Ratio Nirvana
Trigonometry Sec. 01
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Ratio Nirvana The Ratios We Know: Solving triangles is one of the essential promises of trigonometry, and knowing one [or more] of the ratios is one essential tool to accomplish just this. Now, the ratios we have learned thus far are the ratios for 45-45 Triangles, ratios for 30-60 Triangles, and we hesitantly add to the list the ratios for the almost triangles.
45-45 Triangle Ratios s
√ × 2
×
1
√ 3 ×
◦
Almost-Triangle Ratios
45◦
m
≈ 0◦
s
h
≈0
30
s
×2
h
h
30-60 Triangle Ratios
a≈h
Essentially what this means is for any of these triangles, if we know the size of one, just one, non-zero side we can solve all sides of the triangle. This is great news but very restrictive since we know the ratios, thus far, only for these special triangles. If the angles were different, so too would be the shape of the triangle, and thus the ratios of the sides would be different as well. Naturally, this leads to a different question, could we somehow determine the ratios for other triangles besides these? Or posed differently, which ratios would we like to know? The ambitious answer is we would like to know the ratios for all triangles. Just imagine, if we knew all the ratios for all the triangles we could possible solve any and every triangle just with a mere glance at the angle and one of the sides. This translates into big, very big ideas. We could position the moon at a corner of a triangle, glance at it, compute the ratios and there, just like that, we could compute the distance to the moon, the size of the moon, the height of a mountain, etc. etc. It is no exaggeration to say that if you knew all the ratios for all the triangles, well..., everyone would want to be your friend. Visually, we summarize: Ratios we would like to know
23-67 Triangle Ratios
41.45612◦ Triangle Ratios
Any & every angle θ ?
r
b
r
b
?
b
r
?
?
23◦
? 41.45612◦
a
a
? θ◦ a
Ratio Nirvana: Through sheer audacity, creativity, perseverance, pride, by bits and pieces, and over the centuries, some of the greatest thinkers of our species who have ever lived started putting the ideas together. The chinese from 4,000 years ago, the arabic/persian cultures from 2,000 years ago, the asian, the greeks, the indian, and everyone in-between collectively generation after generation each added a small piece to the puzzle, and in the end we triumphed over this challenge with grace and elegance. Today, we can find the ratios for all Euclidean Triangles with ease ad efficiency. We even have names and symbols for these ratios. The names for these ratios are sine, cosine and tangent. These
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are more than just names, they are indeed functions that describe for us all the ratios for all the triangles. Thanks to these functions and our ability to compute them, we could say that we now live in ratio nirvana, where all ratios are known. Let us now formally introduce the main authors of ratio-nirvana, the trigonometric functions. The Trigonometric Functions: Let us consider the first famous ratio. As before, we will continue to name the sides with respective to the corner, or angle from which we stand. Right triangles usually have two smaller angles [smaller than 90◦ ]. We will usually stand next to one of these two and for now, never next to the right angle.
opp
p hy
θ adj opp Now, we consider a ratio, let us say, opposite to hypothenuse, or hyp , and we give a name. We will call it the sine ratio. Of course this ratio depends on the shape of the triangle which depends on the angle θ, so really the description of the ratio should be more like a function, such as f (θ), rather than just ’sine’. But this function is so famous f (θ) is not special enough, thus we compromise and we call it sin(θ). In other words, we define1
sin(θ) =
opp hyp
p hy
× 21
opp
for the appropriate triangle. We are now ready to compute our very first sine values, let us compute sin(30◦ )
sin(30◦ ) =
opp hyp
sin(30◦ ) =
1 2
30◦ adj
10 p, hy
opp, 5
Usually, we want the ratios so we can find the sides, but often we will know the sides from these derive the ratios:
sin(30◦ ) =
opp hyp
sin(30◦ ) =
5 1 = 10 2
30◦ adj 1 we
will extend the definition of sine later in a much bigger and broader sense.
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Note the sine function is not affected by how big the triangle is. The above triangle could have hypothenuse 10, 20 or 100, the ratio opp/hyp would still be 1/2, thus sin(30◦ ) = 21 This ratio is old news. What we would really like to know is other ratios, we want to know all ratios. Unfortunately, the complete answer to this would quickly take us outside the scope of this course, and we will never [in this class] learn how the intellectual giants of our past resolved this issue. We will have to settle for the pride-less and undecorated method, using a calculator. Go on and try your calculator, check that it is in degrees mode and try sin(30◦ ) = .5, the expected ratio. Now go on and compute sin(23◦ ) ≈ 0.3907, at last, we have arrived! Ratio nirvana, we can compute the ratios for all Euclidean right triangles.
×0.6620
×0.3907
r
× sin(θ)
b
r
r
Generic Triangle Ratios
23◦
41.45612◦ a
θ
a
a
But wait, there is more. Just like we defined the sine function as the ratio of the opp/hyp sides of a triangle we also define2 the cosine function to be the ratio of adjacent side to hypothenuse. So we have cos(θ) =
adj hyp
opp, med,
hy p, 2
√
3
For example:
1
×2
cos(60◦ ) =
adj hyp
cos(60◦ ) =
1 2
60◦ adj, small, 1
And now using our calculator we have ratio nirvana. The adj/hyp ratio for any eucledian triangle is given by the cosine function: 2 we
will later extend this definition to a broader idea
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b
41.45612◦ Triangle Ratios b
23-67 Triangle Ratios
Trigonometry Sec. 01
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23-67 Triangle Ratios
41.62◦ Triangle Ratios
Generic Triangle Ratios r b
b
r
41.62◦
θ
6
05 a
a
θ) os(
.92
23◦
×c
×0
747 ×0.
b
r
a
Similarly, the tangent function is well known to represent the ratio: tan(θ) =
opp hyp
These three are the most famous trigonometric functions describing the most famous ratios for right triangles. The acronym SOH-CAH-TOA is often used to remember each of the ratios. That is SOH helps some students remember Opp Adj , then CAH reminds us that Cos(θ) = Hyp while TOA reminds us that T an(θ) = Opp that Sin(θ) = Hyp Ad We summarize this in:
SOH-CAH-TOA
p hy
θ os( ×c )
θ
× sin(θ)
×
n( ta
Sin(θ) =
Opp Hyp
Cos(θ) =
Adj Hyp
T an(θ) =
Opp Adj
opp
Generic Triangle Ratios
θ)
adj
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1. Find the three famous ratios and write them on the diagram
opp
hyp
20◦ adj
hyp
opp
2. Find the three famous ratios and write them on the diagram
20◦ adj
3. Find the three famous ratios and write them on the diagram
opp
hyp
20◦ adj
p hy
opp
4. Find the three famous ratios and write them on the diagram
30◦ adj
5. Find the three famous ratios and write them on the diagram
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p hy
opp
Trigonometry Sec. 01
40◦ adj
opp
hyp
6. Find the three famous ratios and write them on the diagram
70◦ adj
hyp
opp
7. Find the three famous ratios and write them on the diagram
10◦ adj
8. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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m. 10 c
20◦
9. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
m. 12 c
20◦
10. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
cm. 6.8
20◦
11. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
. cm 10
32◦
12. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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. cm 12
25◦
13. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
m. 6.8 c
15◦
7.5 cm.
14. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
69◦
15. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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10 cm.
Trigonometry Sec. 01
35◦
8 cm.
16. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
42◦
7.5 cm.
17. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
9◦
11 cm.
18. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
12◦
19. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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9 cm.
Trigonometry Sec. 01
23◦
20. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
9◦ 7.5 cm.
21. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
12◦ 11 cm.
22. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
23◦ 9 cm.
23. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
67◦
9 cm.
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24. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
55◦
10 cm.
25. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
50◦
6 cm.
26. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
35◦
7.5 cm.
27. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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4
5
θ 3
28. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 34
5
(May not be drawn to scale.)
θ 3
29. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 26
5
(May not be drawn to scale.)
θ 1
30. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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√ 104
10
Trigonometry Sec. 01
θ 2
31. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 45
6
(May not be drawn to scale.)
θ 3
32. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
1
2
θ √
3
33. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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1
√ 2
θ 1
34. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
0
5
θ 5
35. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
x2
4
6x
θ x
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Ratio Nirvana : SOLUTIONS 1. Find the three famous ratios and write them on the diagram sin(20◦ ) ≈ 0.3420
× 93 0. 97
20
◦
×0.3420
×0.3
cos(20◦ ) ≈ 0.9397
opp
hyp
640
tan(20◦ ) ≈ 0.3640
adj
2. Find the three famous ratios and write them on the diagram sin(20◦ ) ≈ 0.3420
× 93 0.
20
◦
×0.3420
97
×0.3
opp
hyp
640
cos(20◦ ) ≈ 0.9397 tan(20◦ ) ≈ 0.3640
adj
3. Find the three famous ratios and write them on the diagram
sin(20◦ ) ≈ 0.3420
× 93 0. 97
20◦
×0.3420
opp
hyp
3640 ×0.
cos(20◦ ) ≈ 0.9397 tan(20◦ ) ≈ 0.3640
adj
p hy
×0 6 .86 0
30
◦
×0.5000
×0
.57
74
opp
4. Find the three famous ratios and write them on the diagram
sin(30◦ ) ≈ 0.5000 cos(30◦ ) ≈ 0.8660 tan(30◦ ) ≈ 0.5774
adj
5. Find the three famous ratios and write them on the diagram
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766 ×0.
×
0
40
◦
×0.6428
83 0.
opp
p hy
sin(40◦ ) ≈ 0.6428
91
cos(40◦ ) ≈ 0.7660 tan(40◦ ) ≈ 0.8391
adj
hyp
×0 .93 97
475
×0.3420 70
opp
6. Find the three famous ratios and write them on the diagram
×2.7
sin(70◦ ) ≈ 0.9397 cos(70◦ ) ≈ 0.3420 tan(70◦ ) ≈ 2.7475
◦
adj
sin(10◦ ) ≈ 0.1736
7. Find the three famous ratios and write them on the diagram
10
◦
×0
.98 4 adj
×0.1736
8
×0.1763
opp
hyp
cos(10◦ ) ≈ 0.9848 tan(10◦ ) ≈ 0.1763
8. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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m. 10 c
opp ≈ (10)0.3420
opp, 3.42cm.
Trigonometry Sec. 01
×0.3420
×
= 3.42 adj ≈ (10)0.9397
93 0.
= 9.397
97
20◦
adj, 9.397cm.
m. 12 c
opp, 4.104cm.
9. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.3420
opp ≈ (12)0.3420
×
= 4.104 adj ≈ (12)0.9397
93 0.
= 11.276
97
20◦
adj, 11.276cm.
10. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.3420
× 93 0.
= 2.326 adj ≈ (6.8)0.9397 = 6.39
97
20
◦
opp, 2.326cm.
cm. 6.8
opp ≈ (6.8)0.3420
adj, 6.39cm.
. cm 10
×0.5299
opp, 5.299cm.
11. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
opp ≈ (10)0.5299
×0 8 .84
= 5.299 adj ≈ (10)0.8480 = 8.48
0
32◦ adj, 8.48cm.
12. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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. cm 12
×0.4226
opp, 5.071cm.
Trigonometry Sec. 01
opp ≈ (12)0.4226
×0 .90
= 5.071 adj ≈ (12)0.9063 = 10.876
63
25◦
adj, 10.876cm.
m. 6.8 c
15◦
×0
.96
×0.2588
59
opp, 1.76cm.
13. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters. opp ≈ (6.8)0.2588
= 1.76 adj ≈ (6.8)0.9659 = 6.568
adj, 6.568cm.
×0 .93 36
×0.3584 69◦
opp, 7.5 cm.
hyp , 8. 034 cm.
14. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
hyp ≈ (7.5) ÷ 0.9336 = 8.034 adj ≈ (8.034)0.3584 = 2.879
adj, 2.879cm.
15. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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.
opp, 10 cm.
m 4c .43 7 1 p, hy
×0.5736
hyp ≈ (10) ÷ 0.5736
×0
= 17.434 adj ≈ (17.434)0.8192
9 .81 2
= 14.281
35◦ adj, 14.281cm.
16. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
.
opp, 8 cm.
cm 56 9 . 11 p, y h
×0.6691
hyp ≈ (8) ÷ 0.6691
×0.7
= 11.956 adj ≈ (11.956)0.7431
431
= 8.885
42◦ adj, 8.885cm.
. .943cm hyp, 47
9◦
×0.9
×0.1564
877
adj, 47.353cm.
opp, 7.5 cm.
17. Determine the size for each of the three sides of the triangle. Assume all length units arehyp centimeters. ≈ (7.5) ÷ 0.1564 = 47.943 adj ≈ (47.943)0.9877 = 47.353
cm. 2.907 hyp, 5
12◦
×0
.97
×0.2079
opp, 11 cm.
18. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
hyp ≈ (11) ÷ 0.2079 = 52.907 adj ≈ (52.907)0.9781 = 51.751
81
adj, 51.751cm.
19. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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. 4cm .03 3 2 , hyp
opp, 9 cm.
Trigonometry Sec. 01
×0.3907
×0
adj ≈ (23.034)0.9205 = 21.203
.92 05
23◦
hyp ≈ (9) ÷ 0.3907 = 23.034
adj, 21.203cm.
93cm. hyp, 7.5
9◦
×0.9
opp, 1.188cm.
20. Determine the size for each of the three sides of the triangle. Assume all length units arehyp centimeters. ≈ (7.5) ÷ 0.9877 = 7.593 ×0.1564
877
adj, 7.5 cm.
opp ≈ (7.593)0.1564 = 1.188
cm. 1.246 hyp, 1
×0
12◦
.97
opp, 2.338cm.
21. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.2079
81
hyp ≈ (11) ÷ 0.9781
= 11.246 opp ≈ (11.246)0.2079 = 2.338
adj, 11 cm.
. 7cm .77 9 , hyp
opp, 3.820cm.
22. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.3907
×0 .92
opp ≈ (9.777)0.3907 = 3.820
05
23◦
hyp ≈ (9) ÷ 0.9205 = 9.777
adj, 9 cm.
23. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
sin(67◦ ) = 0.9205
. 7cm .77 9 , hyp
×0 .92 05 opp, 9 cm.
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×0.3907
adj, 3.820cm.
67◦
hyp ≈ (9) ÷ 0.9205 = 9.777 cos(67◦ ) = 0.3907 opp ≈ (9.777)0.3907 = 3.820
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Trigonometry Sec. 01
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24. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
m 8c .20 2 1 p, hy
. ×0.5736
adj, 7.002cm.
55◦
sin(55◦ ) = 0.8192 hyp ≈ (10) ÷ 0.8192 = 12.208
×0 9 .81 2
cos(55◦ ) = 0.5736 opp ≈ (12.208)0.5736 = 7.002
opp, 10 cm.
25. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
p, hy
. cm 32 ×0.6428 8 7.
766 ×0.
adj, 5.034cm.
50◦
sin(50◦ ) = 0.7660 hyp ≈ (6) ÷ 0.7660 = 7.832 cos(50◦ ) = 0.6428
0
opp ≈ (7.832)0.6428 = 5.034
opp, 6 cm.
26. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.81 92
×0.57 36 opp, 7.5 cm.
adj, 10.711cm.
hy p, 13 .07 6c m
.
35◦
sin(35◦ ) = 0.5736 hyp ≈ (7.5) ÷ 0.5736 = 13.076
cos(35◦ ) = 0.8192 opp ≈ (13.076)0.8192 = 10.711
27. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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5
. yp ,h
4 , opp.
Trigonometry Sec. 01
θ 3 , adj.
sin(θ) =
opp 4 = hyp 5
cos(θ) =
3 adj = hyp 5
tan(θ) =
opp 4 = adj 3
28. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
sin(θ) =
opp 5 =√ hyp 34
cos(θ) =
3 adj =√ hyp 34
tan(θ) =
5 opp = adj 3
sin(θ) =
opp 5 =√ hyp 26
cos(θ) =
1 adj =√ hyp 26
tan(θ) =
opp 5 = adj 1
5 , opp.
√ 34
. yp ,h
θ 3 , adj.
29. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 26
. yp ,h
θ 1 , adj.
5 , opp.
(May not be drawn to scale.)
30. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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√ 104
. yp ,h
10 , opp.
Trigonometry Sec. 01
θ 2 , adj.
sin(θ) =
opp 10 = √ hyp 104
cos(θ) =
2 adj = √ hyp 104
tan(θ) =
opp 10 = adj 2
31. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
√ 45
sin(θ) =
opp 6 =√ hyp 45
cos(θ) =
3 adj =√ hyp 45
tan(θ) =
6 opp = adj 3
6 , opp.
. yp ,h
θ 3 , adj.
32. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
,
sin(θ) = 1 , opp.
2
p. hy
√ adj 3 = cos(θ) = hyp 2
θ √ 3 , adj.
opp 1 = hyp 2
tan(θ) =
1 opp = √ adj 3
33. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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√ 2
. yp ,h
sin(θ) =
opp 1 =√ hyp 2
cos(θ) =
1 adj =√ hyp 2
tan(θ) =
opp 1 = adj 1
1 , opp.
Trigonometry Sec. 01
θ 1 , adj.
34. Find the three ratios, sin(θ), cos(θ), and tan(θ)
5
. yp ,h
0 , opp.
(May not be drawn to scale.)
θ 5 , adj.
sin(θ) =
opp 0 = hyp 5
cos(θ) =
5 adj = hyp 5
tan(θ) =
opp 0 = adj 5
35. Find the three ratios, sin(θ), cos(θ), and tan(θ)
4
. yp ,h
6x
x2 , opp.
(May not be drawn to scale.)
sin(θ) =
opp x2 = 4 hyp 6x
cos(θ) =
x adj = 4 hyp 6x
tan(θ) =
opp x2 = adj x
θ x , adj.
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Trigonometry Sec. 02
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INVERSE FUNCTIONS The Idea Now we turn our attention to solving for the angles on a right triangle. We also take the opportunity to emphasize what it means for us to completely solve a triangle. We see triangles as 6 pieces of information, 3 sides and 3 corresponding angles. To solve the triangle will mean for us to determine all 6 items, the 3 sides and the 3 angles. The last section, using sin(θ), cos(θ) and tan(θ), provided for us all the ratios for all the triangles, and using the ratios and just one side, we can easily solve all sides. That was a a major milestone, and we are now ready to complete the solving of right triangles challenge by learning to solve for angles. The Inverse Functions
?
opp ratio to find hyp we used the ratio sin(23◦ )
23◦
hyp
×0.3907
opp
hyp
opp
To solve for angles, we introduce the idea of inverse trig functions. Let us take an example from the past. For a 23-67 triangle we used the sin(23◦ ) to determine the opp/hyp ratio, approximately .3907.
23◦ adj
adj
×0.3907
opp ratio is If the hyp known, 0.3907, we use sin−1 (0.3907)
??
hyp
×0.3907
opp
hyp
opp
The question we now contemplate is what if we knew the ratio, but not the angle?. How then could we determine the angles. One primitive approach would be to use a protractor, assuming we can draw such a triangle to correct scale, a protractor would help us get a working approximation of the angle. But there is more. A more accurate answer is to use the sine inverse function. The sine inverse function take ratios as input values and it gives angles as output values. The sine inverse function is often written as sin−1 (k), where k is the ratio for which we want to find the angle. The sine inverse function1 is sometimes also known as the arc-sine function and written as arcsin(k) both names and notations are common and most scientific calculators have these functions built-in. You should become familiar with such functions on your calculator [also with the radians-degree settings on the calculator]. While we are on the calculator topic, while you use the calculator to compute these values, your self-conscious may feel a bit uneasy, it should feel as though you are not really learning how or where these values come from. You should embrace such feelings as these will foster an enduring anticipation for the day when you really learn where and how these values are computed. For now we make do with our calculators and at least begin to understand what these inverse functions do for us.
23◦ adj
adj
Here is another look at the a similar problem. There is a famous triangle going around, the 3, 4, 5 triangle. It happens to be a right triangle, and the sides are famous with 32 + 42 = 52 . The sides are famous, but here we ask, what are the angles? Have you ever wondered? 1 this
definition will be extended to a broader sense later
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math hands
pg. 1
Trigonometry Sec. 02
MathHands.com M´ arquez
5
??
4
×4
5
5
×4
4
5
opp If the hyp ratio is 4 known, 5 , we use sin−1 45
53.13◦ 3
3
But wait, there is more. Now we can extend these ideas to other ratios. For example if we know the can use the arc-cosine function also known as the inverse cosine to compute the angle. For example:
0. 95
18◦
11
hyp
×
0. 92
adj
ratio, we
opp
×
??
opp
hyp
adj ratio is If the hyp known, 0.9511, we use cos−1 (0.9511)
adj hyp
05
adj
Suppose we knew the sides, adjacent and hypothenuse, then we compute the ratio, then the angle using the cosine inverse function.
hy p, 10
hy p, 10
adj ratio is If the hyp 7 , known, 10 we use 7 cos−1 10
× 10
7
7
× 10 ??
45.57◦ adj, 7
adj, 7
adj, 7
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2
×7
15.94◦
opp, 2
2
×7
??
opp, 2
Similarly for arctangent, if we know the ratio, either by knowing the ratio or by knowing the sides, opp & adj, we can find the respective angle. If the opp adj ratio is known, 27 , we use tan−1 27
adj, 7
math hands
pg. 2
Trigonometry Sec. 02
MathHands.com M´ arquez
1. Find the angle in question.
2
1
8
?? ?? 10
7.
2. Find the angle in question.
8. Find the angle in question.
?? 6
4
12
?? 4
3. Find the angle in question.
9. Find the angle in question.
?? 12
1
8
?? 4
4. Find the angle in question.
10. Find the angle in question.
??
??
13
11
1
7
5. Find the angle in question.
?? 9
2
10
11. Find the angle in question.
?? 5
6. Find the angle in question.
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12. Find the angle in question.
math hands
pg. 3
Trigonometry Sec. 02
MathHands.com M´ arquez
1
2
??
?? 10
8
5
18. Find the angle in question.
2
13. Find the angle in question.
??
?? 15
8
14. Find the angle in question.
19. Find the angle in question.
5
15
2
?? 8
??
15. Find the angle in question.
20. Find the angle in question.
7
4
?? 12
?? 11
16. Find the angle in question.
21. Find the angle in question.
4
15
??
?? 2
17. Find the angle in question.
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12
22. Find the angle in question.
math hands
pg. 4
Trigonometry Sec. 02
MathHands.com M´ arquez ?? 7
2
11
?? 15
23. Find the angle in question. 25. Find the angle in question. ??
5
5
?? 8
8
24. Find the angle in question.
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pg. 5
Trigonometry Sec. 02
MathHands.com M´ arquez
INVERSE FUNCTIONS : SOLUTIONS
1. To find the angle in question:
6. To find the angle in question:
, 8 hyp
opp, 2
= 0.2 so we The ratio −1 use tan (0.2)
×0
11.31◦
.2
opp, 1
opp = 0.125 so The ratio hyp −1 we use sin (0.125)
opp adj
×0.125
7.18◦
adj, 10
7.
