Solving Trig Equations: The Easy Ones

Trigonometry Sec. 01 exercises

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Solving Trig Equations: The Easy Ones

1. Solve sin (x) =

2 5

Solution: First we graph y = 0.4 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (0.4) ≈ 23.578◦. This solution is the only one provided by the sin−1 function.

2.5 2.0 y=

2 5

y = sin(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.578

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0

≈ −696.422◦

360◦

≈ −336.23◦

360◦

-1.5 ≈ 23.21◦

360◦

-2.0 ≈ −563

◦

360

◦

≈ −203

◦

360◦ -2.5

≈ 157◦

≈ 384.1◦ 360◦

360◦ ≈ 517◦

360◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 23.578◦ + k360◦

for k ∈ Z

xk ≈ 156.422◦ + k360◦

for k ∈ Z

OR

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2. Solve

4 5

sin (x) =

Solution: First we graph y = 0.8 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (0.8) ≈ 53.13◦. This solution is the only one provided by the sin−1 function.

2.5 2.0 y=

4 5

y = sin(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.13

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0 -1.5 ≈ −666.87◦

≈ −306.8◦

360◦

360◦ -2.0

≈ −593

◦

360

◦

≈ −233

◦

≈ 53.6◦

360◦ -2.5

≈ 127◦

360◦

≈ 414◦ 360◦

≈ 487◦

360◦ 360◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 53.13◦ + k360◦

for k ∈ Z

xk ≈ 126.87◦ + k360◦

for k ∈ Z

OR

3. Solve sin (x) =

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pg. 8

Trigonometry Sec. 01 exercises

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Solution: First we graph y = 0.857 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (0.857) ≈ 58.997◦. This solution is the only one provided by the sin−1 function.

2.5 2.0 y=

6 7

y = sin(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.997

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0 -1.5 ≈ −661.003◦

≈ −301◦

360◦

360◦ -2.0

≈ −598◦

360◦

≈ −238◦

≈ 59◦

360◦ -2.5

≈ 122◦

360◦

≈ 419◦ 360◦

≈ 482◦

360◦ 360◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 58.997◦ + k360◦

for k ∈ Z

xk ≈ 121.003◦ + k360◦

for k ∈ Z

OR

4. Solve sin (x) = −

5 11

Solution: First we graph y = −0.455 then we graph y = sin(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate sin−1 (−0.455) ≈ −27.036◦. This solution is the only one provided by the sin−1 function.

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Trigonometry Sec. 01 exercises

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2.5 2.0 y= −

5 11

y = sin(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.964

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0

≈ −747.036◦

360◦

≈ −387.7◦

-1.5 ≈ −27◦

360◦

360◦

-2.0 ≈ −512

◦

360

◦

≈ −152

360◦

◦

-2.5

≈ 208◦

≈ 333◦ 360◦

360◦ ≈ 568◦

360◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ −27.036◦ + k360◦

for k ∈ Z

xk ≈ 207.036◦ + k360◦

for k ∈ Z

OR

5. Solve cos (x) =

7 10

Solution: First we graph y = 0.7 then we graph y = cos(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate cos−1 (0.7) ≈ 45.573◦. This solution is the only one provided by the cos−1 function.

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Trigonometry Sec. 01 exercises

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2.5 2.0 y=

7 10

y = cos(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.573

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0

≈ −674.427◦

360◦

≈ −314.28◦

360◦

-1.5 ≈ 45.26◦

360◦

-2.0 ≈ −405

◦

360

◦

≈ −45◦ -2.5

360◦

≈ 406.6◦ ≈ 315◦

360◦ 360◦

≈ 675◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 45.573◦ + k360◦

for k ∈ Z

xk ≈ 314.427◦ + k360◦

for k ∈ Z

OR

6. Solve cos (x) = −

7 10

Solution: First we graph y = −0.7 then we graph y = cos(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate cos−1 (−0.7) ≈ 134.427◦. This solution is the only one provided by the cos−1 function.

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360

Trigonometry Sec. 01 exercises

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2.5 2.0 y= −

7 10

y = cos(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.427

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0

≈ −585.573◦

360◦

-1.5 360◦

≈ −225.74◦

≈ 134.72◦

-2.0 ≈ −494

◦

360

◦

≈ −134

360◦

◦

-2.5

≈ 226◦

360◦

≈ 495.2◦ 360◦

360◦

≈ 586◦

360◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 134.427◦ + k360◦

for k ∈ Z

xk ≈ 225.573◦ + k360◦

for k ∈ Z

OR

7. Solve cos (x) =

3 5

Solution: First we graph y = 0.6 then we graph y = cos(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate cos−1 (0.6) ≈ 53.13◦ . This solution is the only one provided by the cos−1 function.

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2.5 2.0 y=

3 5

y = cos(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.13

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0 -1.5 ≈ −666.87◦

360◦

≈ −306.8◦

360◦ -2.0

≈ −413

◦

360

◦

≈ 53.6◦

≈ −53◦ -2.5

360◦

360◦ ≈ 307◦

≈ 414◦ 360◦

360◦ ≈ 667◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 53.13◦ + k360◦

for k ∈ Z

xk ≈ 306.87◦ + k360◦

for k ∈ Z

OR

8. Solve cos (x) =

9 10

Solution: First we graph y = 0.9 then we graph y = cos(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate cos−1 (0.9) ≈ 25.842◦. This solution is the only one provided by the cos−1 function.

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pg. 13

360◦

Trigonometry Sec. 01 exercises

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2.5 2.0 y=

9 10

y = cos(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.842

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0

≈ −694.158◦

360◦

≈ −334.59◦

360◦

-1.5 ≈ 25.57◦

360◦

-2.0 ≈ −385

◦

360

◦

≈ −25◦ -2.5

360◦

≈ 386.7◦ ≈ 335◦

360◦ 360◦

≈ 695◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 25.842◦ + k360◦

for k ∈ Z

xk ≈ 334.158◦ + k360◦

for k ∈ Z

OR

9. Solve cos (x) = −

8 15

Solution: First we graph y = −0.533 then we graph y = cos(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate cos−1 (−0.533) ≈ 122.231◦. This solution is the only one provided by the cos−1 function.

