar cordinates (−5, −36.87◦) is one way to represent the point (−4, 3)
math hands
pg. 3
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
Converting Complex numbers From and To EULER FORM ◦
1. Convert the Euler Form, 5ei240 to Standard form of the complex number. Solution: 5ei(240
◦
)
= 5 cos (240◦) + i5 sin (240◦ ) √ −5 −5 3 + i = 2 2 ≈ −2.5 + −4.33i
(Euler, Baby!!)
◦
2. Convert the Euler Form, 7ei210 to Standard form of the complex number. Solution: 7ei(210
◦
)
= 7 cos (210◦) + i7 sin (210◦ ) √ −7 3 −7 = + i 2 2 ≈ −6.06 + −3.5i
(Euler, Baby!!)
◦
3. Convert the Euler Form, 7ei330 to Standard form of the complex number. Solution: 7ei(330
◦
)
= 7 cos (330◦) + i7 sin (330◦ ) √ 7 3 −7 + i = 2 2 ≈ 6.06 + −3.5i
(Euler, Baby!!)
◦
4. Convert the Euler Form, 5ei150 to Standard form of the complex number. Solution: 5ei(150
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2007-2009 MathHands.com
◦
)
= 5 cos (150◦) + i5 sin (150◦ ) √ −5 3 5 + i = 2 2 ≈ −4.33 + 2.5i
math hands
(Euler, Baby!!)
pg. 4
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez ◦
5. Convert the Euler Form, 2ei240 to Standard form of the complex number. Solution: 2ei(240
◦
)
= 2 cos (240◦) + i2 sin (240◦ ) √ −2 −2 3 = + i 2 2 ≈ −1 + −1.73i
(Euler, Baby!!)
◦
6. Convert the Euler Form, 7ei135 to Standard form of the complex number. Solution: 7ei(135
◦
)
= 7 cos (135◦) + i7 sin (135◦ ) ≈ −4.95 + 4.95i
(Euler, Baby!!)
◦
7. Convert the Euler Form, 3ei−30 to Standard form of the complex number. Solution: ◦
3ei(−30
)
= 3 cos (−30◦ ) + i3 sin (−30◦ ) √ 3 3 −3 + i = 2 2 ≈ 2.6 + −1.5i
(Euler, Baby!!)
◦
8. Convert the Euler Form, 2ei−90 to Standard form of the complex number. Solution: ◦
2ei(−90
)
= 2 cos (−90◦ ) + i2 sin (−90◦ ) √ 0 −2.31 3 = + i 2 2 ≈ 0 + −2i
(Euler, Baby!!)
◦
9. Convert the Euler Form, 4ei180 to Standard form of the complex number.
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math hands
pg. 5
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez
Solution: 4ei(180
◦
)
= 4 cos (180◦) + i4 sin (180◦ ) √ −8 0 3 + i = 2 2 ≈ −4 + 0i
(Euler, Baby!!)
◦
10. Convert the Euler Form, 7ei225 to Standard form of the complex number. Solution: 7ei(225
◦
)
= 7 cos (225◦) + i7 sin (225◦ )
(Euler, Baby!!)
≈ −4.95 + −4.95i
◦
11. Convert the Euler Form, 5ei−150 to Standard form of the complex number. Solution: ◦
5ei(−150
)
= 5 cos (−150◦) + i5 sin (−150◦) √ −5 3 −5 = + i 2 2 ≈ −4.33 + −2.5i
(Euler, Baby!!)
◦
12. Convert the Euler Form, 5ei−300 to Standard form of the complex number. Solution: ◦
5ei(−300
)
= 5 cos (−300◦) + i5 sin (−300◦) √ 5 5 3 = + i 2 2 ≈ 2.5 + 4.33i
(Euler, Baby!!)
◦
13. Convert the Euler Form, 2ei400 to Standard form of the complex number. Solution: 2ei(400
◦
)
= 2 cos (400◦) + i2 sin (400◦ )
(Euler, Baby!!)
≈ 1.53 + 1.29i
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pg. 6
Trigonometry Sec. 05 exercises
MathHands.com M´ arquez ◦
14. Convert the Euler Form, 3ei360 to Standard form of the complex number. Solution: 3ei(360
◦
)
= 3 cos (360◦) + i3 sin (360◦ ) √ 6 −0 3 = + i 2 2 ≈ 3 + −0i
(Euler, Baby!!)
◦
15. Convert the Euler Form, 2ei3600 to Standard form of the complex number. Solution: 2ei(3600
◦
)
= 2 cos (3600◦) + i2 sin (3600◦) √ 4 −0 3 i = + 2 2 ≈ 2 + −0i
(Euler, Baby!!)
◦
16. Convert the Euler Form, 4ei−180 to Standard form of the complex number. Solution: ◦
4ei(−180
)
= 4 cos (−180◦) + i4 sin (−180◦) √ −8 −0 3 + i = 2 2 ≈ −4 + −0i
(Euler, Baby!!)
◦
17. Convert the Euler Form, 5ei45 to Standard form of the complex number. Solution: ◦
5ei(45
)
= 5 cos (45◦ ) + i5 sin (45◦ ) ≈ 3.54 + 3.54i
(Euler, Baby!!)
◦
18. Convert the Euler Form, 5ei225 to Standard form of the complex number.
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Solution: 5ei(225
◦
)
= 5 cos (225◦) + i5 sin (225◦ ) ≈ −3.54 + −3.54i
(Euler, Baby!!)
19. Convert the complex number 4 + 5i to EULER form. Solution: r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
reb iθ
120◦
6
4+ 5i b
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
-6 270◦
tan θ = 5/4
(note: eq has infinite solutions)
θ ≈ 51.34◦ (note: this is just one [acceptable] value for θ) r2 =x2 + y 2 note: if we use the angle 51.34◦, then we must choose the positive value of r, r ≈ 6.4, thus the Euler Form
r2 =(4)2 + (5)2 r2 =41 r =±
√
◦
6.4ei(51.34
41 ≈ ±6.4
)
is one way to represent the complex number 4 + 5i
20. Convert the complex number −4 + 1i to EULER form. Solution:
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r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2
reb iθ
−4b + 1i
1
0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
tan θ = 1/ − 4
270◦
(note:-6 eq has infinite solutions)
θ ≈ −14.04◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2
note: if we use the angle −14.04◦, then we must choose the negative value of r, r ≈ −4.12, thus the polar cordinates
r2 =(−4)2 + (1)2 r2 =17 r =±
√ 17 ≈ ±4.12
◦
−4.12ei(−14.04
)
is one way to represent the complex number −4 + 1i
21. Convert the complex number −3 + −4i to EULER form. Solution:
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r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦ 210◦
-3
reb iθ
−3 b + −4i
-4 -5
300◦ 240◦
-6 270◦
tan θ = −4/ − 3
(note: eq has infinite solutions)
θ ≈ 53.13◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 note: if we use the angle 53.13◦, then we must choose the negative value of r, r ≈ −5, thus the polar cordinates ◦ −5ei(53.13 )
r2 =(−3)2 + (−4)2 r2 =25 r =±
√ 25 ≈ ±5
is one way to represent the complex number −3+−4i
22. Convert the complex number −1 + 5i to EULER form. Solution:
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r 2 = x2 + y 2 tan θ = y/x
90◦
120◦
60◦
reb iθ
6
−1b + 5i 5 4
30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4 -5 300◦ 240◦
tan θ = 5/ − 1
270◦
(note:-6 eq has infinite solutions)
θ ≈ −78.69◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2
note: if we use the angle −78.69◦, then we must choose the negative value of r, r ≈ −5.1, thus the polar cordinates
r2 =(−1)2 + (5)2 r2 =26 r =±
√
◦
−5.1ei(−78.69
26 ≈ ±5.1
)
is one way to represent the complex number −1 + 5i
23. Convert the complex number 1 + −5i to EULER form. Solution:
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r 2 = x2 + y 2 tan θ = y/x
90◦ 60◦
6
120◦
5 4 30◦
3
150◦
2 1 0◦
-6
180◦
-5
-4
-3
-2
-1
1
2
3
4
5
6
-1 -2 330◦
-3
210◦
-4
reb iθ
-5 300◦
1 +b −5i
240◦
-6 270◦
tan θ = −5/1
(note: eq has infinite solutions)
θ ≈ −78.69◦ (note: this is just one [acceptable] value for θ)
r2 =x2 + y 2 r2 =(1)2 + (−5)2 r2 =26 r =±
√
26 ≈ ±5.1
note: if we use the angle −78.69◦, then we must choose the positive value of r, r ≈ 5.1, thus the Euler Form ◦ 5.1ei(−78.69 ) is one way to represent the complex number 1 + −5i
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Trigonometry Sec. 5
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More onComplex Numbers The Idea Last section was ’out of this world’. We introduced the imaginary number, their standard form ’a + bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide. The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3
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MathHands.com M´ arquez 5i − 3 5i − 3 = ·1 i+3 i+3 = =
5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi = 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 − 11i −3 11 = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so.. 2i + 3 − 5i − 2 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator. 2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
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Instructions 1. Divide
7i + 1 7i + 3
Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
2. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
3. Divide
4i + 2 5i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
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Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
4. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
5. Divide
3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
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Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
6. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
7. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
8. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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2007-2009 MathHands.com
Trigonometry Sec. 5
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
9. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 10. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 11. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 12. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2 3
= (1) · i2 =i
(just add exponents) (setting up to use fact i4 = 1) (yipei-kae-yeh...)