2. To find the angle in question:
8. To find the angle in question: 41.81
, 6 hyp
◦
opp The ratio hyp = 0.3333 so −1 we use sin (0.3333)
×0
2 , 1 hyp
×0.3333
.66
opp, 4
opp The ratio hyp = 0.6667 so −1 we use sin (0.6667)
67
19.47◦
opp, 4
3. To find the angle in question: 9. To find the angle in question:
adj The ratio hyp = 0.125 so −1 we use cos (0.125)
82.82◦
adj, 1
, 8 hyp
adj = 0.3333 so The ratio hyp we use cos−1 (0.3333)
×0.125
2 , 1 hyp
×0 .33 33
70.53◦
adj, 4
4. To find the angle in question: opp = 0.0769 so The ratio hyp −1 we use sin (0.0769)
10. To find the angle in question:
4.41◦
3 , 1 hyp
opp = 0.6364 so The ratio hyp −1 we use sin (0.6364)
1 , 1 hyp
64
69
.63
.07
×0
×0 opp, 1
opp, 7
5. To find the angle in question:
11. To find the angle in question: opp = 0.5556 so The ratio hyp −1 we use sin (0.5556)
×0.2
, 9 hyp
opp, 2
opp = 0.2 so we The ratio hyp −1 use sin (0.2)
0 , 1 hyp
39.52◦
33.75◦
×0 .55 56
11.54◦
opp, 5
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pg. 1
Trigonometry Sec. 02
MathHands.com M´ arquez
The ratio opp adj = 5 so we use tan−1 (5)
opp, 1
The ratio opp adj = 0.125 so we use tan−1 (0.125)
12. To find the angle in question:
adj, 2
78.69◦
7.13
×5
×0
◦
5 .12
adj, 8
opp, 10
18. To find the angle in question: 13. To find the angle in question:
×0
7.59◦
.1
333
opp, 2
The ratio opp adj = 0.1333 so we use tan−1 (0.1333)
×0
32.01◦
opp, 5
The ratio opp adj = 0.625 so we use tan−1 (0.625)
5 .62
adj, 8
adj, 15
19. To find the angle in question: adj = 0.1333 so The ratio hyp we use cos−1 (0.1333)
opp hyp −1
= 0.625 so (0.625) , 8 hyp
5 , 1 hyp
opp, 5
The ratio we use sin
×0.625
82.34◦ ×0.1333
adj, 2
14. To find the angle in question:
38.68◦ 20. To find the angle in question:
15. To find the angle in question:
70.53◦ ×0.3333
adj, 4
2 , 1 hyp
32.47
×0
◦
6 .63
4
opp, 7
The ratio opp adj = 0.6364 so we use tan−1 (0.6364)
adj The ratio hyp = 0.3333 so we use cos−1 (0.3333)
adj, 11
adj = 0.1333 so The ratio hyp −1 we use cos (0.1333)
21. To find the angle in question: The ratio opp adj = 0.3333 so we use tan−1 (0.3333)
5 , 1 hyp
×0
18.43
.13 33
82.34
◦
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3 .33
3
adj, 12
adj, 2
17. To find the angle in question:
×0
◦
opp, 4
16. To find the angle in question:
22. To find the angle in question:
math hands
pg. 2
Trigonometry Sec. 02
MathHands.com M´ arquez
1 , 1 hyp
adj, 2
82.41◦
opp = 0.6364 so The ratio hyp −1 we use sin (0.6364)
×7
×0.6364
opp, 7
The ratio opp adj = 7.5 so we use tan−1 (7.5)
.5 39.52◦
opp, 15
23. To find the angle in question: 25. To find the angle in question:
adj, 5
57.99◦
×1
adj = 0.625 so The ratio hyp we use cos−1 (0.625)
, 8 hyp
.6
51.32◦ ×0.625
adj, 5
The ratio opp adj = 1.6 so we use tan−1 (1.6)
opp, 8
24. To find the angle in question:
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pg. 3
Trigonometry Sec. 03
MathHands.com M´ arquez
MISSION ACCOMPLISHED The Idea We have done it! At this point, we have developed enough ideas to fulfill one of the fundamental promises of trigonometry, to be able to measure things by simply looking at them. That is, roughly speaking, we now have enough tools to be able to measure distances by looking at a couple points, measuring one or a couple angles and a side or a piece of a side. This is one of the essential goals of trigonometry, and we are ready to declare this mission accomplished. We introduce no new ideas here, rather, we simply show a few examples of how to use the ideas we have developed. Look at a tree and measure it: Example A: In this example, suppose we want to measure the height of a palm tree. Furthermore suppose we complete access to the area around the palm tree. We start at the trunk and walk away from the palm tree 20 ft. We lay down look up, and there it is, we determine the hight of the palm-tree.
To solve this, we need to do more than just look at the tree. We need to measure the angle from the ground to the top of the tree. Moreover, we make some extraordinary assumptions. We assume, for example that the ground is perfectly flat and the tip of the palm-tree lies exactly over the trunk, etc, etc. ... We then turn the unknown quantity, the hight of the palm-tree, into the side of a right triangle. Suppose upon laying down and looking at the tree we measure and angle of 35◦ . Then we note:
tree
p hy
35◦ 20ft.
We then invoke our ratio functions, in this case, we know the side adjacent to 35◦ and want the side opposite. We recall the tangent function is taylor made to describe such ratio, opp/adj. We make note of it on the diagram to obtain:
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pg. 1
Trigonometry Sec. 03
MathHands.com M´ arquez
×
35◦
f irst, we note tan(35◦ ) ≈ 0.7002
tree
p hy
0 0.7
then, tree ≈ (20f t.)(0.7002)
02
≈ 14f t.
20ft.
Look at a tree and measure it: Example B: Suppose we change things just a little. This time we don’t even come near the trunk of the palm-tree. This time we simply glance from far away measure the angle and then determine the height of the palm-tree. This time we make use of a fence directly next to the tree. It just so happens that we know this type of fence is exactly 6 ft tall and at exactly eye level. We measure the following angles, from that we can determine the heigh of the tree.
15◦ 5◦
Solutions: We will solve this by looking at the diagram, assuming we have two right triangles, and solving these two triangles. Let us look at the bottom triangles first.
15◦ 5◦
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6 ft.
math hands
pg. 2
Trigonometry Sec. 03
MathHands.com M´ arquez
We use the tangent function to determine the opp/adj ratio, then determine the adj side. In this step we effectively figure out how far away from the palm-tree we are standing. Note tan(5◦ ) ≈ 0.0875
15◦
68.58 ft. 5◦
×0.08
75
6 ft.
We now turn our attention to the other right triangle. We find have the adjacent side, want the opposite side, thus we use the tangent function tan(15◦ ) ≈ 0.2679 Thus...
”AF”
.2 ×0 15◦
679
68.58 ft. 5◦
×0.08
75
6 ft.
So we have the palm-tree divided into two sections the section below the fence and above the fence. The fence is 6ft. tall so we need only find the section above the fence, let us call it ”AF” for above fence. Then... AF ≈ (68.58)(0.2679) ≈ 18.38f t.
total tree ≈ 24.38 f t. Look at a tree and measure it: Example C:
That as a nice challenge for about 2 minutes. We now make it a bit more interesting. What if there was not fence and we could not come close to the palm-tree? It turns out we can again overcome this challenge. We simply measure an angle from far away, as illustrated below, then we move away some fixed distance, such as 8 ft., we measure the angle again, and voila! done! we can then tell the hight of the tree.
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pg. 3
Trigonometry Sec. 03
MathHands.com M´ arquez
12◦ 8 f t.
15◦
solutions: To solve this we will look for two right triangles. Note we denote the hight of the palm-tree as tree, and the initial distance to the tree as x. Then we analyze the inner triangle.
tree = (x)(0.2679) 79 .26 ×0 12◦ 8 f t.
15◦ x f t.
Then the outer triangle.
tree = (x + 8)(0.2126) 26 .21 ×0
12◦ 8 f t.
15◦ x f t. x+8
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pg. 4
Trigonometry Sec. 03
MathHands.com M´ arquez
We now have two expression for ”tree”. On the one hand tree = (x)(0.2679) while on the other hand, tree = (x + 8)(0.2126) . We now set these equal to each other and solve for x, the first distance to the tree, then ultimately solve for the height of the tree.
(x + 8)(0.2126) = (x)(0.2679) x(0.2126) + (8)(0.2126) = x(0.2679) (8)(0.2126) = x(0.2679) − x(0.2126) (8)(0.2126) = x(0.2679 − 0.2126) 1.7005 = x(0.0554) 1.7005 =x 0.0554 x ≈ 30.6949 f t.
and finally, tree = (x)(0.2679) = (30.6949)(0.2679) ≈ 8.2247 f t.
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pg. 5
Trigonometry Sec. 03
MathHands.com M´ arquez
1. Assume upon walking 20 ft. you lay down and measure the angle to the top of the tree to be 35◦ . Find the height of the tree.
(You may also assume the line of sight, bottom of the tree and top of the tree form a right triangle. ) 2. Suppose the fence as well as eye-level are 6 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
11◦ 4◦
3. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
12◦ 8 f t.
15◦
4. Suppose you know the height of your neighbor’s palm treed is 17 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
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pg. 6
Trigonometry Sec. 03
MathHands.com M´ arquez
75◦
5◦
5. Assume upon walking 30 ft. you lay down and measure the angle to the top of the tree to be 19◦ . Find the height of the tree.
(You may also assume the line of sight, bottom of the tree and top of the tree form a right triangle. ) 6. Suppose the fence as well as eye-level are 5.5 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
9◦ 3◦
7. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
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pg. 7
Trigonometry Sec. 03
MathHands.com M´ arquez
16◦ 10 f t.
18◦
8. Suppose you know the height of your neighbor’s palm treed is 20 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
69◦
3◦
9. Solve for the angle θ
5 in. θ
12 in.
2 in.
10. How long is the belt?
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pg. 8
Trigonometry Sec. 03
MathHands.com M´ arquez
5 in. θ
r
11.
x
37◦
15 x
50◦
r
13.
17 x
23◦
r
14.
49◦
2 in.
12
r
12.
12 in.
8 x
15. Just by looking at him, can you figure out how big Diego’s head is? See picture for angles & measurements.
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pg. 9
Trigonometry Sec. 03
MathHands.com M´ arquez
40◦
49◦
2 in.
16. Determine the distance from the tree trunk to the opposite side of the river as shown.
river 17◦
25◦
10 yd.
17. (a) Determine the distance from the tree trunk to the opposite side of the river as shown.
river 20◦
23◦
5 yd. (b) (con’t from previous exercise) How far away is the tree from the river? Use the diagram below.
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pg. 10
Trigonometry Sec. 03
MathHands.com M´ arquez
d
river 18◦
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pg. 11
Trigonometry Sec. 03
MathHands.com M´ arquez
MISSION ACCOMPLISHED : SOLUTIONS 1. Assume upon walking 20 ft. you lay down and measure the angle to the top of the tree to be 35◦ . Find the height of the tree.
f irst, we note tan(35◦ ) ≈ 0.7002 then,
tree
p hy
×0
35◦
.7
tree ≈ (20f t.)(0.7002) ≈ 14f t.
2 00
20ft.
We then invoke our ratio functions, in this case, we know the side adjacent to 35◦ and want the side opposite. We recall the tangent function is taylor made to describe such ratio, opp/adj. We make note of it on the diagram to obtain: 2. Suppose the fence as well as eye-level are 6 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
”AF”
.1 ×0 11◦
944
85.8 ft. 4
◦
×0.06
99
6 ft.
First we find tan(4◦ ) ≈ 0.0699 and we use the heigh of the fence 6 ft. to find the distance to the tree. Then we use the tan(11◦ ) ≈ 0.1944 ration to go from distance to the tree to distance above the fence ”AF”... thus.. AF ≈ (85.8 f t.)(0.1944) ≈ 16.68f t.
total tree ≈ 22.68 f t. 3. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
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pg. 1
Trigonometry Sec. 03
MathHands.com M´ arquez
tree = (x + 8)(0.2126)
26 79 .21 26 ×0 0. ×
12◦ 8 f t.
15◦ x f t.
tree = (x)(0.2679) then set equal and solve (x)(0.2679)=(x + 8)(0.2126) x ≈ 30.6949 f t. and tree ≈ 8.2247f t.
x+8
4. Suppose you know the height of your neighbor’s palm treed is 17 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
75◦
5◦
distance to tree = 17 ÷ tan(5◦ ) ≈ 194.3109 f t. then.. use the outer triangle...
height of the tree = (194.3109)(tan(75◦ )) ≈ 725.1782 f t. 5. Assume upon walking 30 ft. you lay down and measure the angle to the top of the tree to be 19◦ . Find the height of the tree.
tree
p hy
.3 ×0
19◦
44
3
f irst, we note tan(19◦ ) ≈ 0.3443
then, tree ≈ (30f t.)(0.3443) ≈ 14f t.
30ft.
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pg. 2
Trigonometry Sec. 03
MathHands.com M´ arquez
We then invoke our ratio functions, in this case, we know the side adjacent to 19◦ and want the side opposite. We recall the tangent function is taylor made to describe such ratio, opp/adj. We make note of it on the diagram to obtain: 6. Suppose the fence as well as eye-level are 5.5 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
.1 ×0 9◦
584
104.95 ft. 3
◦
×0.05
24
First we find tan(3◦ ) ≈ 0.0524 and we use the ”AF” heigh of the fence 5.5 ft. to find the distance to the tree. Then we use the tan(9◦ ) ≈ 0.1584 ration to go from distance to the tree to distance above the fence ”AF”... thus.. 5.5 ft.
AF ≈ (104.95 f t.)(0.1584)
≈ 16.62f t. total tree ≈ 22.12 f t. 7. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
tree = (x + 10)(0.2867)
67 49 .28 32 ×0 0. ×
16◦ 10 f t.
18◦ x f t.
tree = (x)(0.3249) then set equal and solve (x)(0.3249)=(x + x ≈ 10)(0.2867) 75.0654 f t. and tree ≈ 24.3902f t.
x+8
8. Suppose you know the height of your neighbor’s palm treed is 20 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
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pg. 3
Trigonometry Sec. 03
MathHands.com M´ arquez
69◦
3◦
distance to tree = 17 ÷ tan(3◦ ) ≈ 381.6227 f t. then.. use the outer triangle...
height of the tree = (381.6227)(tan(69◦ )) ≈ 994.1611 f t. 9. Solve for the angle θ
5 in.
θ
12 in.
2 in.
Solution: to solve for θ we could look at it as one of the angles on a right triangle, by drawing a line parallel to the top portion of the belt between the pulleys. the difference in radii is 3, this is one of the sides, the hypothenuse is 12, thus cos θ =
3 12
, from here we get θ = cos−1 (3/12) ≈ 75.5◦ 10. How long is the belt?
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pg. 4
Trigonometry Sec. 03
MathHands.com M´ arquez
5 in.
θ
12 in.
2 in.
Solution:
once the angle is found.. you should proceed to find the belt one piece at a time.. to find the portions around d the pulleys you may need that s = πθ (in radians) or s = 360 · 2πr in degrees.. where s is the arc-length...
r x
37◦
11.
12
Solution: tan 37◦ = x=
12 tan 37◦
also.. sin 37◦ =
12 r
Therefore.. r =
15 tan 50◦
sin 50◦ =
≈ 19.94
15
Solution: tan 50◦ = also..
12 sin 37◦
x
50◦
x=
Therefore..
≈ 15.92
r 12.
12 x
15 x
Therefore..
≈ 12.59
15 r
Therefore.. r =
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15 sin 50◦
≈ 19.58
math hands
pg. 5
Trigonometry Sec. 03
MathHands.com M´ arquez
r x
23◦
13.
17
Solution: tan 23◦ = x=
17 tan 23◦
also.. sin 23◦ =
17 r
Therefore.. r =
≈ 43.51
8
Solution: tan 49◦ = 8 tan 49◦
also.. sin 49◦ =
17 sin 23◦
x
49◦
x=
Therefore..
≈ 40.05
r 14.
17 x
8 r
8 x
Therefore..
≈ 6.95 Therefore.. r =
8 sin 49◦
≈ 10.60
15. Just by looking at him, can you figure out how big Diego’s head is? See picture for angles & measurements.
40◦
49◦
2 in.
Solution: We first call the size of his head, h and the lower side of the inner triangle, k. Then, we look at the inner triangle to conclude h tan 49◦ = k h , this means that k = tan 49◦ . Similarly, we look at the outer triangle to conclude tan 40◦ =
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h 2+k
pg. 6
Trigonometry Sec. 03
MathHands.com M´ arquez
this means 2 + k = tanh40◦ Or k = −2 + then solve for h, the sought quantity.
h tan 40◦ .
Having solved for k in two different ways, we set the two equal,
h h = −2 + tan 49◦ tan 40◦ then... we move all h terms to one side factor the h and solve...... h ≈ 6.202 in.
16. Determine the distance from the tree trunk to the opposite side of the river as shown.
river
17◦
25◦
10 yd.
Solution: We first call the sought distance, h and the lower side of the inner triangle, k. Then, we look at the inner triangle to conclude h tan 25◦ = k , this means that k =
h tan 25◦ .
Similarly, we look at the outer triangle to conclude tan 17◦ =
h 10 + k
this means 10 + k = tanh17◦ Or k = −10 + tanh17◦ . Having solved for k in two different ways, we set the two equal, then solve for h, the sought quantity. h h = −10 + ◦ tan 25 tan 17◦ then... we move all h terms to one side factor the h and solve...... h ≈ 8.878 yd.
17. (a) Determine the distance from the tree trunk to the opposite side of the river as shown.
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pg. 7
Trigonometry Sec. 03
MathHands.com M´ arquez
river
20◦
23◦
5 yd.
Solution: We first call the sought distance, h and the lower side of the inner triangle, k. Then, we look at the inner triangle to conclude h tan 23◦ = k , this means that k =
h tan 23◦ .
Similarly, we look at the outer triangle to conclude tan 20◦ =
h 5+k
this means 5 + k = tanh20◦ Or k = −5 + tanh20◦ . Having solved for k in two different ways, we set the two equal, then solve for h, the sought quantity. h h = −5 + tan 23◦ tan 20◦ then... we move all h terms to one side factor the h and solve...... h ≈ 12.767 yd. (b) (con’t from previous exercise) How far away is the tree from the river? Use the diagram below.
d river
18
◦
Solution: from the above we can determine h, the distance from the trunk to the opposite side of the river. h ≈ 12.767 yd. If we can determine the width of the river we can subtract the quantities and obtain the distance, d. To figure out the width of the river, we can calculate k, the lower side of the triangle, from part a)... tan 23◦ =
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h k
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Trigonometry Sec. 03 thus... k =
MathHands.com M´ arquez h tan 23◦
≈
12.767 tan 23◦ ◦
≈ 30.111
Then, we can use tan 18 ≈
′
river ′ , k
so.. river′ k river = k ∗ tan 18◦ river ≈ 30.111 ∗ 0.325 ′
tan 18◦ =
river ≈ 9.786 ..etc...etc.. solve for ’river’ and subtract from h to get
”trunk − to − river” = h − river ≈ 12.767 − 9.786 ≈ 2.981
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Trigonometry Sec. 04
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Reference Triangles The MAIN IDEA: The last couple sections are the cornerstone of trigonometry. These established the foundation for this chapter and for essential trigonometric ideas. A deep and meaningful understanding the trigonometry will depend on a deep and profound understanding of the definitions of the trigonometric functions. In the last section, we introduced these functions. We described them as the ratios of sides of a right triangle. We summarize these ideas with this:
SOH-CAH-TOA
p hy
× sin(θ)
θ os( ×c
×
)
θ
n( ta
Sin(θ) =
Opp Hyp
Cos(θ) =
Adj Hyp
T an(θ) =
Opp Adj
opp
Generic Triangle Ratios
θ)
adj
With these functions and our ability to calculate them using some calculator, we could effectively made good on one of the essential promises of trigonometry, to be able solve any right triangle given enough information. However, once this is accomplished, the trig functions take a life of their own. They become important in their own right, and the become in some sense bigger than the original problem they were made to solve. In this section, we turn such corner, we begin to study the trig function, not so much to solve right triangles, but simply to study their properties and behavior. The usefulness and implications of these functions are much bigger than this course. We will be content, and we will feel fortunate just to get to know these function just a tiny little bit. For example, at this point we are well versed with the meaning of sin(23◦ ) as the ratio of opp/hyp on the corresponding right triangle, we never did ponder the meaning of something like sin(275◦ ) In fact, it may seem a little wild to utter such words since our right triangles usually contained angles measuring less than 90◦ . Said differently, we defined our functions for little angles, angles between 0 and 90◦ , but as implied earlier, the function grew to mean much bigger and deeper ideas than that. Specifically, we now begin the study of reference triangles which will eventually lead to the extension of the trig functions to angles larger than 90◦ or even negative angles. Moreover, this will be consistent with the ideas we have thus far. Talk is cheap! Let us begin to define the very important concept of a reference triangle and the idea that every angle has one [at least]. Without further ado, here is the definition of a reference triangle. EVERY ANGLE has a REFERENCE TRIANGLE: Given any angle, θ: 1. We choose a line segment of any positive length, r, and place the segment one endpoint on origin of the xy − plane, the other endpoint on the positive side of the x-axis. c
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Trigonometry Sec. 04
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2. Rotate the segment, maintaining the endpoint at the origin fixed, θ units (degrees or radians). Rotate positive angles counterclockwise, rotate negative angles clockwise. 3. From the endpoint of the segment not on the origin, draw a perpendicular to the x-axis. This will create a triangle or an almost triangle. This is called and defined to be the reference triangle for angle θ 4. The fourth step is to label the sides of the reference triangle, including signs where appropriate. This means, up is positive down is negative, right positive, left is negative, as usual, for now we will keep the r positive. In later chapters, we will have a meaningful interpretation for negative r’s. One should note this last step will be possible, with our current level of knowledge, only for famous angles, such as 0◦ , 30◦ , 45◦ , 60◦ , etc...
STEP 1. Choose any positive length, r
STEP 3. Draw a Perp to x-axis
STEP 2. Rotate θ
y
y 2
r=2
STEP 4. Label sides/signs
y positive angles
θ
2
y
x
2
?
θ x
θ x
?
negative angles
Examples: Example: Draw the reference triangle for 150◦ with a line segment of length 2.
STEP 1. Choose any positive length, r STEP 2. Rotate 150◦
y
STEP 3. Draw a Perp to x-axis
y 2
r=2
STEP 4. Label sides/signs
y positive angles
θ
2
y 1
150◦
x
2
150◦
√ − 3
x negative angles
x
Example: Draw the reference triangle for 150◦ with a line segment of length 10.