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36

Trigonometry Sec. 01 exercises

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2.5 2.0 y= −

8 15

y = cos(x)

1.5 1.0 .5 .231

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0

≈ −597.769◦

360◦

≈ −237.70◦

-1.5 360◦

≈ 122.68◦

-2.0 ≈ −482

◦

360

◦

≈ −122

360◦

◦

-2.5

360◦ ≈ 238◦

≈ 483.8◦ 360◦

360◦ ≈ 598◦

360◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 122.231◦ + k360◦

for k ∈ Z

xk ≈ 237.769◦ + k360◦

for k ∈ Z

OR

10. Solve cos (x) = −

3 4

Solution: First we graph y = −0.75 then we graph y = cos(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate cos−1 (−0.75) ≈ 138.59◦. This solution is the only one provided by the cos−1 function.

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Trigonometry Sec. 01 exercises

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2.5 2.0 y= −

3 4

y = cos(x)

1.5 1.0 .5

-720◦ -630◦ -540◦ -450◦ -360◦ -270◦ -180◦ -90◦

.59

90◦

180◦ 270◦ 360◦ 450◦ 540◦ 630◦ 720◦

-.5 -1.0

≈ −581.41◦

-1.5 360◦

≈ −221.2◦

360◦

≈ 138◦

-2.0 ≈ −498

◦

360

◦

≈ −138

360◦

◦

-2.5

≈ 222◦

360◦

≈ 498◦ 360◦

360◦

≈ 582◦

360◦

We conclude the solution by describing all possible solutions. We summarize, each solution occurs either in the ’uphill bucket’ or in the ’downhill bucket’ thus if x is a solution, it must be one of the following values: xk ≈ 138.59◦ + k360◦

for k ∈ Z

xk ≈ 221.41◦ + k360◦

for k ∈ Z

OR

11. Solve tan (x) = −

3 4

Solution: First we graph y = −0.75 then we graph y = tan(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate tan−1 (−0.75) ≈ −36.87◦. This solution is the only one provided by the tan−1 function.

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Trigonometry Sec. 01 exercises

y= −

MathHands.com M´ arquez

5

3 4

y = tanx

4 3 2 1 -360◦ -315◦ -270◦ -225◦ -180◦ -135◦ -90◦

-45◦

.13

45◦

90◦

135◦ 180◦ 225◦ 270◦ 315◦ 360◦

-1 -2 -3 -4 ≈ −396.87◦

180◦

≈ −216.8◦

180◦

≈ −36◦ -5

180◦

≈ 144◦

180◦

We conclude the solution by describing all possible values of x. xk ≈ −36.87◦ + k180◦

12. Solve tan (x) =

for k ∈ Z

3 5

Solution: First we graph y = 0.6 then we graph y = tan(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate tan−1 (0.6) ≈ 30.964◦. This solution is the only one provided by the tan−1 function.

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Trigonometry Sec. 01 exercises

y=

MathHands.com M´ arquez

5

3 5

y = tanx

4 3 2 1 -360◦ -315◦ -270◦ -225◦ -180◦ -135◦ -90◦

-45◦

.964

45◦

90◦

135◦ 180◦ 225◦ 270◦ 315◦ 360◦

-1 -2 -3 -4 ≈ −329.036◦

180◦

≈ −149.7◦

180◦

-5

≈ 30.5◦

180◦

≈ 211◦

180◦

We conclude the solution by describing all possible values of x. xk ≈ 30.964◦ + k180◦

13. Solve tan (x) =

for k ∈ Z

13 5

Solution: First we graph y = 2.6 then we graph y = tan(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate tan−1 (2.6) ≈ 68.962◦. This solution is the only one provided by the tan−1 function.

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Trigonometry Sec. 01 exercises

MathHands.com M´ arquez

5

13 5

y=

y = tanx

4 3 2 1 -360◦ -315◦ -270◦ -225◦ -180◦ -135◦ -90◦

-45◦

.962

45◦

90◦

135◦ 180◦ 225◦ 270◦ 315◦ 360◦

-1 -2 -3 -4 ≈ −291.038◦

180◦

≈ −111.9◦

180◦ -5

≈ 68.7◦

180◦

≈ 249◦

180◦

We conclude the solution by describing all possible values of x. xk ≈ 68.962◦ + k180◦

14. Solve tan (x) = −

for k ∈ Z

12 5

Solution: First we graph y = −2.4 then we graph y = tan(x), then we mark the intersections. These are the solutions. Clear from the graph is that we have infinite many of them. Of these, the first one is determined by using a calculator to estimate tan−1 (−2.4) ≈ −67.38◦. This solution is the only one provided by the tan−1 function.

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pg. 19

Trigonometry Sec. 01 exercises

y= −

MathHands.com M´ arquez

5

12 5

y = tanx

4 3 2 1 -360◦ -315◦ -270◦ -225◦ -180◦ -135◦ -90◦

-45◦

.62

45◦

90◦

135◦ 180◦ 225◦ 270◦ 315◦ 360◦

-1 -2 -3 -4 ≈ −427.38◦

180◦

≈ −247.9◦

180◦

≈ −67.0◦

-5

180◦

≈ 113◦

180◦

We conclude the solution by describing all possible values of x. xk ≈ −67.38◦ + k180◦

for k ∈ Z

15. Find all solutions cos x = 0

Solution: The set of all real solutions to cos x = 0 is of the form... x = 90◦ + k180◦ said differently... x = . . . , −90◦ , 90◦ , 270◦ , 450◦, . . . 16. Find all solutions cos θ = −1

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Trigonometry Sec. 01 exercises

MathHands.com M´ arquez The set of all real solutions to cos θ = −1 is of the form...

Solution:

θ = 180.0◦ + k360◦ said differently... θ = . . . , −180.0◦, 180.0◦ , 540.0◦, 900.0◦, . . . 17. Find all solutions cos θ = 1

The set of all real solutions to cos θ = 1 is of the form...