2
= −1
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13. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
14. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
15. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
16. Compute w/ Complex Numbers Calculate and write in standard form. i−5
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Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
17. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
18. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
19. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
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20. Compute w/ Complex Numbers Calculate and write in standard form. i2
Solution: i2 = i0 · i1 0 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
0
(used i4 = 1) (ez as sundays)
= (1) · i =i
21. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
22. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 = i4 ·i = (1)
−1
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
23. Compute w/ Complex Numbers Calculate and write in standard form. i11
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Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1) (ez as sundays)
= (1) · i =i
24. Compute w/ Complex Numbers Calculate and write in standard form. i−6
Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
25. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
8
(used i4 = 1) (ez as sundays)
= (1) · i =i
26. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
·i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
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27. Compute w/ Complex Numbers Calculate and write in standard form.
1 1 √ +√ i 2 2
2
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
=
(FOIL) (used i2 = −1)
28. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 1 3 + i 2 2
(FOIL) (used i2 = −1)
29. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
30. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
31. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
10
= (i)
= −1
(getting ready to use A3 = i) (used A3 = i) (ez as Sundays...)
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32. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 33. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
34. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 35. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 36. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 37. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
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Intro to Complex Numbers The Question In the last section we wondered what might be far to the right and to the left of the real number line. In doing so, we entertained the existence of the very special unreal number ∞. In this section, we continue on this path and now consider what lies above or below the real number line?
? ∞
−∞ -5
-4
Out of Audacity, a number is born
-3
-2
-1
0
1
2
3
4
5
?
Here is the account, directly from the world famous, the almost immortal, mathematician Euler himself.... ”I sat in that soft comfortable chair leaned back and enjoyed the million thoughts dancing inside my head. A blank paper, a pencil and my awesome coffee was all there was on the desk. On the paper the equation x2 = −1 The equation was screaming, enticing, talking trash, challenging me, saying ”you can’t solve me!” Hours went by faster than I would have liked. Days past by, weeks and months...There was no real number that would solve the equation. But the forces were greater. The inspiration divine. I would not be stopped.. and one day it happened. There was no real number solution, I had looked on the positive side on the negative side and all numbers between. Resolved to avoid defeat at all costs, I invented a number. From my own imagination, I gathered all my might, my courage, and my audacity, √ and I thought...I will create a number. I will call it i, and I will solve my problem by declaring i = −1. It’s my number so I can make it behave however I please, just as the artists paints the clouds at his whim... This solves the equation x2 = −1 and marks the birth of a grand elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.” Needless, to say, I have taken some artistic liberties with this account of events. In fact, traces of complex numbers or ’imaginary’ numbers can be found in 9th century’s Al-Khwarizmi’s Algebra text. During the next couple centuries these ideas made their way to Italy and France, as people were learning to solve degree 3 equations.√ By the turn of the 17th century Descartes coined the phrase ”imaginary” numbers, referring √ to numbers such as −1. At last, it was Euler, in the 18th century who named such number i, declaring i = −1. Thus... by definition of i;
i2 = −1 and, i is a solution to x2 = −1 In addition to solving the equation x2 =− 1, and this marks the birth of a grand and elegant family of numbers called the complex numbers, C. With the complex numbers also came a batch of fresh new ideas. These ideas include the meaning of negative radicals, a new family of numbers to add, multiply and divide, and a whole new world that adds perspective to our previous views.
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pg. 1
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
Where do all the i’s Live With the invent of i came other numbers such as 2i, 3i, −5i, or 2 + 3i. Generally, complex numbers are numbers that can be written in the form a + bi where a and b are real numbers. Now, in the previous sections we noted a visual representation of the real numbers using the real number line. Considering this, the natural question is where or how do we represent the complex numbers visually? Over the centuries, the most powerful and common way to represent the complex numbers is to place them as an extension of the real number line, extending it above and below to make what we commonly call the complex plane. In essence, this is done by placing an ’imaginary’ axis perpendicular to the real number line. Then we position every number a+ bi on the plane similarly to placing the ordered point (a, b) on the cartesian [xy]plane. came numbers such as 2i, 3i, 4i, . . . as well as numbers such as 2 + 3i. Here are some visual representations of a few complex numbers along with their position relative to the known real number line. 3i
The Complex Numbers, C
2i b
−4 + i
b
3 + 2i
3
4
1i 2 + 0i
R
b
-5
-4
-3
-2
-1
0
1
2
5
−1i −2i −3i b
4 − 3i
Negative Radicals With the invention of i we can now make sense of radicals (i.e. square roots) of √ negative real numbers. Consider √ the radical −1, the number whose square is −1. Recall when we first defined 4 we did so as ’the number whose number square is 4.’ But there are two such numbers 2 and −2. By default, we declared to radical to mean the positive number √ whose square is 4. We follow a similar logic here, as we are confronted with the same dilemma. If we define −1 as the number whose square is −1, we will find there are two possible choices, i and −i (see examples). By convention, we will define negative radical to be, the positive i rather than −i. Examples 1. Simplify
√ −4
√
√ −4 = i 4 = i2 = 2i
2. Simplify
√ −10
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√ √ −10 = i 10
math hands
(neg rad) (def rad) (CoLM)
(neg rad)
pg. 2
Trigonometry Sec. 04 notes 3. Simplify
4. Simplify
√ −15
MathHands.com M´ arquez √ √ −15 = i 15
(neg rad)
√ −x
Solution: Notice, we do not know the value of x. We don’t know if x is positive √ or negative. This means we don’t know if −x is positive or negative therefore we don’t know if the radical −x is positive or negative.