STEP 1. Choose any positive length, r STEP 2. Rotate 150◦
y
STEP 3. Draw a Perp to x-axis
y 10
r=2
y positive angles
θ
x
10
y 5
150◦ x
negative angles
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STEP 4. Label sides/signs
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10
150◦
√ −5 3
x
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Trigonometry Sec. 04
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Draw the reference triangle for 60◦ . Use a line segment of length 4 units, and do not use calculators.
STEP 1. Choose any positive length, r STEP 2. Rotate 60◦
r=4
STEP 4. Label sides/signs
y
60◦
4
y 4
y 4
y
STEP 3. Draw perp. to x
60◦
√ 2 3
60◦
x
2
Draw the reference triangle for 225◦. Use a line segment of length 10 units, and do not use calculators.
STEP 1. Choose any positive length, r STEP 2. Rotate 225◦
y
STEP 3. Draw perp. to x
y
r = 10
y 225◦
y −10 √ 2
225◦ −10 √ 2
225◦
10
10
10
x
STEP 4. Label sides/signs
Finally, it should be noted that the hypothenuse is always, by definition kept positive, and that it’s length can be any positive real number, thus in reality every angle has many reference triangles, depending on how big the initial segment is drawn.
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1. Draw and label the reference triangle for 210◦ , using a segment of length 5 , and no calculators. y
x
2. Draw and label the reference triangle for 330◦ , using a segment of length 10 , and no calculators. y
x
3. Draw and label the reference triangle for 30◦ , using a segment of length 15 , and no calculators.
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Trigonometry Sec. 04
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x
4. Draw and label the reference triangle for 150◦ , using a segment of length K , and no calculators. y
x
5. Draw and label the reference triangle for −150◦ , using a segment of length R , and no calculators.
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Trigonometry Sec. 04
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x
6. Draw and label the reference triangle for 120◦ , using a segment of length 8 , and no calculators. y
x
7. Draw and label the reference triangle for −120◦ , using a segment of length 12 , and no calculators.
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Trigonometry Sec. 04
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x
8. Draw and label the reference triangle for 390◦ , using a segment of length 9 , and no calculators. y
x
9. Draw and label the reference triangle for −210◦ , using a segment of length 14 , and no calculators.
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Trigonometry Sec. 04
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x
10. Draw and label the reference triangle for −60◦ , using a segment of length 6 , and no calculators. y
x
11. Draw and label the reference triangle for 750◦ , using a segment of length 2 , and no calculators.
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Trigonometry Sec. 04
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x
12. Draw and label the reference triangle for −120◦ , using a segment of length 1 , and no calculators. y
x
13. Draw and label the reference triangle for −30◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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x
14. Draw and label the reference triangle for 45◦ , using a segment of length 7 , and no calculators. y
x
15. Draw and label the reference triangle for −135◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
16. Draw and label the reference triangle for 135◦ , using a segment of length 10 , and no calculators. y
x
17. Draw and label the reference triangle for −45◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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x
18. Draw and label the reference triangle for 225◦ , using a segment of length 10 , and no calculators. y
x
19. Draw and label the reference triangle for −225◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
20. Draw and label the reference triangle for 495◦ , using a segment of length 7 , and no calculators. y
x
21. Draw and label the reference triangle for 315◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
22. Draw and label the reference triangle for −315◦ , using a segment of length 10 , and no calculators. y
x
23. Draw and label the reference triangle for 405◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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x
24. Draw and label the reference triangle for −405◦ , using a segment of length 10 , and no calculators. y
x
25. Draw and label the reference triangle for 675◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
26. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
x
27. Draw and label the reference triangle for 180◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
28. Draw and label the reference triangle for 450◦ , using a segment of length 7 , and no calculators. y
x
29. Draw and label the reference triangle for −270◦ , using a segment of length 3 , and no calculators.
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Trigonometry Sec. 04
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x
30. Draw and label the reference triangle for −90◦ , using a segment of length 8 , and no calculators. y
x
31. Draw and label the reference triangle for 0◦ , using a segment of length 8 , and no calculators.
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Trigonometry Sec. 04
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x
32. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
x
33. Draw and label the reference triangle for −450◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
34. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
x
35. Draw and label the reference triangle for 720◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
36. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
x
37. Draw and label the reference triangle for −630◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
38. Draw and label the reference triangle for 360◦ , using a segment of length 5 , and no calculators. y
x
39. Draw and label the reference triangle for −360◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
40. Draw and label the reference triangle for −810◦ , using a segment of length 5 , and no calculators. y
x
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Trigonometry Sec. 04
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Reference Triangles : SOLUTIONS 1. Draw and label the reference triangle for 210◦ , using a segment of length 5 , and no calculators. y
210◦
√ − 52 · 3 − 25
x
5
Solution:
2. Draw and label the reference triangle for 330◦ , using a segment of length 10 , and no calculators. y
330◦ 10 2
·
√ 3
10
x − 10 2
Solution:
3. Draw and label the reference triangle for 30◦ , using a segment of length 15 , and no calculators.
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Trigonometry Sec. 04
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15 2
15 30◦ 15 2
·
√ 3
x
Solution:
4. Draw and label the reference triangle for 150◦ , using a segment of length K , and no calculators. y
K 2
K 150◦
−K 2
√ · 3
x
Solution:
5. Draw and label the reference triangle for −150◦ , using a segment of length R , and no calculators.
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Trigonometry Sec. 04
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− R2 · − R2
√
−150◦
3
x
R
Solution:
6. Draw and label the reference triangle for 120◦ , using a segment of length 8 , and no calculators. y
8 2
·
√
3
8
120◦
− 82
x
Solution:
7. Draw and label the reference triangle for −120◦ , using a segment of length 12 , and no calculators.
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Trigonometry Sec. 04
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−120◦
x
− 12 2
− 12 2 ·
√ 3
12
Solution:
8. Draw and label the reference triangle for 390◦ , using a segment of length 9 , and no calculators. y
9 2
9 390◦ 9 2
·
√
3
x
Solution:
9. Draw and label the reference triangle for −210◦ , using a segment of length 14 , and no calculators.
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Trigonometry Sec. 04
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14 2
14
− 14 2
−210◦
√ · 3
x
Solution:
10. Draw and label the reference triangle for −60◦ , using a segment of length 6 , and no calculators. y
−60◦
x
6 2
6
− 62 ·
√ 3
Solution:
11. Draw and label the reference triangle for 750◦ , using a segment of length 2 , and no calculators.
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Trigonometry Sec. 04
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2 2
2 750◦ 2 2
·
√
3
x
Solution:
12. Draw and label the reference triangle for −120◦ , using a segment of length 1 , and no calculators. y
−120◦
− 12
− 12 ·
√ 3
x
1
Solution:
13. Draw and label the reference triangle for −30◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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−30◦ 7 2
·
√
3
x − 27
7
Solution:
14. Draw and label the reference triangle for 45◦ , using a segment of length 7 , and no calculators. y
7
√7 2
45◦ √7 2
x
Solution:
15. Draw and label the reference triangle for −135◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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−135◦
10 −√ 2
10 −√ 2
x
10
Solution:
16. Draw and label the reference triangle for 135◦ , using a segment of length 10 , and no calculators. y
10 √ 2
10
135◦ 10 −√ 2
x
Solution:
17. Draw and label the reference triangle for −45◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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−45◦
√7 2
7
x
− √72
Solution:
18. Draw and label the reference triangle for 225◦ , using a segment of length 10 , and no calculators. y
225◦ 10 −√ 2
10 −√ 2
x
10
Solution:
19. Draw and label the reference triangle for −225◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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10 √ 2
10
−225◦
10 −√ 2
x
Solution:
20. Draw and label the reference triangle for 495◦ , using a segment of length 7 , and no calculators. y
√7 2
7
495◦
− √72
x
Solution:
21. Draw and label the reference triangle for 315◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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315◦ 10 √ 2
x
10
10 −√ 2
Solution:
22. Draw and label the reference triangle for −315◦ , using a segment of length 10 , and no calculators. y
10
−315◦
10 √ 2
10 √ 2
x
Solution:
23. Draw and label the reference triangle for 405◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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7
√7 2
405◦ √7 2
x
Solution:
24. Draw and label the reference triangle for −405◦ , using a segment of length 10 , and no calculators. y
−405◦
10 √ 2
x
10
10 −√ 2
Solution:
25. Draw and label the reference triangle for 675◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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675◦ 10 √ 2
x
10
10 −√ 2
Solution:
26. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
≈ 270◦
≈0
x
5−5
Solution:
27. Draw and label the reference triangle for 180◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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≈0
≈ 180◦
5 −5
x
Solution:
28. Draw and label the reference triangle for 450◦ , using a segment of length 7 , and no calculators. y
77
≈ 450◦
≈0
x
Solution:
29. Draw and label the reference triangle for −270◦ , using a segment of length 3 , and no calculators.
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Trigonometry Sec. 04
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33
≈ −270◦
≈0
x
Solution:
30. Draw and label the reference triangle for −90◦ , using a segment of length 8 , and no calculators. y
≈ −90◦
≈0
x
8−8
Solution:
31. Draw and label the reference triangle for 0◦ , using a segment of length 8 , and no calculators.
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Trigonometry Sec. 04
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≈ 0◦
8 8
x
≈0
Solution:
32. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
≈ 270◦
≈0
x
5−5
Solution:
33. Draw and label the reference triangle for −450◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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≈ −450◦
≈0
x
5−5
Solution:
34. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
≈0
5 −5
≈ −540◦
x
Solution:
35. Draw and label the reference triangle for 720◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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≈ 720◦
5 5
x
≈0
Solution:
36. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
≈0
5 −5
≈ −540◦
x
Solution:
37. Draw and label the reference triangle for −630◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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55
≈ −630◦
x
≈0
Solution:
38. Draw and label the reference triangle for 360◦ , using a segment of length 5 , and no calculators. y
≈ 360◦
5 5
x
≈0
Solution:
39. Draw and label the reference triangle for −360◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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≈ −360◦
5 5
x
≈0
Solution:
40. Draw and label the reference triangle for −810◦ , using a segment of length 5 , and no calculators. y
≈ −810◦
≈0
x
5−5
Solution:
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Trigonometry Sec. 05
MathHands.com M´ arquez
The Rest of the Trig Functions THE REST of the SINE and COSINE Functions: This section is yet another immensely critical milestone in our journey to understanding all that is trigonometry. It is here and now that we extend these functions for angles larger than 90◦ or smaller than 0◦ (when possible). We will learn how to evaluate them, graph them, use them to solve triangles, solve equations with such functions in them, and we will study some of their most famous and celebrated properties. Thus, once again, the importance of this section can not be overstated. There are six very famous trig functions, sine, cosine, tangent, cotangent, cosecant, and secant. Sine and Cosine are personal favorites, in some way, all other functions are composed of variations of these. Thus, we will focus most of our attention on these two functions for now. Once comfortable with these, we will move on to the other four functions. Before we dive into the small details, we take a big-picture perspective on what these functions are. These functions are ratios. For any given angle, we construct a reference triangle. These functions are, by definition, ratios of certain sides of the corresponding reference triangle. Once angle is given and the reference triangle constructed and labeled, we stand at the origin and observe which of the sides of the triangle is adjacent, which side is opposite, and which side is the hypothenuse. The cosine function is defined to be the ratio of adjacent/opposite, denoted adj .It should be clear which side of the reference triangle is the hypothenuse, it’s across from the 90◦ angle. cos θ = opp The opposite side and the adjacent sides depend very much on our point of reference. The definitions of all our trig functions all assume the point of reference used is the origin. That is, once we use the origin as a point of reference, there is only one side which is opposite. The other sides are also easily identified, they are the adjacent and the hypothenuse. We state the definition formally: DEF: of sin θ and cos θ For any angle, θ, we construct the reference triangle for θ using any positive length r for the hypothenuse. We label the sides, ’hypothenuse’, ’adjacent’, and ’opposite’, accordingly. sin θ is defined as the ratio of the sides of the reference triangle: sin θ =
opposite hypothenuse
cos θ =
adj hypothenuse
In layman terms, you can think of a function as a cow. The cow eats grass and produces milk. In a similar way, the cosine function ’eats’ angles and produces ratios.
The Cow Function:
grass →
The Cosine Function: adj/hyp
angle 150◦ →
2
1 −
milk cow(grass)=milk
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150◦
30◦ √ 3
ratio√ cos(150◦ ) = −2 3
math hands
The Sine Function: opp/hyp
angle 150◦ →
2
1
150◦
30◦ −
√ 3
ratio sin(150◦ ) = 21
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Trigonometry Sec. 05
MathHands.com M´ arquez
THE REST of the Other Trig FUNCTIONS: We are now ready to meet the other famous trigonometric functions, tangent, cotangent, secant and cosecant. These are very similar to sine and cosine in the sense that they too describe ratios of sides of a corresponding reference triangle. For any given angle, θ, we construct a reference triangle, then we define the functions as follows: DEF: of the SIX trig functions
reference triangle for θ angle θ
→
ratio
hyp opp
θ
→
sin θ =
opp hyp
cos θ =
adj hyp
tan θ =
opp adj
csc θ =
hyp opp
sec θ =
hyp adj
cot θ =
adj opp
adj
It is important to direct your attention momentarily to the denominators on each of the 6 ratios. Sine and cosine each have ’hyp’ as the denominator. Keep in mind the hypothenuse is a positive length, r, of the segment chosen to construct the reference triangle. By definition, such r = hyp will never be zero. Thus, such ratios will be well defined for all possible reference triangles and in turn for all possible angles, θ. Said differently, the sine and cosine functions are defined for all possible real angles. The same can not be said of any of the other functions. Consider, for example, the case when θ = 90◦ . In such case, if hyp = 2 then opposite side is 2 and the adjacent is 0. Since the ratio 20 is not a real number, thus the tangent of 90◦ does not exist as a real number. There are many angles (all which end up in an ’almost triangle’ reference triangle) for which the other four functions are not defined.
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Trigonometry Sec. 05
MathHands.com M´ arquez
For each angle, draw and label a reference triangle, then determine all 6 trig ratios, sine , cosine, tangent, secant, cosecant, and cotangent. Do not use calculators here. 1. 120◦ y axis y axis
x axis x axis
5. 135◦ 2. 150◦ y axis y axis
x axis x axis
6. 225◦
3. −120◦
y axis
y axis
x axis
x axis
4. 330◦
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2007-2009 MathHands.com
7. 315◦
math hands
pg. 3
Trigonometry Sec. 05
MathHands.com M´ arquez 11. −330◦
y axis
y axis
x axis x axis
8. 30◦ 12. −210◦ y axis y axis
x axis x axis
9. −30◦
13. 180◦ y axis
y axis
x axis
10. 690◦
x axis
14. −180◦ y axis
y axis
x axis
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math hands
x axis
pg. 4
Trigonometry Sec. 05
MathHands.com M´ arquez
15. 360◦ y axis
y axis x axis
x axis
18. −90◦ y axis
16. −360◦
x axis
y axis
19. 270◦ x axis
y axis
x axis
17. 90◦
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math hands
pg. 5
Trigonometry Sec. 05
MathHands.com M´ arquez
The Rest of the Trig Functions : SOLUTIONS
For each angle, draw and label a reference triangle, then determine all 6 trig ratios, sine , cosine, tangent, secant, cosecant, and cotangent. Do not use calculators here.
opp hyp hyp csc(120◦ ) = opp adj cos(120◦ ) = hyp hyp sec(120◦ ) = adj sin(120◦ ) =
= = = =
opp tan(120◦ ) = = adj adj = cot(120◦ ) = opp
y axis
√ 2 3 4 4 √ 2 3 −2 4 4 −2 √ 2 3 −2 −2 √ 2 3
y axis opp −2 = hyp 4 hyp 4 csc(330◦ ) = = opp −2 √ 2 3 adj = cos(330◦ ) = hyp 4 hyp 4 ◦ x axis sec(330 )= = √ adj 2 3 opp −2 tan(330◦ ) = = √ adj 2 3 √ 2 3 adj = cot(330◦ ) = opp −2 sin(330◦ ) =
√ 2 3 4
120◦
−2
2 opp = hyp 4 4 hyp ◦ = csc(150 ) = opp 2
y axis
√ adj −2 3 = hyp 4 hyp 4 sec(150◦ ) = = √ adj −2 3 opp 2 √ tan(150◦ ) = = adj −2 3 √ −2 adj 3 = cot(150◦ ) = opp 2 cos(150◦ ) =
√ opp 4/ 2 = hyp 4 4 hyp = √ csc(135◦ ) = opp 4/ 2 √ adj −4/ 2 cos(135◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(135 )= adj −4/ 2 √ opp 4/ 2 tan(135◦ ) = = √ adj −4/ 2 √ −4/ 2 adj = √ cot(135◦ ) = opp 4/ 2
2 4 √ −2 3
150
◦
2.
= = = =
opp = tan(−120◦ ) = adj adj cot(−120◦ ) = = opp
4
y axis
sin(135◦ ) =
sin(150◦ ) =
opp hyp hyp csc(−120◦ ) = opp adj ◦ cos(−120 ) = hyp hyp sec(−120◦ ) = adj
x axis −2
4.
1.
sin(−120◦ ) =
3302◦√3
√ 4/ 2 4 √ −4/ 2
135◦ x axis
5.
√ −2 3 4 4 √ −2 3 −2 4 4 −2 √ −2 3 −2 −2 √ −2 3
y axis
√ −4/ 2 opp = hyp 4 4 hyp = √ csc(225◦ ) = opp −4/ 2 √ 2 adj −4/ cos(225◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(225 )= adj −4/ 2 √ 2 opp −4/ tan(225◦ ) = √ = adj −4/ 2 √ adj −4/ 2 √ cot(225◦ ) = = opp −4/ 2
y axis
sin(225◦ ) =
−120◦
−2
√ −2 3
4
3.
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2007-2009 MathHands.com
√ −4/ 2
225◦ x axis
√ −4/ 2
4
6.
math hands
pg. 1
Trigonometry Sec. 05 √ opp −4/ 2 = hyp 4 4 hyp √ = csc(315◦ ) = opp −4/ 2 √ adj 4/ 2 cos(315◦ ) = = hyp 4 4 hyp = √ sec(315◦ ) = adj 4/ 2 √ opp −4/ 2 tan(315◦ ) = √ = adj 4/ 2 √ 4/ 2 adj ◦ = √ cot(315 ) = opp −4/ 2
MathHands.com M´ arquez y axis
sin(−330◦ ) =
√◦ 315 4/ 2
4
√ −4/ 2
7.
2 opp = hyp 4 hyp 4 csc(30◦ ) = = opp 2 √ adj 2 3 cos(30◦ ) = = hyp 4 hyp 4 sec(30◦ ) = = √ adj 2 3 2 opp = √ tan(30◦ ) = adj 2 3 √ adj 2 3 cot(30◦ ) = = opp 2
y axis
opp 2 = hyp 4 hyp 4 csc(−330◦ ) = = opp 2 √ 2 3 adj = cos(−330◦ ) = hyp 4 4 x axis◦ ) = hyp = √ sec(−330 adj 2 3 2 opp = √ tan(−330◦ ) = adj 2 3 √ 2 3 adj = cot(−330◦ ) = opp 2
sin(315◦ ) =
2 4
◦ −3302√ 3
x axis
11.
y axis
sin(30◦ ) =
y axis
2 opp = hyp 4 hyp 4 csc(−210◦ ) = = opp 2 sin(−210◦ ) =
2 4 30 2 3
√ adj −2 3 = hyp 4 hyp 4 ◦ x axis sec(−210 )= = √ adj −2 3 2 opp ◦ = √ tan(−210 ) = adj −2 3 √ adj −2 3 cot(−210◦ ) = = opp 2
2 4 √ −2 3
cos(−210◦ ) =
◦√
x axis
12.
8.
y axis opp −2 = hyp 4 hyp 4 csc(−30◦ ) = = opp −2 √ 2 3 adj = cos(−30◦ ) = hyp 4 hyp 4 sec(−30◦ ) = = √ adj 2 3 opp −2 tan(−30◦ ) = = √ adj 2 3 √ adj 2 3 cot(−30◦ ) = = opp −2
−210◦
y axis
sin(−30◦ ) =
−302◦√3
≈ −2 ≈0
x axis 4
≈ 180◦ x axis
2
−2
sin(≈ 180◦ ) = csc(≈ 180◦ ) = ◦
cos(≈ 180 ) =
9.
13.
sec(≈ 180◦ ) = tan(≈ 180◦ ) = cot(≈ 180◦ ) =
y axis opp −2 = hyp 4 4 hyp = csc(690◦ ) = opp −2 √ adj 2 3 cos(690◦ ) = = hyp 4 hyp 4 sec(690◦ ) = = √ adj 2 3 opp −2 tan(690◦ ) = = √ adj 2 3 √ 2 3 adj = cot(690◦ ) = opp −2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
y axis
sin(690◦ ) =
6902◦√3
≈ −2 ≈0
x axis 4
−2
−180◦ x axis
2 ◦
sin(−180 ) = csc(−180◦ ) = cos(−180◦ ) =
10.
14.
◦
sec(−180 ) = ◦
tan(−180 ) =
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math hands
cot(−180◦ ) =
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
pg. 2
Trigonometry Sec. 05
MathHands.com M´ arquez y axis
y axis
360◦2 ≈2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
sin(360◦ ) = csc(360◦ ) = ◦
cos(360 ) =
15.
◦
sec(360 ) = tan(360◦ ) =
y axis cot(360◦ ) =
−360◦≈ 2
sin(−360◦ ) = csc(−360◦ ) = ◦
cos(−360 ) = ◦
sec(−360 ) = tan(−360◦ ) =
y axis 2
cot(−360◦ ) =
≈2
−90◦
x axis 0 2 2 0 2 2 2 2 0 2 2 0
= = = = = =
x axis
=0
cos(−90◦ ) =
is NOT REAL 2
sec(−90◦ ) =
≈ −2
◦
sin(−90 ) =
=1 =1
18.
◦
csc(−90 ) =
=0
cot(−90◦ ) =
y axis
= is NOT REAL
≈0
≈0
2
16.
≈0
≈0
tan(−90◦ ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
= = = =
x axis
=0
cos(270◦ ) =
is NOT REAL 2 =1 =1
=
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
270◦
x axis
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
=
≈ −2
sec(270◦ ) = sin(270◦ ) =
19.
=0 = is NOT REAL
csc(270◦ ) = cot(270◦ ) = ◦
tan(270 ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
90◦ x axis
≈0 cos(90◦ ) = sec(90◦ ) = ◦
sin(90 ) =
17.