Solution:

θ = 0.0◦ + k360◦ said differently... θ = . . . , −360.0◦, 0.0◦ , 360.0◦, 720.0◦, . . . 18. Find all solutions cos θ = 1/2

Solution: The set of all real solutions to cos θ = .5 is of the form... θ = 60.0◦ + k360◦

or θ = 300.0◦ + k360◦

said differently... θ = . . . , −60.0◦, 60.0◦ , 300.0◦, 420.0◦, . . . 19. Find all solutions cos θ = .25

Solution: The set of all real solutions to cos θ = .25 is of the form... θ = 75.5◦ + k360◦

or θ = 284.5◦ + k360◦

said differently... θ = . . . , −75.5◦, 75.5◦ , 284.5◦, 435.5◦, . . . NOTE: these are approximations... 20. Find all solutions cos θ = −2 Solution:

There are NO real solutions to cos θ = −2

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Trigonometry Sec. 01 exercises

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21. Find all solutions cos x = 1.01

Solution:

There are NO real solutions to cos x = 1.01

22. Find all solutions cos x = .999

Solution: The set of all real solutions to cos x = .999 is of the form... x = 2.6◦ + k360◦

or x = 357.4◦ + k360◦

said differently... x = . . . , −2.6◦ , 2.6◦ , 357.4◦, 362.6◦, . . . 23. Find all solutions cos x =

−1 2

Solution: The set of all real solutions to cos x = −.5 is of the form... x = 120.0◦ + k360◦

or x = 240.0◦ + k360◦

said differently... x = . . . , −120.0◦, 120.0◦, 240.0◦, 480.0◦, . . . 24. Find all solutions sin x = 0

Solution: The set of all real solutions to sin x = 0 is of the form... x = 180◦k said differently... x = . . . , −180◦ , 0◦ , 180◦ , 360◦, . . . 25. Find all solutions sin θ = −1

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Trigonometry Sec. 01 exercises

MathHands.com M´ arquez The set of all real solutions to sin θ = −1 is of the form...

Solution:

θ = −90.0◦ + 360◦ k said differently... θ = . . . , −450.0◦, −90.0◦, 270.0◦, −810.0◦, . . . 26. Find all solutions sin θ = 1

The set of all real solutions to sin θ = 1 is of the form...

Solution:

θ = 90.0◦ + 360◦ k said differently... θ = . . . , −270.0◦, 90.0◦ , 450.0◦, −630.0◦, . . . 27. Find all solutions sin θ = 1/2

Solution: The set of all real solutions to sin θ = .5 is of the form... θ = 30.0◦ + 360◦ k

or θ = 150.0◦ + 360◦ k

said differently... θ = . . . , −210.0◦, 30.0◦ , 150.0◦, 390.0◦, . . . 28. Find all solutions sin θ = .25

Solution: The set of all real solutions to sin θ = .25 is of the form... θ = 14.5◦ + 360◦ k

or θ = 165.5◦ + 360◦ k

said differently... θ = . . . , −194.5◦, 14.5◦ , 165.5◦, 374.5◦, . . . NOTE: these are approximations... 29. Find all solutions sin θ = −2 Solution:

There are NO real solutions to sin θ = −2

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Trigonometry Sec. 01 exercises

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30. Find all solutions sin x = 1.01

Solution:

There are NO real solutions to sin x = 1.01

31. Find all solutions sin x = .999

Solution: The set of all real solutions to sin x = .999 is of the form... x = 87.4◦ + 360◦ k

or x = 92.6◦ + 360◦ k

said differently... x = . . . , −267.4◦, 87.4◦ , 92.6◦, 447.4◦ , . . . 32. Find all solutions sin x =

−1 2

Solution: The set of all real solutions to sin x = −.5 is of the form... x = −30.0◦ + 360◦ k

or x = −150.0◦ + 360◦k

said differently... x = . . . , −510.0◦, −30.0◦, −150.0◦, 330.0◦, . . . 33. Find all solutions csc x = −2 Solution: this is the same as solving sin x = 1/ − 2

The set of all real solutions to sin x = −.5 is of the form... x = −30.0◦ + 360◦ k

or x = −150.0◦ + 360◦k

said differently... x = . . . , −510.0◦, −30.0◦, −150.0◦, 330.0◦, . . . 34. Find all solutions csc x = 3

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Trigonometry Sec. 01 exercises

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Solution: this is the same as solving sin x = 1/3 The set of all real solutions to sin x = .333333333 is of the form... x = 19.5◦ + 360◦ k

or x = 160.5◦ + 360◦ k

said differently... x = . . . , −199.5◦, 19.5◦ , 160.5◦, 379.5◦, . . . 35. Find all solutions tan x = 1

Solution: x = 45◦ + k180◦ 36. Find all solutions tan x = −1 37. Find all solutions tan x = 5 38. Find all solutions sin x = 5

Solution:

There are NO real solutions to sin x = 5

39. Find all solutions

√ − 3 sin x = 2

40. Find all solutions sin x =

√

5 6

41. Find all solutions cos x = .123

Solution: The set of all real solutions to cos x = .123 is of the form... x = 82.9◦ + k360◦

or x = 277.1◦ + k360◦

said differently... x = . . . , −82.9◦ , 82.9◦, 277.1◦, 442.9◦ , . . . 42. (extra fun points) Find all solutions cos x = sin x

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Trigonometry Sec. 02 exercises

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Solving Trig Equations: The Almost-Easy Ones

1. Solve


7 tan 2x − 45◦ = 3

Solution: From solving the easy version of the equation we obtain 2x − 45◦ ≈ 66.801◦ + k180◦

for k ∈ Z

Therefore:

2x − 45◦ ≈66.801◦ + k180◦ 2x ≈111.801◦ + k180◦ x ≈55.9◦ + k90◦

2. Solve the equation sin x =

1 2

Solution: Solution: We first draw the y = sin x graph, then the line y = 21 . We then point to the solutions and list them.

y = sin x

1.5

solutions

1.0 0.5 -360◦ -270◦ -180◦

-90◦

90◦

180◦

270◦

uphill solutions

x = 30◦ + k360◦

downhill solutions

x = 150◦ + k360◦

360◦

-0.5 -1.0 -1.5

3. Solve the equation sin(2t − 50◦ ) =

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math hands

1 2

pg. 4

Trigonometry Sec. 02 exercises

MathHands.com M´ arquez

Solution: Solution: The key here is to note that this is almost like the previous example. The only difference is x, instead of x, we have the quantity 2t − 50◦ . Thus, the answer will be almost like the previous answer except instead of x we will have the quantity 2t − 50◦ . solutions y = sin x

1.5

uphill solutions substitute

1.0 0.5 -360◦ -270◦ -180◦

-90◦

90◦

180◦

270◦

x = 30◦ + k360◦ instead of x, we have 2t − 50◦ 2t − 50◦ = 30◦ + k360◦ 2t = 80◦ + k360◦