5. Adding in C
3 + 5i + 2 + 3i
Solution:
3 + 5i + 2 + 3i = = (3 + 2) + (5i + 3i)
(given) (ALA)
= 3 + 2 + (5 + 3)i
(DL)
= 5 + 8i
(AT)
6. Multiplying in C (3 + 5i)(2 + 3i)
Solution:
(3 + 5i)(2 + 3i) = = 3 · 2 + 5i · 2 + 3 · 3i + 5i · 3i = 6 + 10i + 9i + 15i2
= 6 + (10 + 9)i + 15i
2
(given) (FOIL) (BI) (DL)
2
(AT) (Def of i)
= 6 + 19i + −15 = −9 + 19i
(BI) (BI)
= 6 + 19i + 15i = 6 + 19i + 15 · −1
7. Multiplying in C (4 +− 5i)(2 + 3i)
Solution:
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pg. 3
Trigonometry Sec. 04 notes
MathHands.com M´ arquez (4 +− 5i)(2 + 3i) =
(given)
= 4 · 2 + 5i · 2 + 4 · 3i + 5i · 3i −
−
= 8 + 10i + 12i + 15i −
−
2
= 8 + ( 10 + 12)i + 15i −
−
= 6 + 2i + 15i −
2
2
(FOIL) (BI) (DL) (BI)
= 6 + 2i + 15 · −1 = 6 + 2i + 15
(Def of i) (NNT)
−
= 21 + 2i
(BI)
8. Multiplying in C (4 + 3i)(2 + 3i)
Solution:
(4 + 3i)(2 + 3i) =
(given)
= 8 + 12i + 6i + 9i
2
= 8 + 12i + 6i + 9 · 1 −
(FOIL) (def i)
= 8 + 12i + 6i + 9
(BI)
= 1 + 18i
(BI)
−
−
9. Multiplying in C i7
Solution:
i7 = iiiiiii 2 2 2
(+Expo)
=i i i i
(+Expo)
= −1 · −1 · −1 · i = −1 · i
(Def of i) (BI)
= −i
(MT)
And Now, Divide We have now introduced the imaginary number, their standard form ’a+bi’, we introduced their home, the complex plane, and we introduced some simple arithmetic operations on them such as adding/multiplying. In this section, we continue on the same theme, adding to that some division skills, we add some some famous terminology, such as ’conjugates’, and we look further into the calculation of many exponential powers of i. How to divide in the C-world The layman way to divide.
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math hands
pg. 4
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
The key lies in the observation that multiplying pairs of conjugate complex numbers always yields real numbers. In a way, it is sort of a way to smack a complex number on its head and turn it into a real number, sort of. Every complex number has a conjugate defined as follows, when written in standard form, the conjugate of a + bi is a − bi. In other words, the conjugate of a complex number is the same number with the sign of the complex part switched. Now, observe how the product of conjugates always yields a real number. Take, for example, the complex number 2 + 3i, its conjugate is 2 − 3i: (2 + 3i)(2 − 3i) = 4 + 6i − 6i − 9 · i2 = 4 + 0 − 9 · (−1)
(FOIL) (BI)
= 13
Now, we see how this will help us divide. Suppose we want to divide Divide
(as promised, a real number)
3+5i 2+3i
5i + 3 3i + 2 5i + 3 5i + 3 = ·1 3i + 2 3i + 2 = =
5i + 3 − 3i + 2 · 3i + 2 − 3i + 2 − 15i2 + i + 6 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
i + 21 13
(BI)
=
21 i + 13 13
(BI)
Here is another example, Divide
5i − 3 i+3 5i − 3 5i − 3 = ·1 i+3 i+3 = =
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5i − 3 − i + 3 · i+3 −i+3 − 5i2 + 18i − 9 − i2 + 9
(MiD) (JOT) (MAT, FOIL)
=
18i − 4 10
(BI)
=
− 4 18i + 10 10
(BI)
math hands
pg. 5
Trigonometry Sec. 04 notes
MathHands.com M´ arquez
How to divide in the C-world As usual, to divide means to multiply by the multiplicative inverse. Thus, we need and want to address this question: for any non-zero complex number, a + bi what is its multiplicative inverse? We claim the inverse is aa−bi 2 +b2 . To check this we simply check that their product is 1. Multiplicative Inverses in C (a + bi)
a − bi a2 + b 2
(a + bi) a − bi 1 a2 + b 2 a2 + abi − abi − bi2 = a2 + b 2 2 a + b2 = = 2 =1 a + b2
=
Example Dividing in C : (3 − 2i) ÷ (1 + 3i) = = =
1 − 3i 12 + 32 3 − 9i − 2i + 6i2 10 −3 11 −3 − 11i = − i 10 10 10
(3 − 2i) ·
Another way to ’divide’ and in essence carry out the same computation is to multiply numerator and denominator by the conjugate of the denominator. For example, if the denominator is a + bi, then multiplying both numerator and denominator will annihilate the i’s on the denominator. This is a very popular method of ’dividing. For example. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
2i + 1 −i+1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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math hands
pg. 6
Trigonometry Sec. 04 notes
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
Compute w/ Complex Numbers Calculate and write in standard form. 2i + 3 5i − 2 There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 5i − 2 2i + 3 = 2i + 3 · 5i − 2 − 22 + 52 =
2i + 3 − 5i − 2 29
=
− 19i + 4 29
=
4 − 19 + i 29 29
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 3 2i + 3 = ·1 5i − 2 5i − 2 =
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2i + 3 − 5i − 2 · 5i − 2 − 5i − 2
=
− 19i + 4 29
=
4 − 19 + i 29 29
math hands
pg. 7
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Intro to Complex Numbers 1. Plot the following points. (a) 2 + 3i (b) 5 + 1i (c) −4 + −3i
(d) −4i
Solution:
8i 7i 6i 5i 4i b
3i
2 + 3i
2i b
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5 + 1i
5
6
7
8
9 10 11 12
-2i
−4 + −3i b
-3i
0 + −4i
-4i b -5i -6i -7i -8i
2. Plot the following points. (a) −1 + 4i
(b) 2 + 2i
(c) 5 + −5i
(d) −7i
Solution:
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math hands
pg. 8
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
8i 7i 6i 5i
−1 + 4i b 4i 3i b
2i
2 + 2i
1i -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -1i
1
2
3
4
5
6
7
8
9 10 11 12
-2i -3i -4i -5i
5b + −5i
-6i
0 + −7i
-7i b -8i
Simplify
√ −90
Solution: √ −90
4. 3. Simplify
√ = i 90 √ = i 9 · 10 √ √ = i 9 10 √ = i · 3 10 √ = 3i 10
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −150
Solution:
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math hands
pg. 9
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −150
5. Simplify
√ = i 150 √ = i 25 · 6 √ √ = i 25 6 √ = i·5 6 √ = 5i 6
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −375
Solution: √ −375
6. Simplify
√ = i 375 √ = i 25 · 15 √ √ = i 25 15 √ = i · 5 15 √ = 5i 15
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −18
Solution: √ −18
7. Simplify
√ = i 18 √ =i 9·2 √ √ =i 9 2 √ = i·3 2 √ = 3i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −50
Solution:
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math hands
pg. 10
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √ −50
8. Simplify