◦
csc(90 ) = cot(90◦ ) = tan(90◦ ) =
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adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
=0 is NOT REAL =1 =1 =0 = is NOT REAL
math hands
pg. 3
Trigonometry Sec. 05
MathHands.com M´ arquez
The Rest of the Trig Functions : SOLUTIONS
For each angle, draw and label a reference triangle, then determine all 6 trig ratios, sine , cosine, tangent, secant, cosecant, and cotangent. Do not use calculators here.
opp hyp hyp csc(120◦ ) = opp adj cos(120◦ ) = hyp hyp sec(120◦ ) = adj sin(120◦ ) =
= = = =
opp tan(120◦ ) = = adj adj = cot(120◦ ) = opp
y axis
√ 2 3 4 4 √ 2 3 −2 4 4 −2 √ 2 3 −2 −2 √ 2 3
y axis opp −2 = hyp 4 hyp 4 csc(330◦ ) = = opp −2 √ 2 3 adj = cos(330◦ ) = hyp 4 hyp 4 ◦ x axis sec(330 )= = √ adj 2 3 opp −2 tan(330◦ ) = = √ adj 2 3 √ 2 3 adj = cot(330◦ ) = opp −2 sin(330◦ ) =
√ 2 3 4
120◦
−2
2 opp = hyp 4 4 hyp ◦ = csc(150 ) = opp 2
y axis
√ adj −2 3 = hyp 4 hyp 4 sec(150◦ ) = = √ adj −2 3 opp 2 √ tan(150◦ ) = = adj −2 3 √ −2 adj 3 = cot(150◦ ) = opp 2 cos(150◦ ) =
√ opp 4/ 2 = hyp 4 4 hyp = √ csc(135◦ ) = opp 4/ 2 √ adj −4/ 2 cos(135◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(135 )= adj −4/ 2 √ opp 4/ 2 tan(135◦ ) = = √ adj −4/ 2 √ −4/ 2 adj = √ cot(135◦ ) = opp 4/ 2
2 4 √ −2 3
150
◦
2.
= = = =
opp = tan(−120◦ ) = adj adj cot(−120◦ ) = = opp
4
y axis
sin(135◦ ) =
sin(150◦ ) =
opp hyp hyp csc(−120◦ ) = opp adj ◦ cos(−120 ) = hyp hyp sec(−120◦ ) = adj
x axis −2
4.
1.
sin(−120◦ ) =
3302◦√3
√ 4/ 2 4 √ −4/ 2
135◦ x axis
5.
√ −2 3 4 4 √ −2 3 −2 4 4 −2 √ −2 3 −2 −2 √ −2 3
y axis
√ −4/ 2 opp = hyp 4 4 hyp = √ csc(225◦ ) = opp −4/ 2 √ 2 adj −4/ cos(225◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(225 )= adj −4/ 2 √ 2 opp −4/ tan(225◦ ) = √ = adj −4/ 2 √ adj −4/ 2 √ cot(225◦ ) = = opp −4/ 2
y axis
sin(225◦ ) =
−120◦
−2
√ −2 3
4
3.
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2007-2009 MathHands.com
√ −4/ 2
225◦ x axis
√ −4/ 2
4
6.
math hands
pg. 1
Trigonometry Sec. 05 √ opp −4/ 2 = hyp 4 4 hyp √ = csc(315◦ ) = opp −4/ 2 √ adj 4/ 2 cos(315◦ ) = = hyp 4 4 hyp = √ sec(315◦ ) = adj 4/ 2 √ opp −4/ 2 tan(315◦ ) = √ = adj 4/ 2 √ 4/ 2 adj ◦ = √ cot(315 ) = opp −4/ 2
MathHands.com M´ arquez y axis
sin(−330◦ ) =
√◦ 315 4/ 2
4
√ −4/ 2
7.
2 opp = hyp 4 hyp 4 csc(30◦ ) = = opp 2 √ adj 2 3 cos(30◦ ) = = hyp 4 hyp 4 sec(30◦ ) = = √ adj 2 3 2 opp = √ tan(30◦ ) = adj 2 3 √ adj 2 3 cot(30◦ ) = = opp 2
y axis
opp 2 = hyp 4 hyp 4 csc(−330◦ ) = = opp 2 √ 2 3 adj = cos(−330◦ ) = hyp 4 4 x axis◦ ) = hyp = √ sec(−330 adj 2 3 2 opp = √ tan(−330◦ ) = adj 2 3 √ 2 3 adj = cot(−330◦ ) = opp 2
sin(315◦ ) =
2 4
◦ −3302√ 3
x axis
11.
y axis
sin(30◦ ) =
y axis
2 opp = hyp 4 hyp 4 csc(−210◦ ) = = opp 2 sin(−210◦ ) =
2 4 30 2 3
√ adj −2 3 = hyp 4 hyp 4 ◦ x axis sec(−210 )= = √ adj −2 3 2 opp ◦ = √ tan(−210 ) = adj −2 3 √ adj −2 3 cot(−210◦ ) = = opp 2
2 4 √ −2 3
cos(−210◦ ) =
◦√
x axis
12.
8.
y axis opp −2 = hyp 4 hyp 4 csc(−30◦ ) = = opp −2 √ 2 3 adj = cos(−30◦ ) = hyp 4 hyp 4 sec(−30◦ ) = = √ adj 2 3 opp −2 tan(−30◦ ) = = √ adj 2 3 √ adj 2 3 cot(−30◦ ) = = opp −2
−210◦
y axis
sin(−30◦ ) =
−302◦√3
≈ −2 ≈0
x axis 4
≈ 180◦ x axis
2
−2
sin(≈ 180◦ ) = csc(≈ 180◦ ) = ◦
cos(≈ 180 ) =
9.
13.
sec(≈ 180◦ ) = tan(≈ 180◦ ) = cot(≈ 180◦ ) =
y axis opp −2 = hyp 4 4 hyp = csc(690◦ ) = opp −2 √ adj 2 3 cos(690◦ ) = = hyp 4 hyp 4 sec(690◦ ) = = √ adj 2 3 opp −2 tan(690◦ ) = = √ adj 2 3 √ 2 3 adj = cot(690◦ ) = opp −2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
y axis
sin(690◦ ) =
6902◦√3
≈ −2 ≈0
x axis 4
−2
−180◦ x axis
2 ◦
sin(−180 ) = csc(−180◦ ) = cos(−180◦ ) =
10.
14.
◦
sec(−180 ) = ◦
tan(−180 ) =
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2007-2009 MathHands.com
math hands
cot(−180◦ ) =
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
pg. 2
Trigonometry Sec. 05
MathHands.com M´ arquez y axis
y axis
360◦2 ≈2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
sin(360◦ ) = csc(360◦ ) = ◦
cos(360 ) =
15.
◦
sec(360 ) = tan(360◦ ) =
y axis cot(360◦ ) =
−360◦≈ 2
sin(−360◦ ) = csc(−360◦ ) = ◦
cos(−360 ) = ◦
sec(−360 ) = tan(−360◦ ) =
y axis 2
cot(−360◦ ) =
≈2
−90◦
x axis 0 2 2 0 2 2 2 2 0 2 2 0
= = = = = =
x axis
=0
cos(−90◦ ) =
is NOT REAL 2
sec(−90◦ ) =
≈ −2
◦
sin(−90 ) =
=1 =1
18.
◦
csc(−90 ) =
=0
cot(−90◦ ) =
y axis
= is NOT REAL
≈0
≈0
2
16.
≈0
≈0
tan(−90◦ ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
= = = =
x axis
=0
cos(270◦ ) =
is NOT REAL 2 =1 =1
=
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
270◦
x axis
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
=
≈ −2
sec(270◦ ) = sin(270◦ ) =
19.
=0 = is NOT REAL
csc(270◦ ) = cot(270◦ ) = ◦
tan(270 ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
90◦ x axis
≈0 cos(90◦ ) = sec(90◦ ) = ◦
sin(90 ) =
17.
◦
csc(90 ) = cot(90◦ ) = tan(90◦ ) =
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2007-2009 MathHands.com
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
=0 is NOT REAL =1 =1 =0 = is NOT REAL
math hands
pg. 3
Trigonometry Sec. 06
MathHands.com M´ arquez
The Rest of the Inverse Functions Main Idea: We have already introduced the idea of inverse functions earlier. While the sin(θ), cos(θ) and tan(θ) functions are great to describe the ratios for a given angle θ, and its respective triangle, the inverse functions sin−1 (k), cos−1 (k) and tan−1 (k) are great for giving the angle when the ratio is known.
hyp
×.40
4, opp
10,
opp If the hyp ratio is known, 4 = .40, we use 10 −1 hyp sin (.400) 10,
×.40
4, opp
Here is an example to help us recall our previous look at the inverse functions:
23.58◦
??
That was an important introduction to the inverse functions. Here we take the opportunity to reconcile how these functions behave once we have extended the trigonometric functions to just about all angles. The complete discussion of these functions requires a reasonable understanding of ono-to-one and onto functions. While this may not be the appropriate course for such a discussion we will engage in such ideas just enough to gain a broader and adequate perspective of the trigonometric functions. Inverse Functions: Thus far we have seen the trigonometric functions describe for each angle a particular ratio. In this section we take a moment to contemplate this question but backwards. Suppose we know a particular ratio, can we then find the corresponding angle? The answer is a definitive yes. The taylor-made medicine for the job is the study of the inverse functions. The inverse cosine is often called arc-cosine and is usually denoted as arccos(z) or cos−1 (z). Similarly of arc-sine and arc-tangent. These are defined as you would expect, given a ratio z, arccos(z) is defined as the angle θ such that cos θ = z, of course this demands some cautionary and insightful comments, which we will add soon. The following diagram should help solidify the concept of the inverse functions.
The Cow Function:
The Cow−1 Function:
milk grass → grass milk cow(grass)=milk c
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cow−1 (milk)=grass
math hands
pg. 1
Trigonometry Sec. 06
MathHands.com M´ arquez
Now, we apply the idea to cosine function at 60◦
The Cosine Function:
The Cosine Inverse Function:
ratio
angle 60◦ → cos ratio = cos 60◦ =
1 2
cos−1
1 2
cos−1
1 2
angle 60◦ 1 2
= 60◦
The Inverse Functions: Here is yet another way to illustrate the role of the inverse trigonometric functions. Let us continue with the same example.
les an g
cos
ios rat 1 2
60◦
cos−1 [domain]
[co-domain]
Consider the possibility that feeding the cow ”hot-dogs” also produces ’milk’. This adds a small complication when calculating what the inverse function, cow−1 , does to ’milk’. Namely, it’s tricky to decide wether cow −1 = grass or cow −1 = hot − dogs. In fact, in such a situation the cow function does not have an inverse. In general, when a function is not one-to-one it does not have an inverse. Such is the case for all our trigonometric functions. Although a complete discussion of one-to-one functions may not be appropriate here, we are obliged to at least scratch the surface of the topic, at least enough to understand the basic intricacies of trigonometric inverse functions. With that in mind consider the following angles, all of which result in a cosine value of 12 .
cos
les an g
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ios rat
300◦ ◦ 60 420◦
1 2
[domain]
[co-domain] math hands
pg. 2
Trigonometry Sec. 06
MathHands.com M´ arquez
Now, consider the inverse cosine function, that is if the ratio of adjacent to hypothenuse is 21 , can we recover the angle? The answer is a resounding no!. There are too many [more than one] angles for which such ratio is 12 . When this happens we say that the function is not one-to-one.
cos−1
les an g
ios rat
300◦ ?◦ 60 ? 420◦ ?
1 2
[domain]
[co-domain]
In some way, this is a remediable problem. The popular and almost universally accepted approach is to modify the domain on the cosine function as to limit angles to only angles between 0◦ and 180◦ . At the same time, the ratios have to be restricted to values between -1 and 1.
es ngl 80◦ a 1 y on l ≤ θ ≤ 0
cos
os ati 1 r y ≤ on l 1 ≤ r − 1 2
60◦
cos−1 [domain]
[co-domain]
Note how the inverse problem is resolved:
es ngl 80◦ a 1 y on l ≤ θ ≤ 0 300◦ ?
cos−1
os ati 1 r y ≤ on l 1 ≤ r − 1 2
60◦ 420◦ ?
[domain]
[co-domain]
Of the many different angles for which the cosine is 1/2, only one is in the adequate range, thus there is no ambiguity as to which angle the ratio is assigned to. SImilarly, the domain for the sine function, as well as the tangent function can be restricted so that an inverse function can be constructed for each one of these.
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math hands
pg. 3
Trigonometry Sec. 06
MathHands.com M´ arquez es ◦ ngl ≤ 90 a θ y onl 90◦ ≤ −
os ati 1 r y ≤ on l 1 ≤ r −
sin
r
θ
sin−1 [domain]
[co-domain]
For tangent,
es ◦ ngl < 90 a y θ onl 90◦ < −
l rea al l tios ra
tan
r
θ
tan−1 [domain]
[co-domain]
You should know that most scientific calculators are programed to compute inverse trig functions angles according to these ranges. Moreover, you should be aware of alternative names for these functions. arccos(x) is sometimes used to denote cos−1 (x), and similarly arctan(x) = tan−1 (x), and arcsin(x) = sin−1 (x) Examples: Example: Find the angle θ
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 θ = tan−1 2
5
≈ 1.19
θ
≈ 68.199
2
In this case, we know the opposite and the adjacent sides. The function describing such ratio is the tangent, thus 5 tan θ = 2
(in radians [see calculator mode]... OR...) ◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation tan θ = 52 under a different context, where we will solve it completely, not limited to the domain and codomain of the arctan function.
Example:
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math hands
pg. 4
Trigonometry Sec. 06
MathHands.com M´ arquez
Find the angle θ
3
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 −1 θ = sin 3
2
≈ 0.73
≈ 41.81
θ In this case, we know the opposite and the hypothenuse sides. The function describing such ratio is the sine, thus 2 sin θ = 3
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2007-2009 MathHands.com
math hands
(in radians [see calculator mode]... OR...) ◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation sin θ = 32 under a different context, where we will solve it completely, not limited to the domain and codomain of the arcsin function.
pg. 5
Trigonometry Sec. 06
MathHands.com M´ arquez
1. Find the angle θ 6
5
θ
2. Find the angle θ 3 θ
2
3. Find the angle θ 11
2
θ
4. Find the angle θ 5 θ
2
5. Find the angle θ 5 θ
1
6. Find the angle θ 4 θ
3
7. Find the angle θ 9 θ
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math hands
2
pg. 6
Trigonometry Sec. 06
MathHands.com M´ arquez
8. cos−1 (cos(30◦ )) = 30◦ A. TRUE
B. FALSE
9. cos−1 (cos(−30◦ )) = −30◦ A. TRUE
B. FALSE
10. tan−1 (tan(30◦)) = 30◦ A. TRUE
B. FALSE
11. tan−1 (tan(210◦)) = 210◦ A. TRUE
B. FALSE
12.
−1
1 1 = 3 3
−1
5 5 = 3 3
tan tan A. TRUE
B. FALSE
13.
tan tan A. TRUE
B. FALSE
14.
5 5 −1 = cos cos 13 13 A. TRUE
B. FALSE
15. Compute sin cos−1 (.3456) 16. Compute tan cos−1 (.1234)
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The Rest of the Inverse Functions : SOLUTIONS 1. Find the angle θ Solution: In this case, we know the opposite and the hypothenuse sides. The function describing such ratio is the sine, thus sin θ =
6
5 6
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 θ = sin−1 6
5
θ
≈ 0.985 (in radians [see calculator mode]... OR...) ≈ 56.443◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation sin θ = 56 under a different context, where we will solve it completely, not limited to the domain and codomain of the arcsin function.
2. Find the angle θ Solution: In this case, we know the adjacent and the hypothenuse sides. The function describing such ratio is the cosine, thus cos θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 θ = cos−1 3
3 θ
2 3
2
≈ 0.841 (in radians [see calculator mode]... OR...) ≈ 48.19◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation cos θ = 23 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
3. Find the angle θ 11
2
θ
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Trigonometry Sec. 06
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Solution: In this case, we know the opposite and the hypothenuse sides. The function describing such
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Trigonometry Sec. 06
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ratio is the sine, thus sin θ =
2 11
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 θ = sin−1 11 ≈ 0.183 (in radians [see calculator mode]... OR...) ≈ 10.476◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equa2 tion sin θ = 11 under a different context, where we will solve it completely, not limited to the domain and codomain of the arcsin function.
4. Find the angle θ Solution: In this case, we know the opposite and the adjacent sides. The function describing such ratio is the tangent, thus tan θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 θ = tan−1 2 ≈ 1.19 (in radians [see calculator mode]... OR...)
5 θ
5 2
2
≈ 68.199◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation tan θ = 52 under a different context, where we will solve it completely, not limited to the domain and codomain of the arctan function.
5. Find the angle θ Solution: In this case, we know the opposite and the adjacent sides. The function describing such ratio is the tangent, thus 5 tan θ = 1 Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 −1 θ = tan 1
5 θ
1
≈ 1.373 (in radians [see calculator mode]... OR...)
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≈ 78.69◦
(in degrees [see calculator mode])
pg. 3
Trigonometry Sec. 06
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... it should be noted that we will revisit the equation tan θ = 15 under a different context, where we will solve it completely, not limited to the domain and codomain of the arctan function.
6. Find the angle θ Solution: In this case, we know the adjacent and the hypothenuse sides. The function describing such ratio is the cosine, thus cos θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 3 −1 θ = cos 4 ≈ 0.723 (in radians [see calculator mode]... OR...)
4 θ
3 4
3
≈ 41.41◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation cos θ = 34 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
7. Find the angle θ Solution: In this case, we know the adjacent and the hypothenuse sides. The function describing such ratio is the cosine, thus cos θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 θ = cos−1 9
9 θ
2 9
2
≈ 1.347 (in radians [see calculator mode]... OR...) ≈ 77.16◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation cos θ = 29 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
8. cos−1 (cos(30◦ )) = 30◦
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Trigonometry Sec. 06 A. TRUE
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B. FALSE
9. cos−1 (cos(−30◦ )) = −30◦ A. TRUE
B. FALSE
10. tan−1 (tan(30◦)) = 30◦ A. TRUE
B. FALSE
11. tan−1 (tan(210◦)) = 210◦ A. TRUE
B. FALSE
12.
−1
1 1 = 3 3
−1
5 5 = 3 3
tan tan A. TRUE
B. FALSE
13.
tan tan A. TRUE
B. FALSE
14.
5 5 −1 cos cos = 13 13 A. TRUE
B. FALSE
15. Compute sin cos−1 (.3456)
Solution: First, the stuff inside, cos−1 (.3456) ≈ 69.78◦ then...
sin (69.78◦) ≈ 0.9384 16. Compute
tan cos−1 (.1234)
Solution: First, the stuff inside, cos−1 (.1234) ≈ 82.91◦ then...
tan (82.91◦ ) ≈ 8.0418
note: the self quiz on ”mission accomplished” prob #2 outlines a different approach to this same question.
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math hands Trigonometry: Ch2 guide
Reference Triangles
Every angle has a reference triangle. The recipe is as follows:
Def: Inverse Cosine Function STEP 1. Choose any + r
STEP 2. Rotate θ
y
es ◦ ngl 80 ya ≤1 onl θ ◦ ≤ 0
y positive angles
2
r
θ
θ
r=2
os ati 1 yr r≤ onl ≤ −1
cos
cos
−1
x [domain] negative angles STEP 3. Draw a Perp to x-axis STEP 4. Label sides/signs
y 2
Example: Find the angle θ
to estimate the sought angle. „ « 5 θ = cos−1 7
y
x
7
2
?
θ
θ
θ x
?
Def: of THE Trig Functions
≈ 0.775 (in radians [see calculator mode]... OR...) In this case, we know the adjacent ≈ 44.415◦ and the hypothenuse sides. The (in degrees [see calculator mode]) function describing such ratio is the cosine, thus
5
cos θ = ref. triangle for θ angle θ
[co-domain]
ratio of sides
hyp opp
θ
5 7
Since this is not a famous ratio, we allow ourselves use of a calculator
... it should be noted that we will revisit the equation cos θ = 75 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
adj
sin θ =
opp hyp
cos θ =
adj hyp
tan θ =
opp adj
csc θ =
hyp opp
sec θ =
hyp adj
cot θ =
adj opp
Def: Inverse Tangent Function LOP Diagram: ◦ es ngl 90 ya θ< onl ◦ < 0 −9
l rea al l ios rat
tan
• The essential concept 1: once the ratios are known, one side is enough to determine the other sides
r
θ
• The essential concept 2: For each angle, the ratios are described by the trig functions where defined.
−1
tan [domain]
[co-domain]
Def: Inverse Sine Function ◦ es ngl 90 ya θ≤ onl ◦ ≤ 0 −9
sin
p hy
r
θ
sin [domain]
os ati yr ≤1 onl ≤r 1 −
× sin θ os θ ×c
−1
θ
[co-domain]
math hands
×
opp
nθ ta
adj
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Solving Euclidean Right Triangles Solve the triangle [may not be drawn to scale assume typical units for length]:
β 23
34◦
b
a
Solution: Notice, this proposition is typical, three items are given; the right angle, the 34 angle, and the hypothenuse side, while three items are missing; sides a, b, and angle β. It should be noted that the angle β is relatively easy to determine β + 34◦ = 90◦ Therefore, β = 56◦ . Now, we solve for b, the side opposite of the 34◦ angle. Since we know the hypothenuse is 23 units, and we want to know the sine function describes this ratio, sin 34◦ ≈ 0.559, similarly to solve for a we use the cosine ratio, cos 34◦ ≈ 0.829. Thus, we illustrate the ratios on the triangle:
56◦ 23
34◦
×0.559
12.861
×0.829
19.068
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Ratio Nirvana The Ratios We Know: Solving triangles is one of the essential promises of trigonometry, and knowing one [or more] of the ratios is one essential tool to accomplish just this. Now, the ratios we have learned thus far are the ratios for 45-45 Triangles, ratios for 30-60 Triangles, and we hesitantly add to the list the ratios for the almost triangles.
45-45 Triangle Ratios s
√ × 2
×
1
√ 3 ×
◦
Almost-Triangle Ratios
45◦
m
≈ 0◦
s
h
≈0
30
s
×2
h
h
30-60 Triangle Ratios
a≈h
Essentially what this means is for any of these triangles, if we know the size of one, just one, non-zero side we can solve all sides of the triangle. This is great news but very restrictive since we know the ratios, thus far, only for these special triangles. If the angles were different, so too would be the shape of the triangle, and thus the ratios of the sides would be different as well. Naturally, this leads to a different question, could we somehow determine the ratios for other triangles besides these? Or posed differently, which ratios would we like to know? The ambitious answer is we would like to know the ratios for all triangles. Just imagine, if we knew all the ratios for all the triangles we could possible solve any and every triangle just with a mere glance at the angle and one of the sides. This translates into big, very big ideas. We could position the moon at a corner of a triangle, glance at it, compute the ratios and there, just like that, we could compute the distance to the moon, the size of the moon, the height of a mountain, etc. etc. It is no exaggeration to say that if you knew all the ratios for all the triangles, well..., everyone would want to be your friend. Visually, we summarize: Ratios we would like to know
23-67 Triangle Ratios
41.45612◦ Triangle Ratios
Any & every angle θ ?
r
b
r
b
?
b
r
?