360◦

-0.5 -1.0 -1.5

t = 40◦ + k180◦ x = 150◦ + k360◦

downhill solutions substitute

4. Find all solutions

instead of x, we have 2t − 50◦ 2t − 50◦ = 150◦ + k360◦

−1 sin(2θ) = 2

2t = 200◦ + k360◦

t = 100◦ + k180◦

Solution: FIRST we solve... The set of all real solutions to sin 2θ = −.5 is of the form... 2θ = −30.0◦ + 360◦ k

or 2θ = −150.0◦ + 360◦ k

said differently... 2θ = . . . , −510.0◦, −30.0◦ , −150.0◦, 330.0◦ , . . . THEN we solve for θ by dividing everything by 2... thus.. . θ = −15◦ + 180◦ k OR θ = −75◦ + 180◦ k 5. Find all solutions sin(3θ + 90◦ ) = 0

Solution: FIRST we solve... The set of all real solutions to sin (3θ + 90◦ ) = 0 is of the form... (3θ + 90◦ ) = 180◦ k said differently... (3θ + 90◦ ) = . . . , −180◦, 0◦ , 180◦ , 360◦ , . . . THEN we solve for θ by subtracting 90◦ then dividing by 3..... thus.. . θ = −30◦ + 60◦ k

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pg. 5

Trigonometry Sec. 02 exercises

MathHands.com M´ arquez

6. Find all solutions sin(2x − 40◦ ) = 7. Find all solutions cos(5x + π) =

1 3

−1 2

Solution: FIRST we solve... The set of all real solutions to cos (5x + 180◦ ) = −.5 is of the form... (5x + 180◦) = 120.0◦ + k360◦

or (5x + 180◦) = 240.0◦ + k360◦

said differently... (5x + 180◦) = . . . , −120.0◦, 120.0◦, 240.0◦, 480.0◦, . . . THEN we solve for x by subtracting 180◦ then dividing by 5..... thus.. . x=

−60◦ + 360◦ k 5

OR x=

60◦ + 360◦ k 5

8. Find all solutions cos(2t − π) = 9. Find all solutions csc

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2x + π 3

math hands



−1 3

= −2

pg. 6

Trigonometry Sec. 03 exercises

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Solving Trig Equations: The Others 1. Solve 3sin2 (x) + sin(x) = 0

Solution:

sin(x) = 0

3sin2 (x) + 1sin(x) = 0   sin(x) 3sin(x) + 1 = 0 OR

sin(x) = 0

Solve

(given) (factor)

3sin(x) + 1 = 0

OR

(Zero Fact Thm)

sin(x) = −

1 3

(algebra)

Solution: sin (x) = 0

Solution: xk = 0◦ + k180◦ Solve sin (x) = −

for k ∈ Z

xk ≈ −19.471◦ + k360◦

for k ∈ Z

xk ≈ 199.471◦ + k360◦

for k ∈ Z

OR

1 3

2. Solve 2sin2 (x) + 5sin(x) = 0

Solution:

sin(x) = 0

2sin2 (x) + 5sin(x) = 0   sin(x) 2sin(x) + 5 = 0 OR

sin(x) = 0

(given) (factor)

2sin(x) + 5 = 0

OR

(Zero Fact Thm)

sin(x) = −

5 2

(algebra)

Solve sin (x) = 0

Solve sin (x) = −

Solution: xk = 0◦ + k180◦

for k ∈ Z

5 2

no real solution for x

3. Solve 5cos2 (x) + cos(x) = 0

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pg. 3

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution:

cos(x) = 0

5cos2 (x) + 1cos(x) = 0   cos(x) 5cos(x) + 1 = 0 OR

cos(x) = 0

Solve

(given) (factor)

5cos(x) + 1 = 0

OR

cos(x) = −

(Zero Fact Thm) 1 5

(algebra)

Solution: cos (x) = 0

Solution: xk = 90◦ + k180◦ Solve cos (x) = −

for k ∈ Z

xk ≈ 101.537◦ + k360◦

for k ∈ Z

xk ≈ 258.463◦ + k360◦

for k ∈ Z

OR

1 5

4. Solve 6cos2 (x) + 2cos(x) = 0

Solution:

cos(x) = 0

6cos2 (x) + 2cos(x) = 0   cos(x) 6cos(x) + 2 = 0 OR

cos(x) = 0

Solve

6cos(x) + 2 = 0

OR

(given) (factor) (Zero Fact Thm)

1 cos(x) = − 3

(algebra)

Solution: cos (x) = 0

Solution: xk = 90◦ + k180◦ Solve cos (x) = −

for k ∈ Z

xk ≈ 109.471◦ + k360◦

for k ∈ Z

xk ≈ 250.529◦ + k360◦

for k ∈ Z

OR

1 3

5. Solve 8sin2 (x) + − 2sin(x) + − 3 = 0

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pg. 4

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution:



8sin2 (x) + − 2sin(x) + − 3 = 0    2sin(x) + 1 · 4sin(x) + − 3 = 0

2sin(x) + 1 = 0

OR 1 sin(x) = − 2

Solve

(given) (factor)

4sin(x) + − 3 = 0 3 OR sin(x) = 4

(Zero Fact Thm) (algebra)

Solve

1 sin (x) = − 2

sin (x) =

Solution:

3 4

Solution: xk ≈ −30◦ + k360◦

for k ∈ Z

xk ≈ 210◦ + k360◦

for k ∈ Z

OR

xk ≈ 48.59◦ + k360◦

for k ∈ Z

xk ≈ 131.41◦ + k360◦

for k ∈ Z

OR

6. Solve 4sin2 (x) + 0sin(x) + − 1 = 0 Solution: 4sin2 (x) + 0sin(x) + − 1 = 0    2sin(x) + 1 · 2sin(x) + − 1 = 0 2sin(x) + 1 = 0 OR 2sin(x) + − 1 = 0 1 1 OR sin(x) = sin(x) = − 2 2

(given)



Solve sin (x) = −

(factor) (Zero Fact Thm) (algebra)

Solve

1 2

sin (x) =

Solution:

1 2

Solution: xk ≈ −30◦ + k360◦

for k ∈ Z

xk ≈ 210◦ + k360◦

for k ∈ Z

OR

xk ≈ 30◦ + k360◦

for k ∈ Z

xk ≈ 150◦ + k360◦

for k ∈ Z

OR

7. Solve 4cos2 (x) + 0cos(x) + − 1 = 0

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pg. 5

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution: 4cos2 (x) + 0cos(x) + − 1 = 0     2cos(x) + 1 · 2cos(x) + − 1 = 0

2cos(x) + 1 = 0

OR 1 cos(x) = − 2

Solve

(given) (factor)

2cos(x) + − 1 = 0 1 OR cos(x) = 2

(Zero Fact Thm) (algebra)