√ = i 50 √ = i 25 · 2 √ √ = i 25 2 √ = i·5 2 √ = 5i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −8
Solution: √ −8
9. Simplify
√ =i 8 √ = i 4·2 √ √ =i 4 2 √ = i·2 2 √ = 2i 2
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ −125
Solution: √ −125
10. Simplify
√ = i 125 √ = i 25 · 5 √ √ = i 25 5 √ = i·5 5 √ = 5i 5
(given) (def neg rad) (TT) (RP=PR) (def rad) (BI)
√ 3 −8
Solution: −2 since (−2)3 = −8 11. Simplify
√ 3 −125
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math hands
pg. 11
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: −5 since (−5)3 = −125 12. Simplify
√ 3 −1
Solution: −1 since (−1)3 = −1 13. Add 3i + 2 + 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 5i) = 2 + (3i + 4) + 5i = 2 + (4 + 3i) + 5i = (2 + 4) + (3i + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i
(ALA) (CoLA) (ALA) (DL) (BI)
14. Add 2i + 7 + 9i + 3 Solution:
by convention, we write in ’standard form’, ”a + bi” (7 + 2i) + (3 + 9i) = 7 + (2i + 3) + 9i = 7 + (3 + 2i) + 9i = (7 + 3) + (2i + 9i) = (7 + 3) + (2 + 9)i = 10 + 11i
(ALA) (CoLA) (ALA) (DL) (BI)
15. Add 3i − 2 + − 5i + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (−2 + 3i) + (4 + −5i) = −2 + (3i + 4) + −5i
= −2 + (4 + 3i) + −5i = (−2 + 4) + (3i + −5i)
= (−2 + 4) + (3 + −5)i = 2 + −2i
(ALA) (CoLA) (ALA) (DL) (BI)
16. Add 30i + 11 + 5i + 5
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math hands
pg. 12
Trigonometry Sec. 04 exercises Solution:
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” (11 + 30i) + (5 + 5i) = 11 + (30i + 5) + 5i = 11 + (5 + 30i) + 5i = (11 + 5) + (30i + 5i) = (11 + 5) + (30 + 5)i = 16 + 35i
(ALA) (CoLA) (ALA) (DL) (BI)
17. Add i + 1 + i + 2 Solution:
by convention, we write in ’standard form’, ”a + bi” (1 + 1i) + (2 + 1i) = 1 + (1i + 2) + 1i = 1 + (2 + 1i) + 1i = (1 + 2) + (1i + 1i) = (1 + 2) + (1 + 1)i = 3 + 2i
(ALA) (CoLA) (ALA) (DL) (BI)
18. Add 3i + 2 + 4 Solution:
by convention, we write in ’standard form’, ”a + bi” (2 + 3i) + (4 + 0i) = 2 + (3i + 4) + 0i = 2 + (4 + 3i) + 0i = (2 + 4) + (3i + 0i) = (2 + 4) + (3 + 0)i = 6 + 3i
(ALA) (CoLA) (ALA) (DL) (BI)
19. Add 3i + i Solution:
by convention, we write in ’standard form’, ”a + bi” (0 + 3i) + (0 + 1i) = 0 + (3i + 0) + 1i = 0 + (0 + 3i) + 1i = (0 + 0) + (3i + 1i) = (0 + 0) + (3 + 1)i = 0 + 4i
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math hands
(ALA) (CoLA) (ALA) (DL) (BI)
pg. 13
Trigonometry Sec. 04 exercises 20. Multiply 3i + 2
MathHands.com M´ arquez 5i + 4
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 5i + 4 = (2)(4) + (2)(5i) + (3i)(4) + (3i)(5i) = 8 + 10i + 12i + 15i
2
9i + 3
(BI)
= 8 + 10i + 12i + 15(−1)
(def i)
= (8 + −15) + (10i + 12i) = (8 + −15) + (10 + 12)i
(CoLA, ALA,BI) (DL)
= −7 + 22i
21. Multiply 2i + 7
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 2i + 7 9i + 3 = (7)(3) + (7)(9i) + (2i)(3) + (2i)(9i) 2
= 21 + 63i + 6i + 18i = 21 + 63i + 6i + 18(−1)
(BI) (def i)
= (21 + −18) + (63i + 6i) = (21 + −18) + (63 + 6)i
(CoLA, ALA,BI) (DL)
= 3 + 69i
22. Multiply 3i − 2
− 5i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i − 2 − 5i + 4 = (−2)(4) + (−2)(−5i) + (3i)(4) + (3i)(−5i) = −8 + 10i + 12i + −15i
2
= −8 + 10i + 12i + −15(−1) = (−8 + 15) + (10i + 12i)
= (−8 + 15) + (10 + 12)i = 7 + 22i
23. Multiply 30i + 11
5i + 5
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(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
math hands
pg. 14
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: by convention, we write in ’standard form’, ”a + bi” 30i + 11 5i + 5 = (11)(5) + (11)(5i) + (30i)(5) + (30i)(5i) = 55 + 55i + 150i + 150i
2
= 55 + 55i + 150i + 150(−1) = (55 + −150) + (55i + 150i)
= (55 + −150) + (55 + 150)i = −95 + 205i
24. Multiply i + 1
i+2
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
25. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i)
(BI) (def i)
= (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i
(CoLA, ALA,BI) (DL)
= 8 + −8i + 12i + −12i = 8 + −8i + 12i + −12(−1)
= 20 + 4i
26. Multiply 3i − 1
i−1
(FOIL)
2
(BI)
Solution:
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math hands
pg. 15
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i)
(DL)
= −2 + −4i
i+2
(BI) (def i) (CoLA, ALA,BI)
= (1 + −3) + (−1 + −3)i
27. Multiply i + 1
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” i + 1 i + 2 = (1)(2) + (1)(1i) + (1i)(2) + (1i)(1i) 2
= 2 + 1i + 2i + 1i = 2 + 1i + 2i + 1(−1)
(BI) (def i)
= (2 + −1) + (1i + 2i) = (2 + −1) + (1 + 2)i
(CoLA, ALA,BI) (DL)
= 1 + 3i
28. Multiply 3i + 2
− 4i + 4
(FOIL)
(BI)
Solution: by convention, we write in ’standard form’, ”a + bi” 3i + 2 − 4i + 4 = (2)(4) + (2)(−4i) + (3i)(4) + (3i)(−4i) = 8 + −8i + 12i + −12i
2
= 8 + −8i + 12i + −12(−1) = (8 + 12) + (−8i + 12i) = (8 + 12) + (−8 + 12)i = 20 + 4i
29. Multiply 3i − 1
i−1
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
Solution:
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math hands
pg. 16
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
by convention, we write in ’standard form’, ”a + bi” 3i − 1 i − 1 = (−1)(−1) + (−1)(1i) + (3i)(−1) + (3i)(1i) = 1 + −1i + −3i + 3i
2
= 1 + −1i + −3i + 3(−1) = (1 + −3) + (−1i + −3i) = (1 + −3) + (−1 + −3)i = −2 + −4i
(FOIL) (BI) (def i) (CoLA, ALA,BI) (DL) (BI)
30. multiply i3 Solution: i3 = i2 · i
= −1 · i = −i
(JAE) (def of i) (MT)
31. multiply i4 Solution: i4 = i2 · i2 = −1 · −1 =1
(JAE) (def of i) (NotNot)
32. multiply i6 Solution: i6 = i4 · i2
= 1 · −1 = −1
(JAE) (see previous problem) (MiD)
3
33. (3i + 2)
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pg. 17
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution:
(3i + 2)3
(given) 3
= (3i)
2
+
3
2
3 · (3i) (2)
2
+
3(3i)(2)
+
3
(2)
(famous, PP3)
= 27i + 54i + 36i + 8
(BI)
= 27(−i) + 54(−1) + 36(i) + 8 = −46 + 9i
(BI) (BI)
3
34. (2i + 1)
Solution:
(2i + 1)3
(given) 3
= (2i) 3
2
+
3 · (2i) (1)
2
2
+
3(2i)(1)
+
3
(1)
(famous, PP3)
= 8i + 12i + 6i + 1
(BI)
= 8(−i) + 12(−1) + 6(i) + 1 = −11 + −2i
(BI) (BI)
3
35. (−2i + 1)
Solution:
3
(−2i + 1)
(given) 3
= (−2i)
3
2
+ 2
3 · (−2i) (1)
+
= − 8i + 12i − 6i + 1
= −8(−i) + 12(−1) + −6(i) + 1 = −11 + 2i
2
3(−2i)(1)
+
3
(1)
(famous, PP3) (BI) (BI) (BI)
3
36. (−2i + 3)
Solution:
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pg. 18
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 3