?
23◦
? 41.45612◦
a
a
? θ◦ a
Ratio Nirvana: Through sheer audacity, creativity, perseverance, pride, by bits and pieces, and over the centuries, some of the greatest thinkers of our species who have ever lived started putting the ideas together. The chinese from 4,000 years ago, the arabic/persian cultures from 2,000 years ago, the asian, the greeks, the indian, and everyone in-between collectively generation after generation each added a small piece to the puzzle, and in the end we triumphed over this challenge with grace and elegance. Today, we can find the ratios for all Euclidean Triangles with ease ad efficiency. We even have names and symbols for these ratios. The names for these ratios are sine, cosine and tangent. These
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are more than just names, they are indeed functions that describe for us all the ratios for all the triangles. Thanks to these functions and our ability to compute them, we could say that we now live in ratio nirvana, where all ratios are known. Let us now formally introduce the main authors of ratio-nirvana, the trigonometric functions. The Trigonometric Functions: Let us consider the first famous ratio. As before, we will continue to name the sides with respective to the corner, or angle from which we stand. Right triangles usually have two smaller angles [smaller than 90◦ ]. We will usually stand next to one of these two and for now, never next to the right angle.
opp
p hy
θ adj opp Now, we consider a ratio, let us say, opposite to hypothenuse, or hyp , and we give a name. We will call it the sine ratio. Of course this ratio depends on the shape of the triangle which depends on the angle θ, so really the description of the ratio should be more like a function, such as f (θ), rather than just ’sine’. But this function is so famous f (θ) is not special enough, thus we compromise and we call it sin(θ). In other words, we define1
sin(θ) =
opp hyp
p hy
× 21
opp
for the appropriate triangle. We are now ready to compute our very first sine values, let us compute sin(30◦ )
sin(30◦ ) =
opp hyp
sin(30◦ ) =
1 2
30◦ adj
10 p, hy
opp, 5
Usually, we want the ratios so we can find the sides, but often we will know the sides from these derive the ratios:
sin(30◦ ) =
opp hyp
sin(30◦ ) =
5 1 = 10 2
30◦ adj 1 we
will extend the definition of sine later in a much bigger and broader sense.
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Trigonometry Sec. 01
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Note the sine function is not affected by how big the triangle is. The above triangle could have hypothenuse 10, 20 or 100, the ratio opp/hyp would still be 1/2, thus sin(30◦ ) = 21 This ratio is old news. What we would really like to know is other ratios, we want to know all ratios. Unfortunately, the complete answer to this would quickly take us outside the scope of this course, and we will never [in this class] learn how the intellectual giants of our past resolved this issue. We will have to settle for the pride-less and undecorated method, using a calculator. Go on and try your calculator, check that it is in degrees mode and try sin(30◦ ) = .5, the expected ratio. Now go on and compute sin(23◦ ) ≈ 0.3907, at last, we have arrived! Ratio nirvana, we can compute the ratios for all Euclidean right triangles.
×0.6620
×0.3907
r
× sin(θ)
b
r
r
Generic Triangle Ratios
23◦
41.45612◦ a
θ
a
a
But wait, there is more. Just like we defined the sine function as the ratio of the opp/hyp sides of a triangle we also define2 the cosine function to be the ratio of adjacent side to hypothenuse. So we have cos(θ) =
adj hyp
opp, med,
hy p, 2
√
3
For example:
1
×2
cos(60◦ ) =
adj hyp
cos(60◦ ) =
1 2
60◦ adj, small, 1
And now using our calculator we have ratio nirvana. The adj/hyp ratio for any eucledian triangle is given by the cosine function: 2 we
will later extend this definition to a broader idea
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b
41.45612◦ Triangle Ratios b
23-67 Triangle Ratios
Trigonometry Sec. 01
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23-67 Triangle Ratios
41.62◦ Triangle Ratios
Generic Triangle Ratios r b
b
r
41.62◦
θ
6
05 a
a
θ) os(
.92
23◦
×c
×0
747 ×0.
b
r
a
Similarly, the tangent function is well known to represent the ratio: tan(θ) =
opp hyp
These three are the most famous trigonometric functions describing the most famous ratios for right triangles. The acronym SOH-CAH-TOA is often used to remember each of the ratios. That is SOH helps some students remember Opp Adj , then CAH reminds us that Cos(θ) = Hyp while TOA reminds us that T an(θ) = Opp that Sin(θ) = Hyp Ad We summarize this in:
SOH-CAH-TOA
p hy
θ os( ×c )
θ
× sin(θ)
×
n( ta
Sin(θ) =
Opp Hyp
Cos(θ) =
Adj Hyp
T an(θ) =
Opp Adj
opp
Generic Triangle Ratios
θ)
adj
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Trigonometry Sec. 01
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1. Find the three famous ratios and write them on the diagram
opp
hyp
20◦ adj
hyp
opp
2. Find the three famous ratios and write them on the diagram
20◦ adj
3. Find the three famous ratios and write them on the diagram
opp
hyp
20◦ adj
p hy
opp
4. Find the three famous ratios and write them on the diagram
30◦ adj
5. Find the three famous ratios and write them on the diagram
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p hy
opp
Trigonometry Sec. 01
40◦ adj
opp
hyp
6. Find the three famous ratios and write them on the diagram
70◦ adj
hyp
opp
7. Find the three famous ratios and write them on the diagram
10◦ adj
8. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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Trigonometry Sec. 01
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m. 10 c
20◦
9. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
m. 12 c
20◦
10. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
cm. 6.8
20◦
11. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
. cm 10
32◦
12. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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Trigonometry Sec. 01
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. cm 12
25◦
13. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
m. 6.8 c
15◦
7.5 cm.
14. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
69◦
15. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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10 cm.
Trigonometry Sec. 01
35◦
8 cm.
16. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
42◦
7.5 cm.
17. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
9◦
11 cm.
18. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
12◦
19. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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9 cm.
Trigonometry Sec. 01
23◦
20. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
9◦ 7.5 cm.
21. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
12◦ 11 cm.
22. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
23◦ 9 cm.
23. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
67◦
9 cm.
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Trigonometry Sec. 01
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24. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
55◦
10 cm.
25. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
50◦
6 cm.
26. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
35◦
7.5 cm.
27. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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Trigonometry Sec. 01
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4
5
θ 3
28. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 34
5
(May not be drawn to scale.)
θ 3
29. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 26
5
(May not be drawn to scale.)
θ 1
30. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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√ 104
10
Trigonometry Sec. 01
θ 2
31. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 45
6
(May not be drawn to scale.)
θ 3
32. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
1
2
θ √
3
33. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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Trigonometry Sec. 01
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1
√ 2
θ 1
34. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
0
5
θ 5
35. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
x2
4
6x
θ x
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Ratio Nirvana : SOLUTIONS 1. Find the three famous ratios and write them on the diagram sin(20◦ ) ≈ 0.3420
× 93 0. 97
20
◦
×0.3420
×0.3
cos(20◦ ) ≈ 0.9397
opp
hyp
640
tan(20◦ ) ≈ 0.3640
adj
2. Find the three famous ratios and write them on the diagram sin(20◦ ) ≈ 0.3420
× 93 0.
20
◦
×0.3420
97
×0.3
opp
hyp
640
cos(20◦ ) ≈ 0.9397 tan(20◦ ) ≈ 0.3640
adj
3. Find the three famous ratios and write them on the diagram
sin(20◦ ) ≈ 0.3420
× 93 0. 97
20◦
×0.3420
opp
hyp
3640 ×0.
cos(20◦ ) ≈ 0.9397 tan(20◦ ) ≈ 0.3640
adj
p hy
×0 6 .86 0
30
◦
×0.5000
×0
.57
74
opp
4. Find the three famous ratios and write them on the diagram
sin(30◦ ) ≈ 0.5000 cos(30◦ ) ≈ 0.8660 tan(30◦ ) ≈ 0.5774
adj
5. Find the three famous ratios and write them on the diagram
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Trigonometry Sec. 01
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766 ×0.
×
0
40
◦
×0.6428
83 0.
opp
p hy
sin(40◦ ) ≈ 0.6428
91
cos(40◦ ) ≈ 0.7660 tan(40◦ ) ≈ 0.8391
adj
hyp
×0 .93 97
475
×0.3420 70
opp
6. Find the three famous ratios and write them on the diagram
×2.7
sin(70◦ ) ≈ 0.9397 cos(70◦ ) ≈ 0.3420 tan(70◦ ) ≈ 2.7475
◦
adj
sin(10◦ ) ≈ 0.1736
7. Find the three famous ratios and write them on the diagram
10
◦
×0
.98 4 adj
×0.1736
8
×0.1763
opp
hyp
cos(10◦ ) ≈ 0.9848 tan(10◦ ) ≈ 0.1763
8. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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m. 10 c
opp ≈ (10)0.3420
opp, 3.42cm.
Trigonometry Sec. 01
×0.3420
×
= 3.42 adj ≈ (10)0.9397
93 0.
= 9.397
97
20◦
adj, 9.397cm.
m. 12 c
opp, 4.104cm.
9. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.3420
opp ≈ (12)0.3420
×
= 4.104 adj ≈ (12)0.9397
93 0.
= 11.276
97
20◦
adj, 11.276cm.
10. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.3420
× 93 0.
= 2.326 adj ≈ (6.8)0.9397 = 6.39
97
20
◦
opp, 2.326cm.
cm. 6.8
opp ≈ (6.8)0.3420
adj, 6.39cm.
. cm 10
×0.5299
opp, 5.299cm.
11. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
opp ≈ (10)0.5299
×0 8 .84
= 5.299 adj ≈ (10)0.8480 = 8.48
0
32◦ adj, 8.48cm.
12. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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. cm 12
×0.4226
opp, 5.071cm.
Trigonometry Sec. 01
opp ≈ (12)0.4226
×0 .90
= 5.071 adj ≈ (12)0.9063 = 10.876
63
25◦
adj, 10.876cm.
m. 6.8 c
15◦
×0
.96
×0.2588
59
opp, 1.76cm.
13. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters. opp ≈ (6.8)0.2588
= 1.76 adj ≈ (6.8)0.9659 = 6.568
adj, 6.568cm.
×0 .93 36
×0.3584 69◦
opp, 7.5 cm.
hyp , 8. 034 cm.
14. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
hyp ≈ (7.5) ÷ 0.9336 = 8.034 adj ≈ (8.034)0.3584 = 2.879
adj, 2.879cm.
15. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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Trigonometry Sec. 01
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.
opp, 10 cm.
m 4c .43 7 1 p, hy
×0.5736
hyp ≈ (10) ÷ 0.5736
×0
= 17.434 adj ≈ (17.434)0.8192
9 .81 2
= 14.281
35◦ adj, 14.281cm.
16. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
.
opp, 8 cm.
cm 56 9 . 11 p, y h
×0.6691
hyp ≈ (8) ÷ 0.6691
×0.7
= 11.956 adj ≈ (11.956)0.7431
431
= 8.885
42◦ adj, 8.885cm.
. .943cm hyp, 47
9◦
×0.9
×0.1564
877
adj, 47.353cm.
opp, 7.5 cm.
17. Determine the size for each of the three sides of the triangle. Assume all length units arehyp centimeters. ≈ (7.5) ÷ 0.1564 = 47.943 adj ≈ (47.943)0.9877 = 47.353
cm. 2.907 hyp, 5
12◦
×0
.97
×0.2079
opp, 11 cm.
18. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
hyp ≈ (11) ÷ 0.2079 = 52.907 adj ≈ (52.907)0.9781 = 51.751
81
adj, 51.751cm.
19. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
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. 4cm .03 3 2 , hyp
opp, 9 cm.
Trigonometry Sec. 01
×0.3907
×0
adj ≈ (23.034)0.9205 = 21.203
.92 05
23◦
hyp ≈ (9) ÷ 0.3907 = 23.034
adj, 21.203cm.
93cm. hyp, 7.5
9◦
×0.9
opp, 1.188cm.
20. Determine the size for each of the three sides of the triangle. Assume all length units arehyp centimeters. ≈ (7.5) ÷ 0.9877 = 7.593 ×0.1564
877
adj, 7.5 cm.
opp ≈ (7.593)0.1564 = 1.188
cm. 1.246 hyp, 1
×0
12◦
.97
opp, 2.338cm.
21. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.2079
81
hyp ≈ (11) ÷ 0.9781
= 11.246 opp ≈ (11.246)0.2079 = 2.338
adj, 11 cm.
. 7cm .77 9 , hyp
opp, 3.820cm.
22. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.3907
×0 .92
opp ≈ (9.777)0.3907 = 3.820
05
23◦
hyp ≈ (9) ÷ 0.9205 = 9.777
adj, 9 cm.
23. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
sin(67◦ ) = 0.9205
. 7cm .77 9 , hyp
×0 .92 05 opp, 9 cm.
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×0.3907
adj, 3.820cm.
67◦
hyp ≈ (9) ÷ 0.9205 = 9.777 cos(67◦ ) = 0.3907 opp ≈ (9.777)0.3907 = 3.820
pg. 6
Trigonometry Sec. 01
MathHands.com M´ arquez
24. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
m 8c .20 2 1 p, hy
. ×0.5736
adj, 7.002cm.
55◦
sin(55◦ ) = 0.8192 hyp ≈ (10) ÷ 0.8192 = 12.208
×0 9 .81 2
cos(55◦ ) = 0.5736 opp ≈ (12.208)0.5736 = 7.002
opp, 10 cm.
25. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
p, hy
. cm 32 ×0.6428 8 7.
766 ×0.
adj, 5.034cm.
50◦
sin(50◦ ) = 0.7660 hyp ≈ (6) ÷ 0.7660 = 7.832 cos(50◦ ) = 0.6428
0
opp ≈ (7.832)0.6428 = 5.034
opp, 6 cm.
26. Determine the size for each of the three sides of the triangle. Assume all length units are centimeters.
×0.81 92
×0.57 36 opp, 7.5 cm.
adj, 10.711cm.
hy p, 13 .07 6c m
.
35◦
sin(35◦ ) = 0.5736 hyp ≈ (7.5) ÷ 0.5736 = 13.076
cos(35◦ ) = 0.8192 opp ≈ (13.076)0.8192 = 10.711
27. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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5
. yp ,h
4 , opp.
Trigonometry Sec. 01
θ 3 , adj.
sin(θ) =
opp 4 = hyp 5
cos(θ) =
3 adj = hyp 5
tan(θ) =
opp 4 = adj 3
28. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
sin(θ) =
opp 5 =√ hyp 34
cos(θ) =
3 adj =√ hyp 34
tan(θ) =
5 opp = adj 3
sin(θ) =
opp 5 =√ hyp 26
cos(θ) =
1 adj =√ hyp 26
tan(θ) =
opp 5 = adj 1
5 , opp.
√ 34
. yp ,h
θ 3 , adj.
29. Find the three ratios, sin(θ), cos(θ), and tan(θ)
√ 26
. yp ,h
θ 1 , adj.
5 , opp.
(May not be drawn to scale.)
30. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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√ 104
. yp ,h
10 , opp.
Trigonometry Sec. 01
θ 2 , adj.
sin(θ) =
opp 10 = √ hyp 104
cos(θ) =
2 adj = √ hyp 104
tan(θ) =
opp 10 = adj 2
31. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
√ 45
sin(θ) =
opp 6 =√ hyp 45
cos(θ) =
3 adj =√ hyp 45
tan(θ) =
6 opp = adj 3
6 , opp.
. yp ,h
θ 3 , adj.
32. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
,
sin(θ) = 1 , opp.
2
p. hy
√ adj 3 = cos(θ) = hyp 2
θ √ 3 , adj.
opp 1 = hyp 2
tan(θ) =
1 opp = √ adj 3
33. Find the three ratios, sin(θ), cos(θ), and tan(θ) (May not be drawn to scale.)
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√ 2
. yp ,h
sin(θ) =
opp 1 =√ hyp 2
cos(θ) =
1 adj =√ hyp 2
tan(θ) =
opp 1 = adj 1
1 , opp.
Trigonometry Sec. 01
θ 1 , adj.
34. Find the three ratios, sin(θ), cos(θ), and tan(θ)
5
. yp ,h
0 , opp.
(May not be drawn to scale.)
θ 5 , adj.
sin(θ) =
opp 0 = hyp 5
cos(θ) =
5 adj = hyp 5
tan(θ) =
opp 0 = adj 5
35. Find the three ratios, sin(θ), cos(θ), and tan(θ)
4
. yp ,h
6x
x2 , opp.
(May not be drawn to scale.)
sin(θ) =
opp x2 = 4 hyp 6x
cos(θ) =
x adj = 4 hyp 6x
tan(θ) =
opp x2 = adj x
θ x , adj.
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pg. 10
Trigonometry Sec. 02
MathHands.com M´ arquez
INVERSE FUNCTIONS The Idea Now we turn our attention to solving for the angles on a right triangle. We also take the opportunity to emphasize what it means for us to completely solve a triangle. We see triangles as 6 pieces of information, 3 sides and 3 corresponding angles. To solve the triangle will mean for us to determine all 6 items, the 3 sides and the 3 angles. The last section, using sin(θ), cos(θ) and tan(θ), provided for us all the ratios for all the triangles, and using the ratios and just one side, we can easily solve all sides. That was a a major milestone, and we are now ready to complete the solving of right triangles challenge by learning to solve for angles. The Inverse Functions
?
opp ratio to find hyp we used the ratio sin(23◦ )
23◦
hyp
×0.3907
opp
hyp
opp
To solve for angles, we introduce the idea of inverse trig functions. Let us take an example from the past. For a 23-67 triangle we used the sin(23◦ ) to determine the opp/hyp ratio, approximately .3907.
23◦ adj
adj
×0.3907
opp ratio is If the hyp known, 0.3907, we use sin−1 (0.3907)
??
hyp
×0.3907
opp
hyp
opp
The question we now contemplate is what if we knew the ratio, but not the angle?. How then could we determine the angles. One primitive approach would be to use a protractor, assuming we can draw such a triangle to correct scale, a protractor would help us get a working approximation of the angle. But there is more. A more accurate answer is to use the sine inverse function. The sine inverse function take ratios as input values and it gives angles as output values. The sine inverse function is often written as sin−1 (k), where k is the ratio for which we want to find the angle. The sine inverse function1 is sometimes also known as the arc-sine function and written as arcsin(k) both names and notations are common and most scientific calculators have these functions built-in. You should become familiar with such functions on your calculator [also with the radians-degree settings on the calculator]. While we are on the calculator topic, while you use the calculator to compute these values, your self-conscious may feel a bit uneasy, it should feel as though you are not really learning how or where these values come from. You should embrace such feelings as these will foster an enduring anticipation for the day when you really learn where and how these values are computed. For now we make do with our calculators and at least begin to understand what these inverse functions do for us.
23◦ adj
adj
Here is another look at the a similar problem. There is a famous triangle going around, the 3, 4, 5 triangle. It happens to be a right triangle, and the sides are famous with 32 + 42 = 52 . The sides are famous, but here we ask, what are the angles? Have you ever wondered? 1 this
definition will be extended to a broader sense later
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pg. 1
Trigonometry Sec. 02
MathHands.com M´ arquez
5
??
4
×4
5
5
×4
4
5
opp If the hyp ratio is 4 known, 5 , we use sin−1 45
53.13◦ 3
3
But wait, there is more. Now we can extend these ideas to other ratios. For example if we know the can use the arc-cosine function also known as the inverse cosine to compute the angle. For example:
0. 95
18◦
11
hyp
×
0. 92
adj
ratio, we
opp
×
??
opp
hyp
adj ratio is If the hyp known, 0.9511, we use cos−1 (0.9511)
adj hyp
05
adj
Suppose we knew the sides, adjacent and hypothenuse, then we compute the ratio, then the angle using the cosine inverse function.
hy p, 10
hy p, 10
adj ratio is If the hyp 7 , known, 10 we use 7 cos−1 10
× 10
7
7
× 10 ??
45.57◦ adj, 7
adj, 7
adj, 7
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2
×7
15.94◦
opp, 2
2
×7
??
opp, 2
Similarly for arctangent, if we know the ratio, either by knowing the ratio or by knowing the sides, opp & adj, we can find the respective angle. If the opp adj ratio is known, 27 , we use tan−1 27
adj, 7
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Trigonometry Sec. 02
MathHands.com M´ arquez
1. Find the angle in question.
2
1
8
?? ?? 10
7.
2. Find the angle in question.
8. Find the angle in question.
?? 6
4
12
?? 4
3. Find the angle in question.
9. Find the angle in question.
?? 12
1
8
?? 4
4. Find the angle in question.
10. Find the angle in question.
??
??
13
11
1
7
5. Find the angle in question.
?? 9
2
10
11. Find the angle in question.
?? 5
6. Find the angle in question.
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12. Find the angle in question.
math hands
pg. 3
Trigonometry Sec. 02
MathHands.com M´ arquez
1
2
??
?? 10
8
5
18. Find the angle in question.
2
13. Find the angle in question.
??
?? 15
8
14. Find the angle in question.
19. Find the angle in question.
5
15
2
?? 8
??
15. Find the angle in question.
20. Find the angle in question.
7
4
?? 12
?? 11
16. Find the angle in question.
21. Find the angle in question.
4
15
??
?? 2
17. Find the angle in question.
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12
22. Find the angle in question.
math hands
pg. 4
Trigonometry Sec. 02
MathHands.com M´ arquez ?? 7
2
11
?? 15
23. Find the angle in question. 25. Find the angle in question. ??
5
5
?? 8
8
24. Find the angle in question.
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pg. 5
Trigonometry Sec. 02
MathHands.com M´ arquez
INVERSE FUNCTIONS : SOLUTIONS
1. To find the angle in question:
6. To find the angle in question:
, 8 hyp
opp, 2
= 0.2 so we The ratio −1 use tan (0.2)
×0
11.31◦
.2
opp, 1
opp = 0.125 so The ratio hyp −1 we use sin (0.125)
opp adj
×0.125
7.18◦
adj, 10
7.