Solve

1 cos (x) = − 2

cos (x) =

Solution:

1 2

Solution: xk ≈ 120◦ + k360◦

for k ∈ Z

xk ≈ 240◦ + k360◦

for k ∈ Z

OR

xk ≈ 60◦ + k360◦

for k ∈ Z

xk ≈ 300◦ + k360◦

for k ∈ Z

OR

8. Solve 6cos2 (x) + − 1cos(x) + − 1 = 0 Solution: 6cos2 (x) + − 1cos(x) + − 1 = 0     3cos(x) + 1 · 2cos(x) + − 1 = 0 3cos(x) + 1 = 0 OR 2cos(x) + − 1 = 0 1 1 OR cos(x) = cos(x) = − 3 2

Solve cos (x) = −

(given) (factor) (Zero Fact Thm) (algebra)

Solve

1 3

cos (x) =

Solution:

1 2

Solution:

xk ≈ 109.471◦ + k360◦

for k ∈ Z

xk ≈ 250.529◦ + k360◦

for k ∈ Z

OR

xk ≈ 60◦ + k360◦

for k ∈ Z

xk ≈ 300◦ + k360◦

for k ∈ Z

OR

9. Solve 12sin2(x) + − 5sin(x) + − 3 = 0

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pg. 6

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution:



12sin2 (x) + − 5sin(x) + − 3 = 0    3sin(x) + 1 · 4sin(x) + − 3 = 0

3sin(x) + 1 = 0

OR 1 sin(x) = − 3

Solve

(given) (factor)

4sin(x) + − 3 = 0 3 OR sin(x) = 4

(Zero Fact Thm) (algebra)

Solve

1 sin (x) = − 3

sin (x) =

Solution:

3 4

Solution:

xk ≈ −19.471◦ + k360◦

for k ∈ Z

xk ≈ 199.471◦ + k360◦

for k ∈ Z

OR

xk ≈ 48.59◦ + k360◦

for k ∈ Z

xk ≈ 131.41◦ + k360◦

for k ∈ Z

OR

10. Solve 4sin2 (x) + 8sin(x) + − 5 = 0 Solution: 4sin2 (x) + 8sin(x) + − 5 = 0    2sin(x) + 5 · 2sin(x) + − 1 = 0 2sin(x) + 5 = 0 OR 2sin(x) + − 1 = 0 1 5 OR sin(x) = sin(x) = − 2 2 

Solve sin (x) = −

5 2

no real solution for x Solve

(given) (factor) (Zero Fact Thm) (algebra)

Solution: xk ≈ 30◦ + k360◦

for k ∈ Z

xk ≈ 150◦ + k360◦

for k ∈ Z

OR

1 sin (x) = 2

11. Solve 10cos2 (x) + − 18cos(x) + − 4 = 0

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pg. 7

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution: 10cos2 (x) + − 18cos(x) + − 4 = 0     5cos(x) + 1 · 2cos(x) + − 4 = 0

5cos(x) + 1 = 0

OR 1 cos(x) = − 5

Solve cos (x) = −

1 5

(factor)

2cos(x) + − 4 = 0

OR

(Zero Fact Thm)

cos(x) = 2

(algebra)

Solve

Solution:

OR

(given)

cos (x) = 2

xk ≈ 101.537◦ + k360◦

for k ∈ Z

xk ≈ 258.463◦ + k360◦

for k ∈ Z

no real solution for x

12. Solve 60cos2 (x) + 2cos(x) + − 6 = 0 Solution: 60cos2 (x) + 2cos(x) + − 6 = 0     6cos(x) + 2 · 10cos(x) + − 3 = 0 6cos(x) + 2 = 0 OR 10cos(x) + − 3 = 0 3 1 OR cos(x) = cos(x) = − 3 10

Solve

(given) (factor) (Zero Fact Thm) (algebra)

Solve

1 cos (x) = − 3

cos (x) =

Solution:

3 10

Solution:

xk ≈ 109.471◦ + k360◦

for k ∈ Z

xk ≈ 250.529◦ + k360◦

for k ∈ Z

OR

xk ≈ 72.542◦ + k360◦

for k ∈ Z

xk ≈ 287.458◦ + k360◦

for k ∈ Z

OR

13. Solve 3 + sin(x) = 3cos2 (x)

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pg. 8

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution: 3 + sin(x) = 3cos2 (x)

(given) 2


3 + sin(x) = 3 1 − sin (x)

(famous Pyth identity)

2

3 + sin(x) = 3 − 3sin (x)

2

sin(x) = 0

(famous Pyth identity)

3sin (x) + 1sin(x) = 0   sin(x) 3sin(x) + 1 = 0 OR 3sin(x) + 1 = 0

sin(x) = 0

OR

Solve

sin(x) = −

(given) (factor) (Zero Fact Thm)

1 3

(algebra)

Solve sin (x) = 0

sin (x) = −

Solution:

1 3

Solution: xk = 0◦ + k180◦

for k ∈ Z

xk ≈ −19.471◦ + k360◦

for k ∈ Z

xk ≈ 199.471◦ + k360◦

for k ∈ Z

OR

14. Solve 3 + − cos(x) = 3sin2 (x) Solution: 3 + − cos(x) = 3sin2 (x)

(given) 2


3 + − cos(x) = 3 1 − cos (x)

(famous Pyth identity)

2

2

3 + − cos(x) = 3 − 3cos (x)

(famous Pyth identity)

3cos (x) + − 1cos(x) = 0   cos(x) 3cos(x) + − 1 = 0 cos(x) = 0 OR 3cos(x) + − 1 = 0 1 cos(x) = 0 OR cos(x) = 3

Solve

(given) (factor) (Zero Fact Thm) (algebra)

Solve cos (x) = 0

cos (x) =

Solution:

1 3

Solution: xk = 90◦ + k180◦

for k ∈ Z

xk ≈ 70.529◦ + k360◦

for k ∈ Z

xk ≈ 289.471◦ + k360◦

for k ∈ Z

OR

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pg. 9

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

15. Solve 2 + sin(x) = 2cos2 (x)

Solution: 2 + sin(x) = 2cos2 (x)

(given) 2


2 + sin(x) = 2 1 − sin (x)

(famous Pyth identity)

2

2 + sin(x) = 2 − 2sin (x)

2

sin(x) = 0

(famous Pyth identity)

2sin (x) + 1sin(x) = 0   sin(x) 2sin(x) + 1 = 0 OR

(given) (factor)