(−2i + 3)
(given) 3
= (−2i)
2
+
3
2
3 · (−2i) (3)
+
2
3(−2i)(3)
3
+
(3)
= − 8i + 36i − 54i + 27 = −8(−i) + 36(−1) + −54(i) + 27
(famous, PP3) (BI) (BI)
= −9 + −46i
(BI)
37. (−1i + 1)3 Solution:
3
(−1i + 1)
(given) 3
= (−1i)
+
3
2
2
3 · (−1i) (1)
+
2
3(−1i)(1)
3
+
(1)
= − i + 3i − 3i + 1 = −1(−i) + 3(−1) + −3(i) + 1
(famous, PP3) (BI) (BI)
= −2 + −2i
(BI)
38. (1i + 1)3 Solution:
3
(1i + 1)
(given) 3
= (1i) 3
39. Divide
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2007-2009 MathHands.com
+ 2
2
3 · (1i) (1)
+
2
3(1i)(1)
+
3
(1)
(famous, PP3)
= i + 3i + 3i + 1 = 1(−i) + 3(−1) + 3(i) + 1
(BI) (BI)
= −2 + 2i
(BI)
7i + 1 7i + 3
math hands
pg. 19
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 7i + 1 7i + 1 = ·1 7i + 3 7i + 3 =
7i + 1 − 7i + 3 · 7i + 3 − 7i + 3 − 49i2 + 14i + 3 − 49i2 + 9
=
40. Divide
(MiD) (JOT) (MAT, FOIL)
=
14i + 52 58
(BI)
=
52 14i + 58 58
(BI)
4i + 2 − 5i + 2
Solution: 4i + 2 4i + 2 = ·1 − 5i + 2 − 5i + 2 = =
41. Divide
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2007-2009 MathHands.com
5i + 2 4i + 2 · − 5i + 2 5i + 2
20i2 + 18i + 4 − 25i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
18i − 16 29
(BI)
=
− 16 18i + 29 29
(BI)
5i + 2 4i + 2
math hands
pg. 20
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 4i + 2 4i + 2 = =
5i + 2 − 4i + 2 · 4i + 2 − 4i + 2 − 20i2 + 2i + 4 − 16i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
2i + 24 20
(BI)
=
24 2i + 20 20
(BI)
42. Divide
3i + 1 2i + 1
Solution: 3i + 1 3i + 1 = ·1 2i + 1 2i + 1 = =
43. Divide
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2007-2009 MathHands.com
3i + 1 − 2i + 1 · 2i + 1 − 2i + 1 − 6i2 + i + 1 − 4i2 + 1
(MiD) (JOT) (MAT, FOIL)
=
i+7 5
(BI)
=
7 i + 5 5
(BI)
5i + 2 i+2
math hands
pg. 21
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: 5i + 2 5i + 2 = ·1 i+2 i+2 = =
5i + 2 − i + 2 · i+2 −i+2 − 5i2 + 8i + 4 − i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 9 5
(BI)
=
9 8i + 5 5
(BI)
44. Divide
7i + 2 3i + 2
Solution: 7i + 2 7i + 2 = ·1 3i + 2 3i + 2 = =
7i + 2 − 3i + 2 · 3i + 2 − 3i + 2 − 21i2 + 8i + 4 − 9i2 + 4
(MiD) (JOT) (MAT, FOIL)
=
8i + 25 13
(BI)
=
25 8i + 13 13
(BI)
45. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 i+1
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
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pg. 22
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez −i+1 2i + 1 = 2i + 1 · 2 i+1 1 + 12 =
2i + 1 − i + 1 2
=
i+3 2
=
3 1 + i 2 2
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
2i + 1 2i + 1 = ·1 i+1 i+1 =
2i + 1 − i + 1 · i+1 −i+1
=
i+3 2
=
3 1 + i 2 2
46. Compute w/ Complex Numbers Calculate and write in standard form. 2i + 1 2i + 3
Solution: There are at least a couple ways to go about this.. one way, to note ’divide’ means ’multiply by inverse’... so..
− 2i + 3 2i + 1 = 2i + 1 · 2 2i + 3 3 + 22 =
2i + 1 − 2i + 3 13
=
4i + 7 13
=
7 4 + i 13 13
Another way to do it.. (more popular) is to simply multiply numerator and denominator by the conjugate of the denominator.
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pg. 23
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez 2i + 1 2i + 1 = ·1 2i + 3 2i + 3 =
2i + 1 − 2i + 3 · 2i + 3 − 2i + 3
=
4i + 7 13
=
7 4 + i 13 13
47. Compute w/ Complex Numbers Calculate and write in standard form. 1 i
Solution: hint: the bottom is 0 + 1i 48. Compute w/ Complex Numbers Calculate and write in standard form. 1 −i
Solution: hint: the bottom is 0 − 1i 49. Compute w/ Complex Numbers Calculate and write in standard form. i4
Solution: i4 = i2 · i2 = −1 · −1 = 1 50. Compute w/ Complex Numbers Calculate and write in standard form. i14
Solution: i14 = i12 · i2 3 = i4 · i2
(just add exponents) (setting up to use fact i4 = 1)
3
= (1) · i2 =i
(yipei-kae-yeh...)
2
= −1
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pg. 24
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
51. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
52. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−3 · 1
= i−3 · i4
(see previous problem why i4 = 1)
= i1 =i
(just add exponents) (note: there are many other ways to do this problem)
53. Compute w/ Complex Numbers Calculate and write in standard form. i25
Solution: i25 = i24 · i1 6 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
6
(used i4 = 1) (ez as sundays)
= (1) · i =i
54. Compute w/ Complex Numbers Calculate and write in standard form. i−5
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pg. 25
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
55. Compute w/ Complex Numbers Calculate and write in standard form. i−7
Solution: i−7 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
56. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
57. Compute w/ Complex Numbers Calculate and write in standard form. i−5
Solution: i−5 = i−8 · i1 −2 1 ·i = i4 = (1)
−2
=i
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·i
math hands
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
pg. 26
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
58. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4 = (1) =i
−1
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
59. Compute w/ Complex Numbers Calculate and write in standard form. i−3
Solution: i−3 = i−4 · i1 −1 1 ·i = i4
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
= (1)−1 · i =i
(ez as sundays)
60. Compute w/ Complex Numbers Calculate and write in standard form. i11
Solution: i11 = i8 · i1 2 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
2
(used i4 = 1)
= (1) · i
=i
(ez as sundays)
61. Compute w/ Complex Numbers Calculate and write in standard form. i−6
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Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution: i−6 = i−8 · i1 −2 1 ·i = i4 = (1) =i
−2
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1) (ez as sundays)
·i
62. Compute w/ Complex Numbers Calculate and write in standard form. i33
Solution: i33 = i32 · i1 8 = i4 · i1
(preparing to use i4 = 1) (preparing to use i4 = 1)
= (1)8 · i
(used i4 = 1)
=i
(ez as sundays)
63. Compute w/ Complex Numbers Calculate and write in standard form. i−150
Solution: i−150 = i−152 · i1 −38 1 ·i = i4 = (1)
−38
=i
(preparing to use i4 = 1) (preparing to use i4 = 1) (used i4 = 1)
·i
(ez as sundays)