2. To find the angle in question:
8. To find the angle in question: 41.81
, 6 hyp
◦
opp The ratio hyp = 0.3333 so −1 we use sin (0.3333)
×0
2 , 1 hyp
×0.3333
.66
opp, 4
opp The ratio hyp = 0.6667 so −1 we use sin (0.6667)
67
19.47◦
opp, 4
3. To find the angle in question: 9. To find the angle in question:
adj The ratio hyp = 0.125 so −1 we use cos (0.125)
82.82◦
adj, 1
, 8 hyp
adj = 0.3333 so The ratio hyp we use cos−1 (0.3333)
×0.125
2 , 1 hyp
×0 .33 33
70.53◦
adj, 4
4. To find the angle in question: opp = 0.0769 so The ratio hyp −1 we use sin (0.0769)
10. To find the angle in question:
4.41◦
3 , 1 hyp
opp = 0.6364 so The ratio hyp −1 we use sin (0.6364)
1 , 1 hyp
64
69
.63
.07
×0
×0 opp, 1
opp, 7
5. To find the angle in question:
11. To find the angle in question: opp = 0.5556 so The ratio hyp −1 we use sin (0.5556)
×0.2
, 9 hyp
opp, 2
opp = 0.2 so we The ratio hyp −1 use sin (0.2)
0 , 1 hyp
39.52◦
33.75◦
×0 .55 56
11.54◦
opp, 5
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Trigonometry Sec. 02
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The ratio opp adj = 5 so we use tan−1 (5)
opp, 1
The ratio opp adj = 0.125 so we use tan−1 (0.125)
12. To find the angle in question:
adj, 2
78.69◦
7.13
×5
×0
◦
5 .12
adj, 8
opp, 10
18. To find the angle in question: 13. To find the angle in question:
×0
7.59◦
.1
333
opp, 2
The ratio opp adj = 0.1333 so we use tan−1 (0.1333)
×0
32.01◦
opp, 5
The ratio opp adj = 0.625 so we use tan−1 (0.625)
5 .62
adj, 8
adj, 15
19. To find the angle in question: adj = 0.1333 so The ratio hyp we use cos−1 (0.1333)
opp hyp −1
= 0.625 so (0.625) , 8 hyp
5 , 1 hyp
opp, 5
The ratio we use sin
×0.625
82.34◦ ×0.1333
adj, 2
14. To find the angle in question:
38.68◦ 20. To find the angle in question:
15. To find the angle in question:
70.53◦ ×0.3333
adj, 4
2 , 1 hyp
32.47
×0
◦
6 .63
4
opp, 7
The ratio opp adj = 0.6364 so we use tan−1 (0.6364)
adj The ratio hyp = 0.3333 so we use cos−1 (0.3333)
adj, 11
adj = 0.1333 so The ratio hyp −1 we use cos (0.1333)
21. To find the angle in question: The ratio opp adj = 0.3333 so we use tan−1 (0.3333)
5 , 1 hyp
×0
18.43
.13 33
82.34
◦
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3 .33
3
adj, 12
adj, 2
17. To find the angle in question:
×0
◦
opp, 4
16. To find the angle in question:
22. To find the angle in question:
math hands
pg. 2
Trigonometry Sec. 02
MathHands.com M´ arquez
1 , 1 hyp
adj, 2
82.41◦
opp = 0.6364 so The ratio hyp −1 we use sin (0.6364)
×7
×0.6364
opp, 7
The ratio opp adj = 7.5 so we use tan−1 (7.5)
.5 39.52◦
opp, 15
23. To find the angle in question: 25. To find the angle in question:
adj, 5
57.99◦
×1
adj = 0.625 so The ratio hyp we use cos−1 (0.625)
, 8 hyp
.6
51.32◦ ×0.625
adj, 5
The ratio opp adj = 1.6 so we use tan−1 (1.6)
opp, 8
24. To find the angle in question:
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pg. 3
Trigonometry Sec. 03
MathHands.com M´ arquez
MISSION ACCOMPLISHED The Idea We have done it! At this point, we have developed enough ideas to fulfill one of the fundamental promises of trigonometry, to be able to measure things by simply looking at them. That is, roughly speaking, we now have enough tools to be able to measure distances by looking at a couple points, measuring one or a couple angles and a side or a piece of a side. This is one of the essential goals of trigonometry, and we are ready to declare this mission accomplished. We introduce no new ideas here, rather, we simply show a few examples of how to use the ideas we have developed. Look at a tree and measure it: Example A: In this example, suppose we want to measure the height of a palm tree. Furthermore suppose we complete access to the area around the palm tree. We start at the trunk and walk away from the palm tree 20 ft. We lay down look up, and there it is, we determine the hight of the palm-tree.
To solve this, we need to do more than just look at the tree. We need to measure the angle from the ground to the top of the tree. Moreover, we make some extraordinary assumptions. We assume, for example that the ground is perfectly flat and the tip of the palm-tree lies exactly over the trunk, etc, etc. ... We then turn the unknown quantity, the hight of the palm-tree, into the side of a right triangle. Suppose upon laying down and looking at the tree we measure and angle of 35◦ . Then we note:
tree
p hy
35◦ 20ft.
We then invoke our ratio functions, in this case, we know the side adjacent to 35◦ and want the side opposite. We recall the tangent function is taylor made to describe such ratio, opp/adj. We make note of it on the diagram to obtain:
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Trigonometry Sec. 03
MathHands.com M´ arquez
×
35◦
f irst, we note tan(35◦ ) ≈ 0.7002
tree
p hy
0 0.7
then, tree ≈ (20f t.)(0.7002)
02
≈ 14f t.
20ft.
Look at a tree and measure it: Example B: Suppose we change things just a little. This time we don’t even come near the trunk of the palm-tree. This time we simply glance from far away measure the angle and then determine the height of the palm-tree. This time we make use of a fence directly next to the tree. It just so happens that we know this type of fence is exactly 6 ft tall and at exactly eye level. We measure the following angles, from that we can determine the heigh of the tree.
15◦ 5◦
Solutions: We will solve this by looking at the diagram, assuming we have two right triangles, and solving these two triangles. Let us look at the bottom triangles first.
15◦ 5◦
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6 ft.
math hands
pg. 2
Trigonometry Sec. 03
MathHands.com M´ arquez
We use the tangent function to determine the opp/adj ratio, then determine the adj side. In this step we effectively figure out how far away from the palm-tree we are standing. Note tan(5◦ ) ≈ 0.0875
15◦
68.58 ft. 5◦
×0.08
75
6 ft.
We now turn our attention to the other right triangle. We find have the adjacent side, want the opposite side, thus we use the tangent function tan(15◦ ) ≈ 0.2679 Thus...
”AF”
.2 ×0 15◦
679
68.58 ft. 5◦
×0.08
75
6 ft.
So we have the palm-tree divided into two sections the section below the fence and above the fence. The fence is 6ft. tall so we need only find the section above the fence, let us call it ”AF” for above fence. Then... AF ≈ (68.58)(0.2679) ≈ 18.38f t.
total tree ≈ 24.38 f t. Look at a tree and measure it: Example C:
That as a nice challenge for about 2 minutes. We now make it a bit more interesting. What if there was not fence and we could not come close to the palm-tree? It turns out we can again overcome this challenge. We simply measure an angle from far away, as illustrated below, then we move away some fixed distance, such as 8 ft., we measure the angle again, and voila! done! we can then tell the hight of the tree.
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pg. 3
Trigonometry Sec. 03
MathHands.com M´ arquez
12◦ 8 f t.
15◦
solutions: To solve this we will look for two right triangles. Note we denote the hight of the palm-tree as tree, and the initial distance to the tree as x. Then we analyze the inner triangle.
tree = (x)(0.2679) 79 .26 ×0 12◦ 8 f t.
15◦ x f t.
Then the outer triangle.
tree = (x + 8)(0.2126) 26 .21 ×0
12◦ 8 f t.
15◦ x f t. x+8
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pg. 4
Trigonometry Sec. 03
MathHands.com M´ arquez
We now have two expression for ”tree”. On the one hand tree = (x)(0.2679) while on the other hand, tree = (x + 8)(0.2126) . We now set these equal to each other and solve for x, the first distance to the tree, then ultimately solve for the height of the tree.
(x + 8)(0.2126) = (x)(0.2679) x(0.2126) + (8)(0.2126) = x(0.2679) (8)(0.2126) = x(0.2679) − x(0.2126) (8)(0.2126) = x(0.2679 − 0.2126) 1.7005 = x(0.0554) 1.7005 =x 0.0554 x ≈ 30.6949 f t.
and finally, tree = (x)(0.2679) = (30.6949)(0.2679) ≈ 8.2247 f t.
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pg. 5
Trigonometry Sec. 03
MathHands.com M´ arquez
1. Assume upon walking 20 ft. you lay down and measure the angle to the top of the tree to be 35◦ . Find the height of the tree.
(You may also assume the line of sight, bottom of the tree and top of the tree form a right triangle. ) 2. Suppose the fence as well as eye-level are 6 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
11◦ 4◦
3. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
12◦ 8 f t.
15◦
4. Suppose you know the height of your neighbor’s palm treed is 17 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
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pg. 6
Trigonometry Sec. 03
MathHands.com M´ arquez
75◦
5◦
5. Assume upon walking 30 ft. you lay down and measure the angle to the top of the tree to be 19◦ . Find the height of the tree.
(You may also assume the line of sight, bottom of the tree and top of the tree form a right triangle. ) 6. Suppose the fence as well as eye-level are 5.5 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
9◦ 3◦
7. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
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pg. 7
Trigonometry Sec. 03
MathHands.com M´ arquez
16◦ 10 f t.
18◦
8. Suppose you know the height of your neighbor’s palm treed is 20 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
69◦
3◦
9. Solve for the angle θ
5 in. θ
12 in.
2 in.
10. How long is the belt?
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pg. 8
Trigonometry Sec. 03
MathHands.com M´ arquez
5 in. θ
r
11.
x
37◦
15 x
50◦
r
13.
17 x
23◦
r
14.
49◦
2 in.
12
r
12.
12 in.
8 x
15. Just by looking at him, can you figure out how big Diego’s head is? See picture for angles & measurements.
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pg. 9
Trigonometry Sec. 03
MathHands.com M´ arquez
40◦
49◦
2 in.
16. Determine the distance from the tree trunk to the opposite side of the river as shown.
river 17◦
25◦
10 yd.
17. (a) Determine the distance from the tree trunk to the opposite side of the river as shown.
river 20◦
23◦
5 yd. (b) (con’t from previous exercise) How far away is the tree from the river? Use the diagram below.
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pg. 10
Trigonometry Sec. 03
MathHands.com M´ arquez
d
river 18◦
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pg. 11
Trigonometry Sec. 03
MathHands.com M´ arquez
MISSION ACCOMPLISHED : SOLUTIONS 1. Assume upon walking 20 ft. you lay down and measure the angle to the top of the tree to be 35◦ . Find the height of the tree.
f irst, we note tan(35◦ ) ≈ 0.7002 then,
tree
p hy
×0
35◦
.7
tree ≈ (20f t.)(0.7002) ≈ 14f t.
2 00
20ft.
We then invoke our ratio functions, in this case, we know the side adjacent to 35◦ and want the side opposite. We recall the tangent function is taylor made to describe such ratio, opp/adj. We make note of it on the diagram to obtain: 2. Suppose the fence as well as eye-level are 6 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
”AF”
.1 ×0 11◦
944
85.8 ft. 4
◦
×0.06
99
6 ft.
First we find tan(4◦ ) ≈ 0.0699 and we use the heigh of the fence 6 ft. to find the distance to the tree. Then we use the tan(11◦ ) ≈ 0.1944 ration to go from distance to the tree to distance above the fence ”AF”... thus.. AF ≈ (85.8 f t.)(0.1944) ≈ 16.68f t.
total tree ≈ 22.68 f t. 3. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
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tree = (x + 8)(0.2126)
26 79 .21 26 ×0 0. ×
12◦ 8 f t.
15◦ x f t.
tree = (x)(0.2679) then set equal and solve (x)(0.2679)=(x + 8)(0.2126) x ≈ 30.6949 f t. and tree ≈ 8.2247f t.
x+8
4. Suppose you know the height of your neighbor’s palm treed is 17 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
75◦
5◦
distance to tree = 17 ÷ tan(5◦ ) ≈ 194.3109 f t. then.. use the outer triangle...
height of the tree = (194.3109)(tan(75◦ )) ≈ 725.1782 f t. 5. Assume upon walking 30 ft. you lay down and measure the angle to the top of the tree to be 19◦ . Find the height of the tree.
tree
p hy
.3 ×0
19◦
44
3
f irst, we note tan(19◦ ) ≈ 0.3443
then, tree ≈ (30f t.)(0.3443) ≈ 14f t.
30ft.
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We then invoke our ratio functions, in this case, we know the side adjacent to 19◦ and want the side opposite. We recall the tangent function is taylor made to describe such ratio, opp/adj. We make note of it on the diagram to obtain: 6. Suppose the fence as well as eye-level are 5.5 ft. from the ground. Suppose the following angles are given, then find the height of the palm-tree.
.1 ×0 9◦
584
104.95 ft. 3
◦
×0.05
24
First we find tan(3◦ ) ≈ 0.0524 and we use the ”AF” heigh of the fence 5.5 ft. to find the distance to the tree. Then we use the tan(9◦ ) ≈ 0.1584 ration to go from distance to the tree to distance above the fence ”AF”... thus.. 5.5 ft.
AF ≈ (104.95 f t.)(0.1584)
≈ 16.62f t. total tree ≈ 22.12 f t. 7. Solve for the height of the tree. Use the diagram below, and make the usual assumptions.
tree = (x + 10)(0.2867)
67 49 .28 32 ×0 0. ×
16◦ 10 f t.
18◦ x f t.
tree = (x)(0.3249) then set equal and solve (x)(0.3249)=(x + x ≈ 10)(0.2867) 75.0654 f t. and tree ≈ 24.3902f t.
x+8
8. Suppose you know the height of your neighbor’s palm treed is 20 ft. tall. Suppose you spot a bird flying directly above the palm-tree. The following angles are given. Find the height at which the bird is flying.
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69◦
3◦
distance to tree = 17 ÷ tan(3◦ ) ≈ 381.6227 f t. then.. use the outer triangle...
height of the tree = (381.6227)(tan(69◦ )) ≈ 994.1611 f t. 9. Solve for the angle θ
5 in.
θ
12 in.
2 in.
Solution: to solve for θ we could look at it as one of the angles on a right triangle, by drawing a line parallel to the top portion of the belt between the pulleys. the difference in radii is 3, this is one of the sides, the hypothenuse is 12, thus cos θ =
3 12
, from here we get θ = cos−1 (3/12) ≈ 75.5◦ 10. How long is the belt?
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5 in.
θ
12 in.
2 in.
Solution:
once the angle is found.. you should proceed to find the belt one piece at a time.. to find the portions around d the pulleys you may need that s = πθ (in radians) or s = 360 · 2πr in degrees.. where s is the arc-length...
r x
37◦
11.
12
Solution: tan 37◦ = x=
12 tan 37◦
also.. sin 37◦ =
12 r
Therefore.. r =
15 tan 50◦
sin 50◦ =
≈ 19.94
15
Solution: tan 50◦ = also..
12 sin 37◦
x
50◦
x=
Therefore..
≈ 15.92
r 12.
12 x
15 x
Therefore..
≈ 12.59
15 r
Therefore.. r =
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15 sin 50◦
≈ 19.58
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Trigonometry Sec. 03
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r x
23◦
13.
17
Solution: tan 23◦ = x=
17 tan 23◦
also.. sin 23◦ =
17 r
Therefore.. r =
≈ 43.51
8
Solution: tan 49◦ = 8 tan 49◦
also.. sin 49◦ =
17 sin 23◦
x
49◦
x=
Therefore..
≈ 40.05
r 14.
17 x
8 r
8 x
Therefore..
≈ 6.95 Therefore.. r =
8 sin 49◦
≈ 10.60
15. Just by looking at him, can you figure out how big Diego’s head is? See picture for angles & measurements.
40◦
49◦
2 in.
Solution: We first call the size of his head, h and the lower side of the inner triangle, k. Then, we look at the inner triangle to conclude h tan 49◦ = k h , this means that k = tan 49◦ . Similarly, we look at the outer triangle to conclude tan 40◦ =
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h 2+k
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Trigonometry Sec. 03
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this means 2 + k = tanh40◦ Or k = −2 + then solve for h, the sought quantity.
h tan 40◦ .
Having solved for k in two different ways, we set the two equal,
h h = −2 + tan 49◦ tan 40◦ then... we move all h terms to one side factor the h and solve...... h ≈ 6.202 in.
16. Determine the distance from the tree trunk to the opposite side of the river as shown.
river
17◦
25◦
10 yd.
Solution: We first call the sought distance, h and the lower side of the inner triangle, k. Then, we look at the inner triangle to conclude h tan 25◦ = k , this means that k =
h tan 25◦ .
Similarly, we look at the outer triangle to conclude tan 17◦ =
h 10 + k
this means 10 + k = tanh17◦ Or k = −10 + tanh17◦ . Having solved for k in two different ways, we set the two equal, then solve for h, the sought quantity. h h = −10 + ◦ tan 25 tan 17◦ then... we move all h terms to one side factor the h and solve...... h ≈ 8.878 yd.
17. (a) Determine the distance from the tree trunk to the opposite side of the river as shown.
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river
20◦
23◦
5 yd.
Solution: We first call the sought distance, h and the lower side of the inner triangle, k. Then, we look at the inner triangle to conclude h tan 23◦ = k , this means that k =
h tan 23◦ .
Similarly, we look at the outer triangle to conclude tan 20◦ =
h 5+k
this means 5 + k = tanh20◦ Or k = −5 + tanh20◦ . Having solved for k in two different ways, we set the two equal, then solve for h, the sought quantity. h h = −5 + tan 23◦ tan 20◦ then... we move all h terms to one side factor the h and solve...... h ≈ 12.767 yd. (b) (con’t from previous exercise) How far away is the tree from the river? Use the diagram below.
d river
18
◦
Solution: from the above we can determine h, the distance from the trunk to the opposite side of the river. h ≈ 12.767 yd. If we can determine the width of the river we can subtract the quantities and obtain the distance, d. To figure out the width of the river, we can calculate k, the lower side of the triangle, from part a)... tan 23◦ =
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h k
pg. 8
Trigonometry Sec. 03 thus... k =
MathHands.com M´ arquez h tan 23◦
≈
12.767 tan 23◦ ◦
≈ 30.111
Then, we can use tan 18 ≈
′
river ′ , k
so.. river′ k river = k ∗ tan 18◦ river ≈ 30.111 ∗ 0.325 ′
tan 18◦ =
river ≈ 9.786 ..etc...etc.. solve for ’river’ and subtract from h to get
”trunk − to − river” = h − river ≈ 12.767 − 9.786 ≈ 2.981
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Trigonometry Sec. 04
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Reference Triangles The MAIN IDEA: The last couple sections are the cornerstone of trigonometry. These established the foundation for this chapter and for essential trigonometric ideas. A deep and meaningful understanding the trigonometry will depend on a deep and profound understanding of the definitions of the trigonometric functions. In the last section, we introduced these functions. We described them as the ratios of sides of a right triangle. We summarize these ideas with this:
SOH-CAH-TOA
p hy
× sin(θ)
θ os( ×c
×
)
θ
n( ta
Sin(θ) =
Opp Hyp
Cos(θ) =
Adj Hyp
T an(θ) =
Opp Adj
opp
Generic Triangle Ratios
θ)
adj
With these functions and our ability to calculate them using some calculator, we could effectively made good on one of the essential promises of trigonometry, to be able solve any right triangle given enough information. However, once this is accomplished, the trig functions take a life of their own. They become important in their own right, and the become in some sense bigger than the original problem they were made to solve. In this section, we turn such corner, we begin to study the trig function, not so much to solve right triangles, but simply to study their properties and behavior. The usefulness and implications of these functions are much bigger than this course. We will be content, and we will feel fortunate just to get to know these function just a tiny little bit. For example, at this point we are well versed with the meaning of sin(23◦ ) as the ratio of opp/hyp on the corresponding right triangle, we never did ponder the meaning of something like sin(275◦ ) In fact, it may seem a little wild to utter such words since our right triangles usually contained angles measuring less than 90◦ . Said differently, we defined our functions for little angles, angles between 0 and 90◦ , but as implied earlier, the function grew to mean much bigger and deeper ideas than that. Specifically, we now begin the study of reference triangles which will eventually lead to the extension of the trig functions to angles larger than 90◦ or even negative angles. Moreover, this will be consistent with the ideas we have thus far. Talk is cheap! Let us begin to define the very important concept of a reference triangle and the idea that every angle has one [at least]. Without further ado, here is the definition of a reference triangle. EVERY ANGLE has a REFERENCE TRIANGLE: Given any angle, θ: 1. We choose a line segment of any positive length, r, and place the segment one endpoint on origin of the xy − plane, the other endpoint on the positive side of the x-axis. c
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2. Rotate the segment, maintaining the endpoint at the origin fixed, θ units (degrees or radians). Rotate positive angles counterclockwise, rotate negative angles clockwise. 3. From the endpoint of the segment not on the origin, draw a perpendicular to the x-axis. This will create a triangle or an almost triangle. This is called and defined to be the reference triangle for angle θ 4. The fourth step is to label the sides of the reference triangle, including signs where appropriate. This means, up is positive down is negative, right positive, left is negative, as usual, for now we will keep the r positive. In later chapters, we will have a meaningful interpretation for negative r’s. One should note this last step will be possible, with our current level of knowledge, only for famous angles, such as 0◦ , 30◦ , 45◦ , 60◦ , etc...
STEP 1. Choose any positive length, r
STEP 3. Draw a Perp to x-axis
STEP 2. Rotate θ
y
y 2
r=2
STEP 4. Label sides/signs
y positive angles
θ
2
y
x
2
?
θ x
θ x
?
negative angles
Examples: Example: Draw the reference triangle for 150◦ with a line segment of length 2.
STEP 1. Choose any positive length, r STEP 2. Rotate 150◦
y
STEP 3. Draw a Perp to x-axis
y 2
r=2
STEP 4. Label sides/signs
y positive angles
θ
2
y 1
150◦
x
2
150◦
√ − 3
x negative angles
x
Example: Draw the reference triangle for 150◦ with a line segment of length 10.
STEP 1. Choose any positive length, r STEP 2. Rotate 150◦
y
STEP 3. Draw a Perp to x-axis
y 10
r=2
y positive angles
θ
x
10
y 5
150◦ x
negative angles
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STEP 4. Label sides/signs
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10
150◦
√ −5 3
x
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Trigonometry Sec. 04
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Draw the reference triangle for 60◦ . Use a line segment of length 4 units, and do not use calculators.
STEP 1. Choose any positive length, r STEP 2. Rotate 60◦
r=4
STEP 4. Label sides/signs
y
60◦
4
y 4
y 4
y
STEP 3. Draw perp. to x
60◦
√ 2 3
60◦
x
2
Draw the reference triangle for 225◦. Use a line segment of length 10 units, and do not use calculators.
STEP 1. Choose any positive length, r STEP 2. Rotate 225◦
y
STEP 3. Draw perp. to x
y
r = 10
y 225◦
y −10 √ 2
225◦ −10 √ 2
225◦
10
10
10
x
STEP 4. Label sides/signs
Finally, it should be noted that the hypothenuse is always, by definition kept positive, and that it’s length can be any positive real number, thus in reality every angle has many reference triangles, depending on how big the initial segment is drawn.