2sin(x) + 1 = 0

sin(x) = 0

OR

sin(x) = −

(Zero Fact Thm) 1 2

(algebra)

Solve

Solve sin (x) = 0

sin (x) = −

Solution:

1 2

Solution: xk = 0◦ + k180◦

for k ∈ Z

xk ≈ −30◦ + k360◦

for k ∈ Z

xk ≈ 210◦ + k360◦

for k ∈ Z

OR

16. Solve 2 + − 3sin(x) = 2cos2 (x) Solution: 2 + − 3sin(x) = 2cos2 (x)

(given) 2


2 + − 3sin(x) = 2 1 − sin (x)

(famous Pyth identity)

2

2

2 + − 3sin(x) = 2 − 2sin (x)

(famous Pyth identity)

2sin (x) + − 3sin(x) = 0   sin(x) 2sin(x) + − 3 = 0 sin(x) = 0 OR 2sin(x) + − 3 = 0 3 sin(x) = 0 OR sin(x) = 2

(given) (factor) (Zero Fact Thm) (algebra)

Solve sin (x) = 0 Solve

Solution:

sin (x) = xk = 0◦ + k180◦

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2007-2009 MathHands.com

for k ∈ Z

3 2

no real solution for x

math hands

pg. 10

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

17. Solve 4 + − 3cos(x) = 4sin2 (x) Solution: 4 + − 3cos(x) = 4sin2 (x)

(given) 2


4 + − 3cos(x) = 4 1 − cos (x)

(famous Pyth identity)

2

2

cos(x) = 0

4 + − 3cos(x) = 4 − 4cos (x)

(famous Pyth identity)

4cos (x) + − 3cos(x) = 0   cos(x) 4cos(x) + − 3 = 0 OR

(given) (factor)

4cos(x) + − 3 = 0 3 OR cos(x) = 4

cos(x) = 0

(Zero Fact Thm) (algebra)

Solve

Solve cos (x) = 0

cos (x) =

Solution:

3 4

Solution: xk = 90◦ + k180◦

for k ∈ Z

xk ≈ 41.41◦ + k360◦

for k ∈ Z

xk ≈ 318.59◦ + k360◦

for k ∈ Z

OR

18. Solve 6 + − 5cos(x) = 6sin2 (x) Solution: 6 + − 5cos(x) = 6sin2 (x)

(given) 2


6 + − 5cos(x) = 6 1 − cos (x) 2

2

6 + − 5cos(x) = 6 − 6cos (x)

6cos (x) + − 5cos(x) = 0   cos(x) 6cos(x) + − 5 = 0 cos(x) = 0 OR 6cos(x) + − 5 = 0 5 cos(x) = 0 OR cos(x) = 6

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math hands

(famous Pyth identity) (famous Pyth identity) (given) (factor) (Zero Fact Thm) (algebra)

pg. 11

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez Solve

Solve cos (x) = 0

cos (x) =

Solution:

5 6

Solution: xk = 90◦ + k180◦

for k ∈ Z

xk ≈ 33.557◦ + k360◦

for k ∈ Z

xk ≈ 326.443◦ + k360◦

for k ∈ Z

OR

19. Solve

cos 2x = cos 6x Solution: note we will use the famous identity: cos a − cos b = −2 sin



a+b 2



sin



a−b 2





cos 2x = cos 6x

cos 2x − cos 6x = 0 ! ! 2x + 6x 2x − 6x −2 sin =0 sin 2 2

−2 sin 4x · sin − 2x = 0

sin 4x · sin − 2x = 0

sin 4x = 0 OR sin − 2x = 0

(given) (Bi) (famous id)

(divide by -2) (ZFT)

Solve

Solve
sin 4x = 0

Solution:

4xk = 0◦ + k180◦


sin − 2x = 0

Solution: for k ∈ Z

− 2xk = 0◦ + k180◦

xk ≈ 45◦ k

for k ∈ Z

xk ≈ −90◦ k

finish solving for x and check solutions.. 20. Solve

cos − 2x = cos 6x Solution: note we will use the famous identity: cos a − cos b = −2 sin

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math hands

a+b 2



sin



a−b 2



pg. 12

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

cos − 2x = cos 6x

cos − 2x − cos 6x = 0 ! ! − 2x − 6x − 2x + 6x sin −2 sin =0 2 2

−2 sin 2x · sin − 4x = 0

sin 2x · sin − 4x = 0

sin 2x = 0 OR sin − 4x = 0

(given) (Bi) (famous id)

(divide by -2) (ZFT)

Solve

Solve
sin 2x = 0

Solution:

2xk = 0◦ + k180◦


sin − 4x = 0

Solution: for k ∈ Z

− 4xk = 0◦ + k180◦

xk ≈ 90◦ k

for k ∈ Z

xk ≈ −45◦ k

finish solving for x and check solutions.. 21. Solve

cos − 3x = cos 5x Solution: note we will use the famous identity: cos a − cos b = −2 sin



a+b 2



sin



a−b 2





cos − 3x = cos 5x

cos − 3x − cos 5x = 0 ! ! − 3x + 5x − 3x − 5x −2 sin =0 sin 2 2

−2 sin x · sin − 4x = 0

sin x · sin − 4x = 0

sin x = 0 OR sin − 4x = 0

(given) (Bi) (famous id)

(divide by -2) (ZFT)

Solve

Solve
sin x = 0

Solution:

xk = 0◦ + k180◦

Solution: for k ∈ Z

− 4xk = 0◦ + k180◦

xk ≈ 180◦k

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sin − 4x = 0

for k ∈ Z

xk ≈ −45◦ k

math hands

pg. 13

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

finish solving for x and check solutions.. 22. Solve

cos 3x = cos 4x Solution: note we will use the famous identity: cos a − cos b = −2 sin



a+b 2



sin



a−b 2





cos 3x = cos 4x

cos 3x − cos 4x = 0 ! ! 3x + 4x 3x − 4x −2 sin =0 sin 2 2     1 7 x · sin − x = 0 −2 sin 2 2     7 1 sin x · sin − x = 0 2 2     1 7 x =0 OR sin − x = 0 sin 2 2

Solve sin



Solve

 7 x =0 2

sin

Solution:



(given) (Bi) (famous id)

(divide by -2) (ZFT)