64. Compute w/ Complex Numbers Calculate and write in standard form.
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1 1 √ +√ i 2 2
math hands
2
pg. 28
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
Solution:
1 1 √ +√ i 2 2
2
1 1 1 1 √ +√ i = √ +√ i 2 2 2 2 1 1 1 1 2 = + i+ i+ i 2 2 2 2 1 1 1 1 = + i+ i− 2 2 2 2 =i
(FOIL) (used i2 = −1)
65. Compute w/ Complex Numbers Calculate and write in standard form. !2 √ 3 1 + i 2 2
Solution: !2 √ 3 1 + i = 2 2 = = = =
! √ ! √ 3 1 3 1 + i + i 2 2 2 2 √ √ 3 3 3 1 + i+ i + i2 4 √4 4 4 √ 1 3 3 3 + i+ i− 4 4√ 4 4 2 2 3 + i 4 √4 3 1 + i 2 2
(FOIL) (used i2 = −1)
66. Compute w/ Complex Numbers Calculate and write in standard form. !3 √ 3 1 + i 2 2
Solution: we will use the result from the previous problem:
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pg. 29
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez √
!2 √ 3 1 + i 2 2
!2 √ 3 1 3 1 + i i = + 2 2 2 2 ! ! √ ! √ √ 3 1 3 3 1 1 + i = + i + i 2 2 2 2 2 2 !3 ! √ √ ! √ 1 3 1 3 3 1 + i + i + i = 2 2 2 2 2 2 √ √ 1 3 3 3 2 + i+ i+ i = 4 4 √4 √4 1 3 3 3 = + i+ i− 4 4 4 4 =i
(prev. problem)
(mult both sides by
√
3 2
+ 21 i )
(simplify left side) (FOIL) (use i2 = −1) (yipi-kae-yeh)
67. Compute w/ Complex Numbers Calculate and write in standard form. !6 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3 = (i)2
2
(getting ready to use A3 = i) (used A3 = i)
= −1
(ez as Sundays...)
68. Compute w/ Complex Numbers Calculate and write in standard form. !30 √ 3 1 + i 2 2
√
Solution: just for kicks.. and for the simplicity of writing, let us call this number 23 + 21 i = A. On the previous problem we demonstrated that A3 = i, we will use this fact freely and without inhibitions ... now..
A6 = A3
10
(getting ready to use A3 = i)
10
= (i)
(used A3 = i)
= −1
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(ez as Sundays...)
math hands
pg. 30
Trigonometry Sec. 04 exercises
MathHands.com M´ arquez
69. Inventing Numbers The natural numbers are in many ways natural. In some way, all other numbers are unnatural byproducts of human imagination. Which number was invented just to solve the following equation? 3+x=3
Solution: 0 was innvented 70. Inventing Numbers Which type of numbers were invented to solve the following equation? 3+x=0
Solution: negative numbers were innvented
71. Inventing Numbers Which type of numbers were invented to solve the following equation? 3x = 1
Solution: rational numbers were innvented 72. Inventing Numbers Which type of numbers were invented to solve the following type of equation? x2 = 3
Solution: square roots of numbers were innvented √ 73. Inventing Numbers Contemplate the idea of a world of numbers of the form a + b 3 where a, b are rational numbers. √ √ (a) add 32 + 5 3 + 73 + 53 3 √ √ (b) multiply 32 + 5 3 73 + 53 3 √ √ (c) does 32 + 5 3 have a multiplicative inverse of the form a + b 3 where a, b are rational. √ 74. Is i a complex number? if so can you write it in standard form? Solution: yes.. the answer is on prob #35
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math hands
pg. 31
Trigonometry Sec. 03 notes
MathHands.com M´ arquez
Converting Equations Polar & Cartesian The IDEA In the last couple sections we’ve established, among other things, a ’dictionary’ to convert coordinate points from cartesian to polar and vise-varsa. In this section, we would like to exploit that same dictionary to convert equations. As a reminder, the dictionary has two components, ’from’ polar, and ’to’ polar. The goal in converting, in most cases, is to eliminate all x’s and y’s and replace them with appropriate expressions involving r’s and θ’s if we are translating to polar. When converting from polar, the idea is to eliminate the r’s and θ’s. We summarize the ’dictionary’ below. Translating
Polar
Cartesian
90◦ 120◦
y = r sin θ x = r cos θ
60◦ 30◦
150◦ b
b
(r, θ)
(x, y)
0◦
180◦
210◦
330◦
θ = tan−1 300◦
240◦
r2 = x2 + y
270◦
y x 2
Example: Converting Equation of a line to Polar Convert the following equation to polar coordinates: y=x Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
y y x tan θ θ
=x
(given)
=1
(div by x, setting up to use translating dictionary, tan θ = y/x)
=1 = 45◦
(used tan θ = y/x) (notice: this there are infinite possibilities for θ (ie −135 , 225 ...), this is just one of them.) ◦
◦
Thus, one way to translate the equation y = x to polar is to convert to θ = 45◦ Example: Converting Equation of a circle to Polar Convert the following equation to polar coordinates: x2 + y 2 = 25 Solution: Before we get started, let us recognize that there is not a 1-1 correspondence between polar and cartesian coordinates, that is for each cartesian coordinates, there may be infinite polar coordinates corresponding to the same point. In light of this, the ’translating’ of this equation is not unique. Observe,
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pg. 1
Trigonometry Sec. 03 notes
MathHands.com M´ arquez x2 + y 2 = 25
(given)
2
2
(used the dictionary r = x2 + y 2 )
r = 25
at this point we have eliminated all x’s and y’s from the equation, thus the converting is complete. It is sometimes desirable to solve for r explicitly, thus we could also say r = 5 or r = −5, in fact all three of these equations, r2 = 25, r = 5, AND r = −5 all yield the same graph as x2 + y 2 = 25. Example: Converting the equation of a circle to Polar Convert the following equation to polar coordinates: x2 + 3x + 5 + y 2 − 4y + 2 = 25 Solution:
x2 + 3x + 5 + y 2 − 4y + 2 = 25 2
2
(given)
2
(x + y ) + 3(x) − 4(y) = 18
(just cleaning up, getting ready to use the dictionary)
(r ) + 3(r cos θ) − 4(r sin θ) = 18
(used the dictionary)
2
r + 3r cos θ − 4r sin θ = 18
(algebra)
Example: Converting the equation of a circle from Polar Convert the following equation of a circle to polar coordinates: r = 12 cos θ Solution: r = 12 cos θ r · r = r(12 cos θ)
r2 = 12(r cos θ) 2
2
x + y = 12x 2
2
2
y + x + − 12x = 0 2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
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math hands
(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 2
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Converting Equations Polar & Cartesian 1. Convert the following equation to polar coordinates: y = 2x
Solution: Note: the ’translating’ of this equation is not unique.
y = 2x y =2 x tan θ = 2 θ ≈ 63.435
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) ◦
(notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 63.435◦ 2. Convert the following equation to polar coordinates: 2 y= − x 3
Solution: Note: the ’translating’ of this equation is not unique. 2 y= − x 3 2 y = − x 3 2 tan θ = − 3 θ ≈ −33.69◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ −33.69◦ 3. Convert the following equation to polar coordinates: y=
2 x 5
Solution: Note: the ’translating’ of this equation is not unique.