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1. Draw and label the reference triangle for 210◦ , using a segment of length 5 , and no calculators. y
x
2. Draw and label the reference triangle for 330◦ , using a segment of length 10 , and no calculators. y
x
3. Draw and label the reference triangle for 30◦ , using a segment of length 15 , and no calculators.
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Trigonometry Sec. 04
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x
4. Draw and label the reference triangle for 150◦ , using a segment of length K , and no calculators. y
x
5. Draw and label the reference triangle for −150◦ , using a segment of length R , and no calculators.
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Trigonometry Sec. 04
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x
6. Draw and label the reference triangle for 120◦ , using a segment of length 8 , and no calculators. y
x
7. Draw and label the reference triangle for −120◦ , using a segment of length 12 , and no calculators.
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Trigonometry Sec. 04
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x
8. Draw and label the reference triangle for 390◦ , using a segment of length 9 , and no calculators. y
x
9. Draw and label the reference triangle for −210◦ , using a segment of length 14 , and no calculators.
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Trigonometry Sec. 04
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x
10. Draw and label the reference triangle for −60◦ , using a segment of length 6 , and no calculators. y
x
11. Draw and label the reference triangle for 750◦ , using a segment of length 2 , and no calculators.
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Trigonometry Sec. 04
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x
12. Draw and label the reference triangle for −120◦ , using a segment of length 1 , and no calculators. y
x
13. Draw and label the reference triangle for −30◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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x
14. Draw and label the reference triangle for 45◦ , using a segment of length 7 , and no calculators. y
x
15. Draw and label the reference triangle for −135◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
16. Draw and label the reference triangle for 135◦ , using a segment of length 10 , and no calculators. y
x
17. Draw and label the reference triangle for −45◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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x
18. Draw and label the reference triangle for 225◦ , using a segment of length 10 , and no calculators. y
x
19. Draw and label the reference triangle for −225◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
20. Draw and label the reference triangle for 495◦ , using a segment of length 7 , and no calculators. y
x
21. Draw and label the reference triangle for 315◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
22. Draw and label the reference triangle for −315◦ , using a segment of length 10 , and no calculators. y
x
23. Draw and label the reference triangle for 405◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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x
24. Draw and label the reference triangle for −405◦ , using a segment of length 10 , and no calculators. y
x
25. Draw and label the reference triangle for 675◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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x
26. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
x
27. Draw and label the reference triangle for 180◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
28. Draw and label the reference triangle for 450◦ , using a segment of length 7 , and no calculators. y
x
29. Draw and label the reference triangle for −270◦ , using a segment of length 3 , and no calculators.
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Trigonometry Sec. 04
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x
30. Draw and label the reference triangle for −90◦ , using a segment of length 8 , and no calculators. y
x
31. Draw and label the reference triangle for 0◦ , using a segment of length 8 , and no calculators.
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Trigonometry Sec. 04
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x
32. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
x
33. Draw and label the reference triangle for −450◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
34. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
x
35. Draw and label the reference triangle for 720◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
36. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
x
37. Draw and label the reference triangle for −630◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
38. Draw and label the reference triangle for 360◦ , using a segment of length 5 , and no calculators. y
x
39. Draw and label the reference triangle for −360◦ , using a segment of length 5 , and no calculators.
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Trigonometry Sec. 04
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x
40. Draw and label the reference triangle for −810◦ , using a segment of length 5 , and no calculators. y
x
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Reference Triangles : SOLUTIONS 1. Draw and label the reference triangle for 210◦ , using a segment of length 5 , and no calculators. y
210◦
√ − 52 · 3 − 25
x
5
Solution:
2. Draw and label the reference triangle for 330◦ , using a segment of length 10 , and no calculators. y
330◦ 10 2
·
√ 3
10
x − 10 2
Solution:
3. Draw and label the reference triangle for 30◦ , using a segment of length 15 , and no calculators.
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Trigonometry Sec. 04
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15 2
15 30◦ 15 2
·
√ 3
x
Solution:
4. Draw and label the reference triangle for 150◦ , using a segment of length K , and no calculators. y
K 2
K 150◦
−K 2
√ · 3
x
Solution:
5. Draw and label the reference triangle for −150◦ , using a segment of length R , and no calculators.
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Trigonometry Sec. 04
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− R2 · − R2
√
−150◦
3
x
R
Solution:
6. Draw and label the reference triangle for 120◦ , using a segment of length 8 , and no calculators. y
8 2
·
√
3
8
120◦
− 82
x
Solution:
7. Draw and label the reference triangle for −120◦ , using a segment of length 12 , and no calculators.
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Trigonometry Sec. 04
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−120◦
x
− 12 2
− 12 2 ·
√ 3
12
Solution:
8. Draw and label the reference triangle for 390◦ , using a segment of length 9 , and no calculators. y
9 2
9 390◦ 9 2
·
√
3
x
Solution:
9. Draw and label the reference triangle for −210◦ , using a segment of length 14 , and no calculators.
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Trigonometry Sec. 04
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14 2
14
− 14 2
−210◦
√ · 3
x
Solution:
10. Draw and label the reference triangle for −60◦ , using a segment of length 6 , and no calculators. y
−60◦
x
6 2
6
− 62 ·
√ 3
Solution:
11. Draw and label the reference triangle for 750◦ , using a segment of length 2 , and no calculators.
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Trigonometry Sec. 04
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2 2
2 750◦ 2 2
·
√
3
x
Solution:
12. Draw and label the reference triangle for −120◦ , using a segment of length 1 , and no calculators. y
−120◦
− 12
− 12 ·
√ 3
x
1
Solution:
13. Draw and label the reference triangle for −30◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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−30◦ 7 2
·
√
3
x − 27
7
Solution:
14. Draw and label the reference triangle for 45◦ , using a segment of length 7 , and no calculators. y
7
√7 2
45◦ √7 2
x
Solution:
15. Draw and label the reference triangle for −135◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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−135◦
10 −√ 2
10 −√ 2
x
10
Solution:
16. Draw and label the reference triangle for 135◦ , using a segment of length 10 , and no calculators. y
10 √ 2
10
135◦ 10 −√ 2
x
Solution:
17. Draw and label the reference triangle for −45◦ , using a segment of length 7 , and no calculators.
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Trigonometry Sec. 04
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−45◦
√7 2
7
x
− √72
Solution:
18. Draw and label the reference triangle for 225◦ , using a segment of length 10 , and no calculators. y
225◦ 10 −√ 2
10 −√ 2
x
10
Solution:
19. Draw and label the reference triangle for −225◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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10 √ 2
10
−225◦
10 −√ 2
x
Solution:
20. Draw and label the reference triangle for 495◦ , using a segment of length 7 , and no calculators. y
√7 2
7
495◦
− √72
x
Solution:
21. Draw and label the reference triangle for 315◦ , using a segment of length 10 , and no calculators.
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Trigonometry Sec. 04
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315◦ 10 √ 2
x
10
10 −√ 2
Solution:
22. Draw and label the reference triangle for −315◦ , using a segment of length 10 , and no calculators. y
10
−315◦
10 √ 2
10 √ 2
x
Solution:
23. Draw and label the reference triangle for 405◦ , using a segment of length 7 , and no calculators.
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math hands
pg. 11
Trigonometry Sec. 04
MathHands.com M´ arquez y
7
√7 2
405◦ √7 2
x
Solution:
24. Draw and label the reference triangle for −405◦ , using a segment of length 10 , and no calculators. y
−405◦
10 √ 2
x
10
10 −√ 2
Solution:
25. Draw and label the reference triangle for 675◦ , using a segment of length 10 , and no calculators.
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math hands
pg. 12
Trigonometry Sec. 04
MathHands.com M´ arquez y
675◦ 10 √ 2
x
10
10 −√ 2
Solution:
26. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
≈ 270◦
≈0
x
5−5
Solution:
27. Draw and label the reference triangle for 180◦ , using a segment of length 5 , and no calculators.
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math hands
pg. 13
Trigonometry Sec. 04
MathHands.com M´ arquez y
≈0
≈ 180◦
5 −5
x
Solution:
28. Draw and label the reference triangle for 450◦ , using a segment of length 7 , and no calculators. y
77
≈ 450◦
≈0
x
Solution:
29. Draw and label the reference triangle for −270◦ , using a segment of length 3 , and no calculators.
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math hands
pg. 14
Trigonometry Sec. 04
MathHands.com M´ arquez y
33
≈ −270◦
≈0
x
Solution:
30. Draw and label the reference triangle for −90◦ , using a segment of length 8 , and no calculators. y
≈ −90◦
≈0
x
8−8
Solution:
31. Draw and label the reference triangle for 0◦ , using a segment of length 8 , and no calculators.
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math hands
pg. 15
Trigonometry Sec. 04
MathHands.com M´ arquez y
≈ 0◦
8 8
x
≈0
Solution:
32. Draw and label the reference triangle for 270◦ , using a segment of length 5 , and no calculators. y
≈ 270◦
≈0
x
5−5
Solution:
33. Draw and label the reference triangle for −450◦ , using a segment of length 5 , and no calculators.
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2007-2009 MathHands.com
math hands
pg. 16
Trigonometry Sec. 04
MathHands.com M´ arquez y
≈ −450◦
≈0
x
5−5
Solution:
34. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
≈0
5 −5
≈ −540◦
x
Solution:
35. Draw and label the reference triangle for 720◦ , using a segment of length 5 , and no calculators.
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2007-2009 MathHands.com
math hands
pg. 17
Trigonometry Sec. 04
MathHands.com M´ arquez y
≈ 720◦
5 5
x
≈0
Solution:
36. Draw and label the reference triangle for −540◦ , using a segment of length 5 , and no calculators. y
≈0
5 −5
≈ −540◦
x
Solution:
37. Draw and label the reference triangle for −630◦ , using a segment of length 5 , and no calculators.
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math hands
pg. 18
Trigonometry Sec. 04
MathHands.com M´ arquez y
55
≈ −630◦
x
≈0
Solution:
38. Draw and label the reference triangle for 360◦ , using a segment of length 5 , and no calculators. y
≈ 360◦
5 5
x
≈0
Solution:
39. Draw and label the reference triangle for −360◦ , using a segment of length 5 , and no calculators.
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2007-2009 MathHands.com
math hands
pg. 19
Trigonometry Sec. 04
MathHands.com M´ arquez y
≈ −360◦
5 5
x
≈0
Solution:
40. Draw and label the reference triangle for −810◦ , using a segment of length 5 , and no calculators. y
≈ −810◦
≈0
x
5−5
Solution:
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2007-2009 MathHands.com
math hands
pg. 20
Trigonometry Sec. 05
MathHands.com M´ arquez
The Rest of the Trig Functions THE REST of the SINE and COSINE Functions: This section is yet another immensely critical milestone in our journey to understanding all that is trigonometry. It is here and now that we extend these functions for angles larger than 90◦ or smaller than 0◦ (when possible). We will learn how to evaluate them, graph them, use them to solve triangles, solve equations with such functions in them, and we will study some of their most famous and celebrated properties. Thus, once again, the importance of this section can not be overstated. There are six very famous trig functions, sine, cosine, tangent, cotangent, cosecant, and secant. Sine and Cosine are personal favorites, in some way, all other functions are composed of variations of these. Thus, we will focus most of our attention on these two functions for now. Once comfortable with these, we will move on to the other four functions. Before we dive into the small details, we take a big-picture perspective on what these functions are. These functions are ratios. For any given angle, we construct a reference triangle. These functions are, by definition, ratios of certain sides of the corresponding reference triangle. Once angle is given and the reference triangle constructed and labeled, we stand at the origin and observe which of the sides of the triangle is adjacent, which side is opposite, and which side is the hypothenuse. The cosine function is defined to be the ratio of adjacent/opposite, denoted adj .It should be clear which side of the reference triangle is the hypothenuse, it’s across from the 90◦ angle. cos θ = opp The opposite side and the adjacent sides depend very much on our point of reference. The definitions of all our trig functions all assume the point of reference used is the origin. That is, once we use the origin as a point of reference, there is only one side which is opposite. The other sides are also easily identified, they are the adjacent and the hypothenuse. We state the definition formally: DEF: of sin θ and cos θ For any angle, θ, we construct the reference triangle for θ using any positive length r for the hypothenuse. We label the sides, ’hypothenuse’, ’adjacent’, and ’opposite’, accordingly. sin θ is defined as the ratio of the sides of the reference triangle: sin θ =
opposite hypothenuse
cos θ =
adj hypothenuse
In layman terms, you can think of a function as a cow. The cow eats grass and produces milk. In a similar way, the cosine function ’eats’ angles and produces ratios.
The Cow Function:
grass →
The Cosine Function: adj/hyp
angle 150◦ →
2
1 −
milk cow(grass)=milk
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2007-2009 MathHands.com
150◦
30◦ √ 3
ratio√ cos(150◦ ) = −2 3
math hands
The Sine Function: opp/hyp
angle 150◦ →
2
1
150◦
30◦ −
√ 3
ratio sin(150◦ ) = 21
pg. 1
Trigonometry Sec. 05
MathHands.com M´ arquez
THE REST of the Other Trig FUNCTIONS: We are now ready to meet the other famous trigonometric functions, tangent, cotangent, secant and cosecant. These are very similar to sine and cosine in the sense that they too describe ratios of sides of a corresponding reference triangle. For any given angle, θ, we construct a reference triangle, then we define the functions as follows: DEF: of the SIX trig functions
reference triangle for θ angle θ
→
ratio
hyp opp
θ
→
sin θ =
opp hyp
cos θ =
adj hyp
tan θ =
opp adj
csc θ =
hyp opp
sec θ =
hyp adj
cot θ =
adj opp
adj
It is important to direct your attention momentarily to the denominators on each of the 6 ratios. Sine and cosine each have ’hyp’ as the denominator. Keep in mind the hypothenuse is a positive length, r, of the segment chosen to construct the reference triangle. By definition, such r = hyp will never be zero. Thus, such ratios will be well defined for all possible reference triangles and in turn for all possible angles, θ. Said differently, the sine and cosine functions are defined for all possible real angles. The same can not be said of any of the other functions. Consider, for example, the case when θ = 90◦ . In such case, if hyp = 2 then opposite side is 2 and the adjacent is 0. Since the ratio 20 is not a real number, thus the tangent of 90◦ does not exist as a real number. There are many angles (all which end up in an ’almost triangle’ reference triangle) for which the other four functions are not defined.
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math hands
pg. 2
Trigonometry Sec. 05
MathHands.com M´ arquez
For each angle, draw and label a reference triangle, then determine all 6 trig ratios, sine , cosine, tangent, secant, cosecant, and cotangent. Do not use calculators here. 1. 120◦ y axis y axis
x axis x axis
5. 135◦ 2. 150◦ y axis y axis
x axis x axis
6. 225◦
3. −120◦
y axis
y axis
x axis
x axis
4. 330◦
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7. 315◦
math hands
pg. 3
Trigonometry Sec. 05
MathHands.com M´ arquez 11. −330◦
y axis
y axis
x axis x axis
8. 30◦ 12. −210◦ y axis y axis
x axis x axis
9. −30◦
13. 180◦ y axis
y axis
x axis
10. 690◦
x axis
14. −180◦ y axis
y axis
x axis
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2007-2009 MathHands.com
math hands
x axis
pg. 4
Trigonometry Sec. 05
MathHands.com M´ arquez
15. 360◦ y axis
y axis x axis
x axis
18. −90◦ y axis
16. −360◦
x axis
y axis
19. 270◦ x axis
y axis
x axis
17. 90◦
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math hands
pg. 5
Trigonometry Sec. 05
MathHands.com M´ arquez
The Rest of the Trig Functions : SOLUTIONS
For each angle, draw and label a reference triangle, then determine all 6 trig ratios, sine , cosine, tangent, secant, cosecant, and cotangent. Do not use calculators here.
opp hyp hyp csc(120◦ ) = opp adj cos(120◦ ) = hyp hyp sec(120◦ ) = adj sin(120◦ ) =
= = = =
opp tan(120◦ ) = = adj adj = cot(120◦ ) = opp
y axis
√ 2 3 4 4 √ 2 3 −2 4 4 −2 √ 2 3 −2 −2 √ 2 3
y axis opp −2 = hyp 4 hyp 4 csc(330◦ ) = = opp −2 √ 2 3 adj = cos(330◦ ) = hyp 4 hyp 4 ◦ x axis sec(330 )= = √ adj 2 3 opp −2 tan(330◦ ) = = √ adj 2 3 √ 2 3 adj = cot(330◦ ) = opp −2 sin(330◦ ) =
√ 2 3 4
120◦
−2
2 opp = hyp 4 4 hyp ◦ = csc(150 ) = opp 2
y axis
√ adj −2 3 = hyp 4 hyp 4 sec(150◦ ) = = √ adj −2 3 opp 2 √ tan(150◦ ) = = adj −2 3 √ −2 adj 3 = cot(150◦ ) = opp 2 cos(150◦ ) =
√ opp 4/ 2 = hyp 4 4 hyp = √ csc(135◦ ) = opp 4/ 2 √ adj −4/ 2 cos(135◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(135 )= adj −4/ 2 √ opp 4/ 2 tan(135◦ ) = = √ adj −4/ 2 √ −4/ 2 adj = √ cot(135◦ ) = opp 4/ 2
2 4 √ −2 3
150
◦
2.
= = = =
opp = tan(−120◦ ) = adj adj cot(−120◦ ) = = opp
4
y axis
sin(135◦ ) =
sin(150◦ ) =
opp hyp hyp csc(−120◦ ) = opp adj ◦ cos(−120 ) = hyp hyp sec(−120◦ ) = adj
x axis −2
4.
1.
sin(−120◦ ) =
3302◦√3
√ 4/ 2 4 √ −4/ 2
135◦ x axis
5.
√ −2 3 4 4 √ −2 3 −2 4 4 −2 √ −2 3 −2 −2 √ −2 3
y axis
√ −4/ 2 opp = hyp 4 4 hyp = √ csc(225◦ ) = opp −4/ 2 √ 2 adj −4/ cos(225◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(225 )= adj −4/ 2 √ 2 opp −4/ tan(225◦ ) = √ = adj −4/ 2 √ adj −4/ 2 √ cot(225◦ ) = = opp −4/ 2
y axis
sin(225◦ ) =
−120◦
−2
√ −2 3
4
3.
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2007-2009 MathHands.com
√ −4/ 2
225◦ x axis
√ −4/ 2
4
6.
math hands
pg. 1
Trigonometry Sec. 05 √ opp −4/ 2 = hyp 4 4 hyp √ = csc(315◦ ) = opp −4/ 2 √ adj 4/ 2 cos(315◦ ) = = hyp 4 4 hyp = √ sec(315◦ ) = adj 4/ 2 √ opp −4/ 2 tan(315◦ ) = √ = adj 4/ 2 √ 4/ 2 adj ◦ = √ cot(315 ) = opp −4/ 2
MathHands.com M´ arquez y axis
sin(−330◦ ) =
√◦ 315 4/ 2
4
√ −4/ 2
7.
2 opp = hyp 4 hyp 4 csc(30◦ ) = = opp 2 √ adj 2 3 cos(30◦ ) = = hyp 4 hyp 4 sec(30◦ ) = = √ adj 2 3 2 opp = √ tan(30◦ ) = adj 2 3 √ adj 2 3 cot(30◦ ) = = opp 2
y axis
opp 2 = hyp 4 hyp 4 csc(−330◦ ) = = opp 2 √ 2 3 adj = cos(−330◦ ) = hyp 4 4 x axis◦ ) = hyp = √ sec(−330 adj 2 3 2 opp = √ tan(−330◦ ) = adj 2 3 √ 2 3 adj = cot(−330◦ ) = opp 2
sin(315◦ ) =
2 4
◦ −3302√ 3
x axis
11.
y axis
sin(30◦ ) =
y axis
2 opp = hyp 4 hyp 4 csc(−210◦ ) = = opp 2 sin(−210◦ ) =
2 4 30 2 3
√ adj −2 3 = hyp 4 hyp 4 ◦ x axis sec(−210 )= = √ adj −2 3 2 opp ◦ = √ tan(−210 ) = adj −2 3 √ adj −2 3 cot(−210◦ ) = = opp 2
2 4 √ −2 3
cos(−210◦ ) =
◦√
x axis
12.
8.
y axis opp −2 = hyp 4 hyp 4 csc(−30◦ ) = = opp −2 √ 2 3 adj = cos(−30◦ ) = hyp 4 hyp 4 sec(−30◦ ) = = √ adj 2 3 opp −2 tan(−30◦ ) = = √ adj 2 3 √ adj 2 3 cot(−30◦ ) = = opp −2
−210◦
y axis
sin(−30◦ ) =
−302◦√3
≈ −2 ≈0
x axis 4
≈ 180◦ x axis
2
−2
sin(≈ 180◦ ) = csc(≈ 180◦ ) = ◦
cos(≈ 180 ) =
9.
13.
sec(≈ 180◦ ) = tan(≈ 180◦ ) = cot(≈ 180◦ ) =
y axis opp −2 = hyp 4 4 hyp = csc(690◦ ) = opp −2 √ adj 2 3 cos(690◦ ) = = hyp 4 hyp 4 sec(690◦ ) = = √ adj 2 3 opp −2 tan(690◦ ) = = √ adj 2 3 √ 2 3 adj = cot(690◦ ) = opp −2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
y axis
sin(690◦ ) =
6902◦√3
≈ −2 ≈0
x axis 4
−2
−180◦ x axis
2 ◦
sin(−180 ) = csc(−180◦ ) = cos(−180◦ ) =
10.
14.
◦
sec(−180 ) = ◦
tan(−180 ) =
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2007-2009 MathHands.com
math hands
cot(−180◦ ) =
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
pg. 2
Trigonometry Sec. 05
MathHands.com M´ arquez y axis
y axis
360◦2 ≈2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
sin(360◦ ) = csc(360◦ ) = ◦
cos(360 ) =
15.
◦
sec(360 ) = tan(360◦ ) =
y axis cot(360◦ ) =
−360◦≈ 2
sin(−360◦ ) = csc(−360◦ ) = ◦
cos(−360 ) = ◦
sec(−360 ) = tan(−360◦ ) =
y axis 2
cot(−360◦ ) =
≈2
−90◦
x axis 0 2 2 0 2 2 2 2 0 2 2 0
= = = = = =
x axis
=0
cos(−90◦ ) =
is NOT REAL 2
sec(−90◦ ) =
≈ −2
◦
sin(−90 ) =
=1 =1
18.
◦
csc(−90 ) =
=0
cot(−90◦ ) =
y axis
= is NOT REAL
≈0
≈0
2
16.
≈0
≈0
tan(−90◦ ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
= = = =
x axis
=0
cos(270◦ ) =
is NOT REAL 2 =1 =1
=
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
270◦
x axis
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
=
≈ −2
sec(270◦ ) = sin(270◦ ) =
19.