 1 − x =0 2

Solution: 7 xk = 0◦ + k180◦ 2

1 − xk = 0◦ + k180◦ 2

for k ∈ Z

xk ≈ 51.43◦k

for k ∈ Z

xk ≈ −360◦ k

finish solving for x and check solutions.. 23. Find all solutions cos x = 2 cos2 x Solution: do NOT divide.. dangerous.. instead..

cos x = 2 cos2 x 2

cos x = 0 cos x = 0

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cos x − 2 cos x = 0 cos x(1 − 2 cos x) = 0 1 − 2 cos x = 0

cos x = 1/2

math hands

(given) (algebra) (algebra, factor) (Zero Factor Theorem) (algebra..)

pg. 14

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

then... The set of all real solutions to cos x = 0 is of the form... x = 90◦ + k180◦ said differently... x = . . . , −90◦ , 90◦ , 270◦ , 450◦, . . . AND The set of all real solutions to cos x = .5 is of the form... x = 60.0◦ + k360◦

or x = 300.0◦ + k360◦

said differently... x = . . . , −60.0◦ , 60.0◦, 300.0◦, 420.0◦ , . . .

24. Find all solutions cos x = 1 − sin2 x Solution: ONE way to look at it..

cos x = 1 − sin2 x (given) (famous idea change all to cosines...) cos x = cos2 x 2

cos x = 0 cos x = 0

cos x − cos x = 0 cos x(1 − cos x) = 0 1 − cos x = 0 cos x = 1

(pythagoras ID) (algebra) (algebra, factor) (Zero Factor Theorem) (algebra..)

then... The set of all real solutions to cos x = 0 is of the form... x = 90◦ + k180◦ said differently... x = . . . , −90◦ , 90◦ , 270◦ , 450◦, . . . AND

The set of all real solutions to cos x = 1 is of the form... x = 0.0◦ + k360◦

said differently... x = . . . , −360.0◦, 0.0◦ , 360.0◦, 720.0◦ , . . .

25. Find all solutions tan x = sin x

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pg. 15

Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution: ONE way to look at it..

tan x = sin x

(given)

(famous idea move all to one side, try to factor.., change to sines n cosines.) tan x − sin x = 0 (algebra) sin x − sin x = 0 (IDS) cos  x  1 sin x −1 =0 (algebra, factor) cos x 1 sin x = 0 −1=0 (Zero Factor Theorem) cos x sin x = 0 cos x = 1 (algebra.., note had to mult by cos x)

then... The set of all real solutions to sin x = 0 is of the form... x = 180◦k said differently... x = . . . , −180◦ , 0◦ , 180◦ , 360◦, . . . AND

The set of all real solutions to cos x = 1 is of the form... x = 0.0◦ + k360◦

said differently... x = . . . , −360.0◦, 0.0◦ , 360.0◦, 720.0◦ , . . . HOWEVER, because we mult both by cos x, extraneous solutions may have been introduced so each of these solutions should be checked and extraneous solutions need to be discarded. 26. Find all solutions cos(3x + π) =

−1 2

27. Find all solutions cos(40◦ − 2x) = 28. Find all solutions

−1 3

4 − 1 = sec x sec x − 1

29. Find all solutions csc(2x) = − sin2 +1 30. Find all solutions sin(2x) = cos(2x) 31. Find all solutions sin(4x) = cos(2x) 32. Find all solutions cos(5x) = cos(7x)

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Trigonometry Sec. 03 exercises

MathHands.com M´ arquez

Solution: ONE way to look at it.. is to move all to one side, set to zero.. then try to to change the difference to a product.. use famous identities.. then use zero factor theorem..

33. (xtra fun..take your time on this one) Find all solutions sin(5x) = cos(7x)

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pg. 17

math hands Trigonometry 04 notes

The LAW of COSINES Main Idea At the beginning of the course, we stated the essential concert of trigonometry to be the study and solving of triangles. The general goal is to determine all 3 sides and all 3 angles of a triangles, when only 3 of these are known. Moreover, in the case of right triangles, we have been there, done that. In deed, we promised to return to this question and resolve not just right triangles, but any and all solvable triangles. We are now ready to take on such task. The key ingredient will be The Law of Cosines. One may find it helpful to think of the law of cosines as a generalization of the Pythagoras Theorem. Let us recall, the Pythagoras Theorem. If we stand next to the 90◦ angle of a right triangle and look across, there we will find the hypothenuse, thus we are reminded of: Pythagoras Theorem:

A c

B

b

a

c2 = a2 + b 2 The question of interest now is, what if this is not a right triangle?

A b

c

B

a

C

c2

c

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math hands

? 2 2 a +b =

pg. 1

math hands Trigonometry 04 notes The Law of Cosines is taylor-made for exactly this purpose, to address such triangles. It states how to ’tweak’ the pythagoras theorem to apply it to non-right triangles. Without further delay, here is the law of cosines: Law of Cosines:

A b

c

B

C

a

c2 = a2 + b2 −2ab cos C In fact, the law of cosines can be applied from any corner of the triangle: observe:

a2 = c2 + b2 −2cb cos A A

A b

c

B

C

a

b

c

B

a

C

b2 = c2 + a2 −2ca cos B It can even be applied to right triangles:

A c

B

b

a

c2 = a2 + b2 −2ab cos 90◦ Notice, cos 90◦ = 0, thus in this case, the law of cosines reduces to just pythagoras.

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pg. 2

math hands Trigonometry 04 notes Now, that we have seen the law of cosines, there are two important tasks at hand. One, we would like to prove the law of cosines, and two, we would like to learn how to use it to solve triangles. Let us take the latter first. Count The Unknown Variables: Law of cosines always involves 4 quantities from the triangle. For example, in this case,

A b

c

B

C◦

a

c2 = a2 + b2 −2(a)b cos C ◦ the four quantities involved are c, a, b, and angle C. The use of the law of cosines would be most successful, when out of these 4 quantities, 3 are known, since this would lead to an equation with just one unknown variable. Consider the following example:

A b

12

B

70◦

(10)

122 = (10)2 + b2 −2((10))b cos 70◦ In this case, we could simplify the equation, estimate cos 70◦ and solve the quadratic equation using the quadratic formula, for example. Ultimately, we could solve for b... However consider what would happen if we applied the law of cosines from a different corner.