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pg. 3
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez 2 x 5 2 = 5 2 = 5 ≈ 21.801◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 2x to polar is to convert to θ ≈ 21.801◦ 4. Convert the following equation to polar coordinates: y=
4 x 3
Solution: Note: the ’translating’ of this equation is not unique. 4 x 3 4 = 3 4 = 3 ≈ 53.13◦
(given)
y= y x tan θ θ
(div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = 4x to polar is to convert to θ ≈ 53.13◦ 5. Convert the following equation to polar coordinates: 4 y= − x 3 Solution: Note: the ’translating’ of this equation is not unique. 4 y= − x 3 4 y = − x 3 4 tan θ = − 3 θ ≈ −53.13◦
(given) (div by x, setting up to use translating dictionary, tan θ = y/x) (used tan θ = y/x) (notice: above eq has infinite solutions for θ, this is just one of them.)
Thus, one way to translate the equation y = − 4x to polar is to convert to θ ≈ −53.13◦ 6. Convert the following equation of a circle to polar coordinates: 2x2 + 3x + 2y 2 + − 5y = 7
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pg. 4
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution:
2x2 + 3x + 2y 2 + − 5y = 7 2 x2 + y 2 + 3(x) + − 5(y) = 7 2 r2 + 3(r cos θ) + − 5(r sin θ) = 7
(given) (ready to use the dictionary....) (used the dictionary)
7. Convert the following equation of a circle to polar coordinates: − 5x2 + 2x + − 5y 2 + 7y = 25 Solution:
− 5x2 + 2x + − 5y 2 + 7y = 25 − 5 x2 + y 2 + 2(x) + 7(y) = 25 − 5 r2 + 2(r cos θ) + 7(r sin θ) = 25
(given) (ready to use the dictionary....) (used the dictionary)
8. Convert the following equation of a circle to polar coordinates: 3 4x2 + x + 4y 2 + 1y = 9 2
Solution: 3 4x2 + x + 4y 2 + 1y = 9 2 3 4 x2 + y 2 + (x) + 1(y) = 9 2 3 2 4 r + (r cos θ) + 1(r sin θ) = 9 2
(given) (ready to use the dictionary....) (used the dictionary)
9. Convert the following equation of a circle to polar coordinates: 3x2 + 3x + 3y 2 + 5y = 4
Solution:
3x2 + 3x + 3y 2 + 5y = 4 3 x2 + y 2 + 3(x) + 5(y) = 4 3 r2 + 3(r cos θ) + 5(r sin θ) = 4
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(given) (ready to use the dictionary....) (used the dictionary)
pg. 5
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
10. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 9 25
Solution: y2 x2 + =1 9 25 (r cos θ)2 (r sin θ)2 + =1 9 25 ! cos2 θ sin2 θ 2 + =1 r 9 25
(given) (ready to use the dictionary....) (alebra)
11. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 16
Solution: y2 x2 + =1 4 16 (r cos θ)2 (r sin θ)2 + =1 4 16 ! cos2 θ sin2 θ + =1 r2 4 16
(given) (ready to use the dictionary....) (alebra)
12. Convert the following equation of a circle to polar coordinates: x2 y2 + =1 4 49
Solution: y2 x2 + =1 4 49 (r cos θ)2 (r sin θ)2 + =1 4 49 ! cos2 θ sin2 θ =1 + r2 4 49
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(given) (ready to use the dictionary....) (alebra)
pg. 6
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
13. Convert the following equation of a circle to polar coordinates: r = 2 sin θ
Solution: r = 2 sin θ
(given)
r · r = r(2 sin θ)
(clever little idea, mult both sides by r)
r2 = 2(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 2y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 2y = 0
(optional step: complete the square)
2
x + y + − 2y + 1 = 1 2 x2 + y + − 1 = 1
(optional to step: complete the square) (Shows its a circle, shows center/radius)
14. Convert the following equation of a circle to polar coordinates: r = 5 sin θ
Solution: r = 5 sin θ
(given)
r · r = r(5 sin θ)
(clever little idea, mult both sides by r)
2
r = 5(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 5y = 0 25 25 x2 + y 2 + − 5y + = 4 4 2 25 5 = x2 + y + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
15. Convert the following equation of a circle to polar coordinates: r = 6 sin θ
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pg. 7
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 6 sin θ r · r = r(6 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 6(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 6y 2
2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 6y = 0
(optional step: complete the square)
2
x + y + − 6y + 9 = 9 2 x2 + y + − 3 = 9
(optional to step: complete the square) (Shows its a circle, shows center/radius)
16. Convert the following equation of a circle to polar coordinates: r = 11 sin θ
Solution: r = 11 sin θ r · r = r(11 sin θ)
(given) (clever little idea, mult both sides by r)
r2 = 11(r sin θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 11y 2
(used dictionary, all r’s and θ’s gone!!)
2
x + y + − 11y = 0 121 121 = x2 + y 2 + − 11y + 4 4 2 11 121 x2 + y + − = 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
17. Convert the following equation of a circle to polar coordinates: r = 8 cos θ
Solution: r = 8 cos θ
(given)
r · r = r(8 cos θ)
(clever little idea, mult both sides by r)
2
r = 8(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 8x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 8x = 0
(optional step: complete the square)
2
y + x + − 8x + 16 = 16 2 y 2 + x + − 4 = 16
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(optional to step: complete the square) (Shows its a circle, shows center/radius)
math hands
pg. 8
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
18. Convert the following equation of a circle to polar coordinates: r = 12 cos θ
Solution: r = 12 cos θ
(given)
r · r = r(12 cos θ)
(clever little idea, mult both sides by r)
r2 = 12(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 12x 2
2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 12x = 0
(optional step: complete the square)
2
y + x + − 12x + 36 = 36 2 y 2 + x + − 6 = 36
(optional to step: complete the square) (Shows its a circle, shows center/radius)
19. Convert the following equation of a circle to polar coordinates: r = 5 cos θ
Solution: r = 5 cos θ
(given)
r · r = r(5 cos θ)
(clever little idea, mult both sides by r)
2
r = 5(r cos θ) 2
(algebra, getting ready to use dictionary)
2
x + y = 5x 2
(used dictionary, all r’s and θ’s gone!!)
2
y + x + − 5x = 0 25 25 y 2 + x2 + − 5x + = 4 4 2 25 5 = y2 + x + − 2 4
(optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
20. Convert the following equation of a circle to polar coordinates: r = 11 cos θ
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pg. 9
Trigonometry Sec. 03 exercises
MathHands.com M´ arquez
Solution: r = 11 cos θ r · r = r(11 cos θ) r2 = 11(r cos θ) 2
2
x + y = 11x 2
2
y + x + − 11x = 0 121 121 = y 2 + x2 + − 11x + 4 4 2 121 11 = y2 + x + − 2 4
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math hands
(given) (clever little idea, mult both sides by r) (algebra, getting ready to use dictionary) (used dictionary, all r’s and θ’s gone!!) (optional step: complete the square) (optional to step: complete the square) (Shows its a circle, shows center/radius)
pg. 10
Trigonometry Sec. 02 notes
MathHands.com M´ arquez
Introduction to Polar Graphs Main Idea Plot Points You should be very conformable plotting points in radians as well as degrees. Simply make a table of values with r’s and corresponding θ’s, then start plotting until you can make sense of the the graph. If connecting the points is uneasy, choose smaller angle intervals, i.e. instead of plotting and calculating the r every 30 degrees, plot every 5 degrees. Get to Know Your Calculator I am usually not a big calculator fan, but this may be a good time when we can use it appropriately to graph some of these functions. After all, there is limited beauty and creativity that occurs when plotting points. Get to know Famous Questions 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ? Get to know Famous Graphs Get to know Famous Graphs 1. Can you calculate what is the largest r value on a graph for a given equation? and for which angles does it occur? 2. For which angles θ does the graph go through the origin? 3. How is the graph affected if we restrict the values of θ?
r = sin θ
r = cos θ
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2007-2009 MathHands.com
r = cos(2θ)
r = cos(3θ)
r = 5 + 3 sin θ
r = sin(2θ)
r = sin(4θ)
r = 3 + 5 sin θ
r = sin(3θ)
r = cos(5θ)
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Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
Introduction to Polar Graphs 1. Graph and Understand the graph of r = 5 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 2. Graph and Understand the graph of r = 5 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 3. Graph and Understand the graph of r = −3 cos θ
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pg. 2
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 4. Graph and Understand the graph of r = 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 5. Graph and Understand the graph of r = −6 sin θ
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pg. 3
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 6. Graph and Understand the graph of r = 5 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 7. Graph and Understand the graph of r = 5 sin θ. Limit the study of this graph to the values of θ ranging from −90◦ to 0◦ .