=0 = is NOT REAL
csc(270◦ ) = cot(270◦ ) = ◦
tan(270 ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
90◦ x axis
≈0 cos(90◦ ) = sec(90◦ ) = ◦
sin(90 ) =
17.
◦
csc(90 ) = cot(90◦ ) = tan(90◦ ) =
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2007-2009 MathHands.com
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
=0 is NOT REAL =1 =1 =0 = is NOT REAL
math hands
pg. 3
Trigonometry Sec. 05
MathHands.com M´ arquez
The Rest of the Trig Functions : SOLUTIONS
For each angle, draw and label a reference triangle, then determine all 6 trig ratios, sine , cosine, tangent, secant, cosecant, and cotangent. Do not use calculators here.
opp hyp hyp csc(120◦ ) = opp adj cos(120◦ ) = hyp hyp sec(120◦ ) = adj sin(120◦ ) =
= = = =
opp tan(120◦ ) = = adj adj = cot(120◦ ) = opp
y axis
√ 2 3 4 4 √ 2 3 −2 4 4 −2 √ 2 3 −2 −2 √ 2 3
y axis opp −2 = hyp 4 hyp 4 csc(330◦ ) = = opp −2 √ 2 3 adj = cos(330◦ ) = hyp 4 hyp 4 ◦ x axis sec(330 )= = √ adj 2 3 opp −2 tan(330◦ ) = = √ adj 2 3 √ 2 3 adj = cot(330◦ ) = opp −2 sin(330◦ ) =
√ 2 3 4
120◦
−2
2 opp = hyp 4 4 hyp ◦ = csc(150 ) = opp 2
y axis
√ adj −2 3 = hyp 4 hyp 4 sec(150◦ ) = = √ adj −2 3 opp 2 √ tan(150◦ ) = = adj −2 3 √ −2 adj 3 = cot(150◦ ) = opp 2 cos(150◦ ) =
√ opp 4/ 2 = hyp 4 4 hyp = √ csc(135◦ ) = opp 4/ 2 √ adj −4/ 2 cos(135◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(135 )= adj −4/ 2 √ opp 4/ 2 tan(135◦ ) = = √ adj −4/ 2 √ −4/ 2 adj = √ cot(135◦ ) = opp 4/ 2
2 4 √ −2 3
150
◦
2.
= = = =
opp = tan(−120◦ ) = adj adj cot(−120◦ ) = = opp
4
y axis
sin(135◦ ) =
sin(150◦ ) =
opp hyp hyp csc(−120◦ ) = opp adj ◦ cos(−120 ) = hyp hyp sec(−120◦ ) = adj
x axis −2
4.
1.
sin(−120◦ ) =
3302◦√3
√ 4/ 2 4 √ −4/ 2
135◦ x axis
5.
√ −2 3 4 4 √ −2 3 −2 4 4 −2 √ −2 3 −2 −2 √ −2 3
y axis
√ −4/ 2 opp = hyp 4 4 hyp = √ csc(225◦ ) = opp −4/ 2 √ 2 adj −4/ cos(225◦ ) = = hyp 4 4 hyp ◦ x axis = √ sec(225 )= adj −4/ 2 √ 2 opp −4/ tan(225◦ ) = √ = adj −4/ 2 √ adj −4/ 2 √ cot(225◦ ) = = opp −4/ 2
y axis
sin(225◦ ) =
−120◦
−2
√ −2 3
4
3.
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2007-2009 MathHands.com
√ −4/ 2
225◦ x axis
√ −4/ 2
4
6.
math hands
pg. 1
Trigonometry Sec. 05 √ opp −4/ 2 = hyp 4 4 hyp √ = csc(315◦ ) = opp −4/ 2 √ adj 4/ 2 cos(315◦ ) = = hyp 4 4 hyp = √ sec(315◦ ) = adj 4/ 2 √ opp −4/ 2 tan(315◦ ) = √ = adj 4/ 2 √ 4/ 2 adj ◦ = √ cot(315 ) = opp −4/ 2
MathHands.com M´ arquez y axis
sin(−330◦ ) =
√◦ 315 4/ 2
4
√ −4/ 2
7.
2 opp = hyp 4 hyp 4 csc(30◦ ) = = opp 2 √ adj 2 3 cos(30◦ ) = = hyp 4 hyp 4 sec(30◦ ) = = √ adj 2 3 2 opp = √ tan(30◦ ) = adj 2 3 √ adj 2 3 cot(30◦ ) = = opp 2
y axis
opp 2 = hyp 4 hyp 4 csc(−330◦ ) = = opp 2 √ 2 3 adj = cos(−330◦ ) = hyp 4 4 x axis◦ ) = hyp = √ sec(−330 adj 2 3 2 opp = √ tan(−330◦ ) = adj 2 3 √ 2 3 adj = cot(−330◦ ) = opp 2
sin(315◦ ) =
2 4
◦ −3302√ 3
x axis
11.
y axis
sin(30◦ ) =
y axis
2 opp = hyp 4 hyp 4 csc(−210◦ ) = = opp 2 sin(−210◦ ) =
2 4 30 2 3
√ adj −2 3 = hyp 4 hyp 4 ◦ x axis sec(−210 )= = √ adj −2 3 2 opp ◦ = √ tan(−210 ) = adj −2 3 √ adj −2 3 cot(−210◦ ) = = opp 2
2 4 √ −2 3
cos(−210◦ ) =
◦√
x axis
12.
8.
y axis opp −2 = hyp 4 hyp 4 csc(−30◦ ) = = opp −2 √ 2 3 adj = cos(−30◦ ) = hyp 4 hyp 4 sec(−30◦ ) = = √ adj 2 3 opp −2 tan(−30◦ ) = = √ adj 2 3 √ adj 2 3 cot(−30◦ ) = = opp −2
−210◦
y axis
sin(−30◦ ) =
−302◦√3
≈ −2 ≈0
x axis 4
≈ 180◦ x axis
2
−2
sin(≈ 180◦ ) = csc(≈ 180◦ ) = ◦
cos(≈ 180 ) =
9.
13.
sec(≈ 180◦ ) = tan(≈ 180◦ ) = cot(≈ 180◦ ) =
y axis opp −2 = hyp 4 4 hyp = csc(690◦ ) = opp −2 √ adj 2 3 cos(690◦ ) = = hyp 4 hyp 4 sec(690◦ ) = = √ adj 2 3 opp −2 tan(690◦ ) = = √ adj 2 3 √ 2 3 adj = cot(690◦ ) = opp −2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
y axis
sin(690◦ ) =
6902◦√3
≈ −2 ≈0
x axis 4
−2
−180◦ x axis
2 ◦
sin(−180 ) = csc(−180◦ ) = cos(−180◦ ) =
10.
14.
◦
sec(−180 ) = ◦
tan(−180 ) =
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2007-2009 MathHands.com
math hands
cot(−180◦ ) =
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
pg. 2
Trigonometry Sec. 05
MathHands.com M´ arquez y axis
y axis
360◦2 ≈2
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
sin(360◦ ) = csc(360◦ ) = ◦
cos(360 ) =
15.
◦
sec(360 ) = tan(360◦ ) =
y axis cot(360◦ ) =
−360◦≈ 2
sin(−360◦ ) = csc(−360◦ ) = ◦
cos(−360 ) = ◦
sec(−360 ) = tan(−360◦ ) =
y axis 2
cot(−360◦ ) =
≈2
−90◦
x axis 0 2 2 0 2 2 2 2 0 2 2 0
= = = = = =
x axis
=0
cos(−90◦ ) =
is NOT REAL 2
sec(−90◦ ) =
≈ −2
◦
sin(−90 ) =
=1 =1
18.
◦
csc(−90 ) =
=0
cot(−90◦ ) =
y axis
= is NOT REAL
≈0
≈0
2
16.
≈0
≈0
tan(−90◦ ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
= = = =
x axis
=0
cos(270◦ ) =
is NOT REAL 2 =1 =1
=
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
270◦
x axis
opp hyp hyp opp adj hyp hyp adj opp adj adj opp
=
≈ −2
sec(270◦ ) = sin(270◦ ) =
19.
=0 = is NOT REAL
csc(270◦ ) = cot(270◦ ) = ◦
tan(270 ) =
adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 =0 2 2 is NOT REAL 0 −2 = −1 2 2 = −1 −2 0 =0 −2 −2 = is NOT REAL 0
90◦ x axis
≈0 cos(90◦ ) = sec(90◦ ) = ◦
sin(90 ) =
17.
◦
csc(90 ) = cot(90◦ ) = tan(90◦ ) =
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adj hyp hyp adj opp hyp hyp opp adj opp opp adj
= = = = = =
0 2 2 0 2 2 2 2 0 2 2 0
=0 is NOT REAL =1 =1 =0 = is NOT REAL
math hands
pg. 3
Trigonometry Sec. 06
MathHands.com M´ arquez
The Rest of the Inverse Functions Main Idea: We have already introduced the idea of inverse functions earlier. While the sin(θ), cos(θ) and tan(θ) functions are great to describe the ratios for a given angle θ, and its respective triangle, the inverse functions sin−1 (k), cos−1 (k) and tan−1 (k) are great for giving the angle when the ratio is known.
hyp
×.40
4, opp
10,
opp If the hyp ratio is known, 4 = .40, we use 10 −1 hyp sin (.400) 10,
×.40
4, opp
Here is an example to help us recall our previous look at the inverse functions:
23.58◦
??
That was an important introduction to the inverse functions. Here we take the opportunity to reconcile how these functions behave once we have extended the trigonometric functions to just about all angles. The complete discussion of these functions requires a reasonable understanding of ono-to-one and onto functions. While this may not be the appropriate course for such a discussion we will engage in such ideas just enough to gain a broader and adequate perspective of the trigonometric functions. Inverse Functions: Thus far we have seen the trigonometric functions describe for each angle a particular ratio. In this section we take a moment to contemplate this question but backwards. Suppose we know a particular ratio, can we then find the corresponding angle? The answer is a definitive yes. The taylor-made medicine for the job is the study of the inverse functions. The inverse cosine is often called arc-cosine and is usually denoted as arccos(z) or cos−1 (z). Similarly of arc-sine and arc-tangent. These are defined as you would expect, given a ratio z, arccos(z) is defined as the angle θ such that cos θ = z, of course this demands some cautionary and insightful comments, which we will add soon. The following diagram should help solidify the concept of the inverse functions.
The Cow Function:
The Cow−1 Function:
milk grass → grass milk cow(grass)=milk c
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cow−1 (milk)=grass
math hands
pg. 1
Trigonometry Sec. 06
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Now, we apply the idea to cosine function at 60◦
The Cosine Function:
The Cosine Inverse Function:
ratio
angle 60◦ → cos ratio = cos 60◦ =
1 2
cos−1
1 2
cos−1
1 2
angle 60◦ 1 2
= 60◦
The Inverse Functions: Here is yet another way to illustrate the role of the inverse trigonometric functions. Let us continue with the same example.
les an g
cos
ios rat 1 2
60◦
cos−1 [domain]
[co-domain]
Consider the possibility that feeding the cow ”hot-dogs” also produces ’milk’. This adds a small complication when calculating what the inverse function, cow−1 , does to ’milk’. Namely, it’s tricky to decide wether cow −1 = grass or cow −1 = hot − dogs. In fact, in such a situation the cow function does not have an inverse. In general, when a function is not one-to-one it does not have an inverse. Such is the case for all our trigonometric functions. Although a complete discussion of one-to-one functions may not be appropriate here, we are obliged to at least scratch the surface of the topic, at least enough to understand the basic intricacies of trigonometric inverse functions. With that in mind consider the following angles, all of which result in a cosine value of 12 .
cos
les an g
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ios rat
300◦ ◦ 60 420◦
1 2
[domain]
[co-domain] math hands
pg. 2
Trigonometry Sec. 06
MathHands.com M´ arquez
Now, consider the inverse cosine function, that is if the ratio of adjacent to hypothenuse is 21 , can we recover the angle? The answer is a resounding no!. There are too many [more than one] angles for which such ratio is 12 . When this happens we say that the function is not one-to-one.
cos−1
les an g
ios rat
300◦ ?◦ 60 ? 420◦ ?
1 2
[domain]
[co-domain]
In some way, this is a remediable problem. The popular and almost universally accepted approach is to modify the domain on the cosine function as to limit angles to only angles between 0◦ and 180◦ . At the same time, the ratios have to be restricted to values between -1 and 1.
es ngl 80◦ a 1 y on l ≤ θ ≤ 0
cos
os ati 1 r y ≤ on l 1 ≤ r − 1 2
60◦
cos−1 [domain]
[co-domain]
Note how the inverse problem is resolved:
es ngl 80◦ a 1 y on l ≤ θ ≤ 0 300◦ ?
cos−1
os ati 1 r y ≤ on l 1 ≤ r − 1 2
60◦ 420◦ ?
[domain]
[co-domain]
Of the many different angles for which the cosine is 1/2, only one is in the adequate range, thus there is no ambiguity as to which angle the ratio is assigned to. SImilarly, the domain for the sine function, as well as the tangent function can be restricted so that an inverse function can be constructed for each one of these.
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pg. 3
Trigonometry Sec. 06
MathHands.com M´ arquez es ◦ ngl ≤ 90 a θ y onl 90◦ ≤ −
os ati 1 r y ≤ on l 1 ≤ r −
sin
r
θ
sin−1 [domain]
[co-domain]
For tangent,
es ◦ ngl < 90 a y θ onl 90◦ < −
l rea al l tios ra
tan
r
θ
tan−1 [domain]
[co-domain]
You should know that most scientific calculators are programed to compute inverse trig functions angles according to these ranges. Moreover, you should be aware of alternative names for these functions. arccos(x) is sometimes used to denote cos−1 (x), and similarly arctan(x) = tan−1 (x), and arcsin(x) = sin−1 (x) Examples: Example: Find the angle θ
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 θ = tan−1 2
5
≈ 1.19
θ
≈ 68.199
2
In this case, we know the opposite and the adjacent sides. The function describing such ratio is the tangent, thus 5 tan θ = 2
(in radians [see calculator mode]... OR...) ◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation tan θ = 52 under a different context, where we will solve it completely, not limited to the domain and codomain of the arctan function.
Example:
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math hands
pg. 4
Trigonometry Sec. 06
MathHands.com M´ arquez
Find the angle θ
3
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 −1 θ = sin 3
2
≈ 0.73
≈ 41.81
θ In this case, we know the opposite and the hypothenuse sides. The function describing such ratio is the sine, thus 2 sin θ = 3
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math hands
(in radians [see calculator mode]... OR...) ◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation sin θ = 32 under a different context, where we will solve it completely, not limited to the domain and codomain of the arcsin function.
pg. 5
Trigonometry Sec. 06
MathHands.com M´ arquez
1. Find the angle θ 6
5
θ
2. Find the angle θ 3 θ
2
3. Find the angle θ 11
2
θ
4. Find the angle θ 5 θ
2
5. Find the angle θ 5 θ
1
6. Find the angle θ 4 θ
3
7. Find the angle θ 9 θ
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math hands
2
pg. 6
Trigonometry Sec. 06
MathHands.com M´ arquez
8. cos−1 (cos(30◦ )) = 30◦ A. TRUE
B. FALSE
9. cos−1 (cos(−30◦ )) = −30◦ A. TRUE
B. FALSE
10. tan−1 (tan(30◦)) = 30◦ A. TRUE
B. FALSE
11. tan−1 (tan(210◦)) = 210◦ A. TRUE
B. FALSE
12.
−1
1 1 = 3 3
−1
5 5 = 3 3
tan tan A. TRUE
B. FALSE
13.
tan tan A. TRUE
B. FALSE
14.
5 5 −1 = cos cos 13 13 A. TRUE
B. FALSE
15. Compute sin cos−1 (.3456) 16. Compute tan cos−1 (.1234)
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math hands
pg. 7
Trigonometry Sec. 06
MathHands.com M´ arquez
The Rest of the Inverse Functions : SOLUTIONS 1. Find the angle θ Solution: In this case, we know the opposite and the hypothenuse sides. The function describing such ratio is the sine, thus sin θ =
6
5 6
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 θ = sin−1 6
5
θ
≈ 0.985 (in radians [see calculator mode]... OR...) ≈ 56.443◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation sin θ = 56 under a different context, where we will solve it completely, not limited to the domain and codomain of the arcsin function.
2. Find the angle θ Solution: In this case, we know the adjacent and the hypothenuse sides. The function describing such ratio is the cosine, thus cos θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 θ = cos−1 3
3 θ
2 3
2
≈ 0.841 (in radians [see calculator mode]... OR...) ≈ 48.19◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation cos θ = 23 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
3. Find the angle θ 11
2
θ
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math hands
pg. 1
Trigonometry Sec. 06
MathHands.com M´ arquez
Solution: In this case, we know the opposite and the hypothenuse sides. The function describing such
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math hands
pg. 2
Trigonometry Sec. 06
MathHands.com M´ arquez
ratio is the sine, thus sin θ =
2 11
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 θ = sin−1 11 ≈ 0.183 (in radians [see calculator mode]... OR...) ≈ 10.476◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equa2 tion sin θ = 11 under a different context, where we will solve it completely, not limited to the domain and codomain of the arcsin function.
4. Find the angle θ Solution: In this case, we know the opposite and the adjacent sides. The function describing such ratio is the tangent, thus tan θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 θ = tan−1 2 ≈ 1.19 (in radians [see calculator mode]... OR...)
5 θ
5 2
2
≈ 68.199◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation tan θ = 52 under a different context, where we will solve it completely, not limited to the domain and codomain of the arctan function.
5. Find the angle θ Solution: In this case, we know the opposite and the adjacent sides. The function describing such ratio is the tangent, thus 5 tan θ = 1 Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 5 −1 θ = tan 1
5 θ
1
≈ 1.373 (in radians [see calculator mode]... OR...)
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≈ 78.69◦
(in degrees [see calculator mode])
pg. 3
Trigonometry Sec. 06
MathHands.com M´ arquez
... it should be noted that we will revisit the equation tan θ = 15 under a different context, where we will solve it completely, not limited to the domain and codomain of the arctan function.
6. Find the angle θ Solution: In this case, we know the adjacent and the hypothenuse sides. The function describing such ratio is the cosine, thus cos θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 3 −1 θ = cos 4 ≈ 0.723 (in radians [see calculator mode]... OR...)
4 θ
3 4
3
≈ 41.41◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation cos θ = 34 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
7. Find the angle θ Solution: In this case, we know the adjacent and the hypothenuse sides. The function describing such ratio is the cosine, thus cos θ =
Since this is not a famous ratio, we allow ourselves use of a calculator to estimate the sought angle. 2 θ = cos−1 9
9 θ
2 9
2
≈ 1.347 (in radians [see calculator mode]... OR...) ≈ 77.16◦
(in degrees [see calculator mode])
... it should be noted that we will revisit the equation cos θ = 29 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
8. cos−1 (cos(30◦ )) = 30◦
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math hands
pg. 4
Trigonometry Sec. 06 A. TRUE
MathHands.com M´ arquez
B. FALSE
9. cos−1 (cos(−30◦ )) = −30◦ A. TRUE
B. FALSE
10. tan−1 (tan(30◦)) = 30◦ A. TRUE
B. FALSE
11. tan−1 (tan(210◦)) = 210◦ A. TRUE
B. FALSE
12.
−1
1 1 = 3 3
−1
5 5 = 3 3
tan tan A. TRUE
B. FALSE
13.
tan tan A. TRUE
B. FALSE
14.
5 5 −1 cos cos = 13 13 A. TRUE
B. FALSE
15. Compute sin cos−1 (.3456)
Solution: First, the stuff inside, cos−1 (.3456) ≈ 69.78◦ then...
sin (69.78◦) ≈ 0.9384 16. Compute
tan cos−1 (.1234)
Solution: First, the stuff inside, cos−1 (.1234) ≈ 82.91◦ then...
tan (82.91◦ ) ≈ 8.0418
note: the self quiz on ”mission accomplished” prob #2 outlines a different approach to this same question.
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math hands
pg. 5
math hands Trigonometry: Ch2 guide
Reference Triangles
Every angle has a reference triangle. The recipe is as follows:
Def: Inverse Cosine Function STEP 1. Choose any + r
STEP 2. Rotate θ
y
es ◦ ngl 80 ya ≤1 onl θ ◦ ≤ 0
y positive angles
2
r
θ
θ
r=2
os ati 1 yr r≤ onl ≤ −1
cos
cos
−1
x [domain] negative angles STEP 3. Draw a Perp to x-axis STEP 4. Label sides/signs
y 2
Example: Find the angle θ
to estimate the sought angle. „ « 5 θ = cos−1 7
y
x
7
2
?
θ
θ
θ x
?
Def: of THE Trig Functions
≈ 0.775 (in radians [see calculator mode]... OR...) In this case, we know the adjacent ≈ 44.415◦ and the hypothenuse sides. The (in degrees [see calculator mode]) function describing such ratio is the cosine, thus
5
cos θ = ref. triangle for θ angle θ
[co-domain]
ratio of sides
hyp opp
θ
5 7
Since this is not a famous ratio, we allow ourselves use of a calculator
... it should be noted that we will revisit the equation cos θ = 75 under a different context, where we will solve it completely, not limited to the domain and codomain of the arccos function.
adj
sin θ =
opp hyp
cos θ =
adj hyp
tan θ =
opp adj
csc θ =
hyp opp
sec θ =
hyp adj
cot θ =
adj opp
Def: Inverse Tangent Function LOP Diagram: ◦ es ngl 90 ya θ< onl ◦ < 0 −9
l rea al l ios rat
tan
• The essential concept 1: once the ratios are known, one side is enough to determine the other sides
r
θ
• The essential concept 2: For each angle, the ratios are described by the trig functions where defined.
−1
tan [domain]
[co-domain]
Def: Inverse Sine Function ◦ es ngl 90 ya θ≤ onl ◦ ≤ 0 −9
sin
p hy
r
θ
sin [domain]
os ati yr ≤1 onl ≤r 1 −
× sin θ os θ ×c
−1
θ
[co-domain]
math hands
×
opp
nθ ta
adj
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Solving Euclidean Right Triangles Solve the triangle [may not be drawn to scale assume typical units for length]:
β 23
34◦
b
a
Solution: Notice, this proposition is typical, three items are given; the right angle, the 34 angle, and the hypothenuse side, while three items are missing; sides a, b, and angle β. It should be noted that the angle β is relatively easy to determine β + 34◦ = 90◦ Therefore, β = 56◦ . Now, we solve for b, the side opposite of the 34◦ angle. Since we know the hypothenuse is 23 units, and we want to know the sine function describes this ratio, sin 34◦ ≈ 0.559, similarly to solve for a we use the cosine ratio, cos 34◦ ≈ 0.829. Thus, we illustrate the ratios on the triangle:
56◦ 23
34◦
×0.559
12.861
×0.829
19.068
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