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pg. 4
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 8. Graph and Understand the graph of r = −3 cos θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 9. Graph and Understand the graph of r = 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 180◦.
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pg. 5
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 10. Graph and Understand the graph of r = −6 sin θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 11. Graph and Understand the graph of r = 5 + cos θ
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pg. 6
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 12. Graph and Understand the graph of r = 4 − 2 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 13. Graph and Understand the graph of r = −3 + 2 cos θ
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pg. 7
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 14. Graph and Understand the graph of r = 3 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 15. Graph and Understand the graph of r = 4 − 2 cos θ
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2007-2009 MathHands.com
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pg. 8
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 16. Graph and Understand the graph of r = −2 + 3 cos θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 17. Graph and Understand the graph of r = 3 + 2 sin θ
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2007-2009 MathHands.com
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pg. 9
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 18. Graph and Understand the graph of r = 3 − 2 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 19. Graph and Understand the graph of r = −3 + 2 sin θ
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2007-2009 MathHands.com
math hands
pg. 10
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 20. Graph and Understand the graph of r = 3 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 21. Graph and Understand the graph of r = 2 − 3 sin θ
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pg. 11
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 22. Graph and Understand the graph of r = −2 + 3 sin θ
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 23. Graph and Understand the graph of r = 2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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2007-2009 MathHands.com
math hands
pg. 12
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 24. Graph and Understand the graph of r = 2 − 3 cos θ. Limit the study of this graph to the values of θ ranging from 0◦ to 60◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 25. Graph and Understand the graph of r = −2 + 3 cos θ. Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
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math hands
pg. 13
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 26. Graph and Understand the graph of r = 5 + 5 sin θ. Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 27. Graph and Understand the graph of r = 3 − 2 sin θ. Limit the study of this graph to the values of θ ranging from 135◦ to 225◦ .
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2007-2009 MathHands.com
math hands
pg. 14
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 28. Graph and Understand the graph of r = −3 + 2 sin θ. Limit the study of this graph to the values of θ ranging from 360◦ to 540◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 29. Graph and Understand the graph of r = 5 cos(2θ)
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2007-2009 MathHands.com
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pg. 15
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 30. Graph and Understand the graph of r = 6 cos(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 31. Graph and Understand the graph of r = 5 sin(2θ)
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2007-2009 MathHands.com
math hands
pg. 16
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 32. Graph and Understand the graph of r = 6 sin(4θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 33. Graph and Understand the graph of r = 5 cos(3θ)
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pg. 17
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 34. Graph and Understand the graph of r = 6 cos(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 35. Graph and Understand the graph of r = 5 sin(3θ)
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2007-2009 MathHands.com
math hands
pg. 18
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 36. Graph and Understand the graph of r = 4 sin(5θ)
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 37. Graph and Understand the graph of r = 5 sin(2θ). Limit the study of this graph to the values of θ ranging from 0◦ to 90◦ .
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math hands
pg. 19
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 38. Graph and Understand the graph of r = 6 sin(4θ). Limit the study of this graph to the values of θ ranging from 90◦ to 180◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 39. Graph and Understand the graph of r = 5 cos(3θ). Limit the study of this graph to the values of θ ranging from 0◦ to 180◦ .
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2007-2009 MathHands.com
math hands
pg. 20
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez 90◦ 60◦
120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 40. Graph and Understand the graph of r = 6 cos(5θ). Limit the study of this graph to the values of θ ranging from 180◦ to 270◦ .
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
Solution: 41. Graph and Understand the graph of r2 = 25 sin2 θ √ Solution: Note, we can solve for r as r√= ± 25 sin2 θ... then we can plot each r separately, that is we graph √ r = 25 sin2 θ and then we graph r = − 25 sin2 θ
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math hands
pg. 21
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
42. Graph and Understand the graph of r2 = 25 sin2 3θ √ 25 sin2 3θ... then we can plot each r separately, that is we graph Solution: Note, we can solve for r as r = ± √ √ 2 r = 25 sin 3θ and then we graph r = − 25 sin2 3θ 90◦ 60◦ 120◦
30◦ 150◦
0◦ 180◦
330◦ 210◦
300◦ 240◦ 270◦
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2007-2009 MathHands.com
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pg. 22
Trigonometry Sec. 02 exercises
MathHands.com M´ arquez
43. Understanding The graph of r = 2 − 3 cos θ has an inner loop. What values of θ trace precisely this inner loop? i.e. ?◦ ≤ θ ≤?◦ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center
44. Understanding On the graph of r = 5 sin(3θ) (a) ’sin 3θ’ will take values between 1 and −1; never larger than one, never smaller than −1. Find the angles θ, at which sin 3θ take on its largest value 1, or is smallest value -1. (i.e solve sin 3θ = 1 and solve sin 3θ = −1 )
(b) Graph r = 5 sin(3θ) and mark the points obtained from part a). What can you say about these points? A) nothing B) they are the ’tip of the leaves’ C) they are not on the leaves. (c) Find the angles at which the graph goes through (0, 0) (d) Find the interval/s for theta that trace the second ’leaf’ Solution: hint: solve for the values of θ when r = 0, that is when graph goes through the center, such leaf starts and ends at center.. 45. Understanding On the graph of r = −5 + 2 cos θ
(a) ’cos θ’ will take values between 1 and −1; never larger than one, never smaller than −1. What is the largest or the smallest that the quantity ”−5 + 2 cos θ” can become?
(b) Find the angles at which ”−5 + 2 cos θ” takes on its largest value -3, or is smallest value -7. (c) Graph r = −5 + 2 cos θ and mark the points obtained from part b). What can you say about these points? A) nothing B) its the furthest point from the origin (d) Find the angles at which the graph goes through (0, 0) (e) Find the interval/s for theta that trace ONLY the outer leaf.
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pg. 23
Trigonometry Sec. 01 exercises
MathHands.com M´ arquez
Intro to Polar Coordinates 1. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (3, 120◦), p2 = (−6, 120◦), p3 = (−3, −120◦), p4 = (3, −300◦), p5 = (−3, 135◦) Solution: 90◦ 60◦ 120◦
30◦
p1
150◦
b
p4
p3 b
0◦ 180◦
p5
b
330◦ 210◦
p2 b
300◦
240◦ 270◦
2. The following is just a random sequence of point in polar coordinates. Practice by plotting and labeling each of the points: p1 = (2, −120◦), p2 = (−4, −120◦), p3 = (−4, 120◦), p4 = (4, −60◦ ), p5 = (−4, 135◦) Solution:
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