(10)2 = (12)2 + b2 −2(12)b cos A◦ A ) (12

B

c

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b

(10)

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70◦

pg. 3

math hands Trigonometry 04 notes Note, in this case, the law of cosines leads to an equation with two unknown quantities. Now, consider the choices: solving an equation with one variable OR solving an equation with two variables. If you prefer equations with less variables, you should apply the law of cosines on from the appropriate corner, and always be conscious that applying it at different corners may lead to a different number of unknown variables to solve for. Example: Note, in this example exactly 3 quantities are known, 8, 10 and 70◦ . Our mission is to find the other 3 missing items, if such feat is possible. We first apply the law of cosines:

A b

12

B

70◦

10

122 = 102 + b2 −2(10)b cos 70◦

(12)2 = (10)2 + (b)2 − 2(10)(b) cos(70◦ ) 2

144 = 100 + (b) − 20b cos(70 ) ◦

2

0 = (b) − [20 cos(70 )]b + −44 ◦

2

b ≈ 10.88

OR

0 ≈ (b) + −6.84b + −44 b ≈ − 4.04

(THE LoC) (BI) (BI) (BI) (Quadratic Formula)

Furthermore, if we assume b is is a positive real number, then there is only one choice: namely, b ≈ 10.88, and we have solved for one of the variables, leaving two more to go, B and A. We now apply the law of cosines from the A angle, aware of the number of unknown variables from such, in this case, one.

102 = 122 + (10.88)2−2(12)(10.88) cos A A 88 10.

12

B

c

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pg. 4

math hands Trigonometry 04 notes

102 ≈ 122 + (10.88)2 − 2(12)(10.88) cos A 100 ≈ 144 + 118.37 + −261.12 cos A −162.37 ≈ −261.12 cos A −162.37 ≈ cos A −261.12 0.622 ≈ cos A

(LoC) (bi) (algebra) (algebra) (algebra)

Solve cos (A) ≈ 0.622 Solution: Ak ≈ 51.538◦ + k360◦

for k ∈ Z

Ak ≈ 308.462◦ + k360◦

for k ∈ Z

OR

Now, if we assume A is an interior angle of a triangle, then it has to be more than zero but less than 180◦. Then of all the possible choices above, only A ≈ 51.538◦ works since it it the only one less than 180◦ and larger than 0. Then to solve for B, we use the fact that B + 70◦ + 51.538◦ = 180◦ thus... B ≈ 58.46◦ ... completely solving the triangle..

51.538◦ 88 10.

12

58.46◦

c

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70◦

pg. 5

math hands Trigonometry 04 exercises

The LAW of COSINES 1. Solve

A b

13

B

60◦

10

Solution:

(13)2 = (10)2 + (b)2 − 2(10)(b) cos(60◦ ) 2

169 = 100 + (b) − 20b cos(60 ) ◦

2

0 = (b) − [20 cos(60 )]b + −69 ◦

2

0 ≈ (b) + −10b + −69 b ≈ 14.7

OR

b ≈ − 4.7

(THE LoC) (BI) (BI) (BI) (Quadratic Formula)

Furthermore, if we assume b is is a positive real number, then there is only one choice: namely, b ≈ 14.7, and we have solved for one of the variables, leaving two more to go, B and A. We now apply the law of cosines from the A angle, aware of the number of unknown variables from such, in this case, one.

102 = 132 + (14.7)2 −2(13)(14.7) cos A A 7 14.

13

B

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math hands Trigonometry 04 exercises

102 ≈ 132 + (14.7)2 − 2(13)(14.7) cos A 100 ≈ 169 + 216.09 + −382.2 cos A −285.09 ≈ −382.2 cos A −285.09 ≈ cos A −382.2 0.746 ≈ cos A

(LoC) (bi) (algebra) (algebra) (algebra)

Solve cos (A) ≈ 0.746 Solution: Ak ≈ 41.755◦ + k360◦

for k ∈ Z

Ak ≈ 318.245◦ + k360◦

for k ∈ Z

OR

Now, if we assume A is an interior angle of a triangle, then it has to be more than zero but less than 180◦. Then of all the possible choices above, only A ≈ 41.755◦ works since it it the only one less than 180◦ and larger than 0. Then to solve for B, we use the fact that B + 60◦ + 41.755◦ = 180◦ thus... B ≈ 78.25◦ ... completely solving the triangle..

41.755◦ 7 14.

13

78.25◦

c

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10

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60◦

pg. 7

math hands Trigonometry 04 exercises 2. Solve

A b

9

B

40◦

12

Solution:

(9)2 = (12)2 + (b)2 − 2(12)(b) cos(40◦ ) 2

81 = 144 + (b) − 24b cos(40 ) ◦

2

0 = (b) − [24 cos(40 )]b + 63 ◦

2

0 ≈ (b) + −18.385b + 63 b ≈ 13.83

OR

(THE LoC) (BI) (BI) (BI) (Quadratic Formula)

b ≈4.56

Furthermore, if we assume b is is a positive real number, then there is only one choice: namely, b ≈ 13.83, and we have solved for one of the variables, leaving two more to go, B and A. We now apply the law of cosines from the A angle, aware of the number of unknown variables from such, in this case, one.

122 = 92 + (13.83)2 −2(9)(13.83) cos A A 83 13.

9

B

c

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pg. 8

math hands Trigonometry 04 exercises

122 ≈ 92 + (13.83)2 − 2(9)(13.83) cos A 144 ≈ 81 + 191.27 + −248.94 cos A −128.27 ≈ −248.94 cos A −128.27 ≈ cos A −248.94 0.515 ≈ cos A

(LoC) (bi) (algebra) (algebra) (algebra)

Solve cos (A) ≈ 0.515 Solution: Ak ≈ 59.003◦ + k360◦

for k ∈ Z

Ak ≈ 300.997◦ + k360◦

for k ∈ Z

OR

Now, if we assume A is an interior angle of a triangle, then it has to be more than zero but less than 180◦. Then of all the possible choices above, only A ≈ 59.003◦ works since it it the only one less than 180◦ and larger than 0. Then to solve for B, we use the fact that B + 40◦ + 59.003◦ = 180◦ thus... B ≈ 81◦ ... completely solving the triangle..

59.003◦ 83 13.

9

81◦

c

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pg. 9

math hands Trigonometry 04 exercises 3. Solve

A

13 B

b 50◦

7

Solution:

(13)2 = (7)2 + (b)2 − 2(7)(b) cos(50◦ ) 2

(THE LoC)

169 = 49 + (b) − 14b cos(50 )

(BI)

◦

2

0 = (b) − [14 cos(50 )]b + −120

(BI)

◦

2

(BI)

b ≈ − 7.34

(Quadratic Formula)

0 ≈ (b) + −8.999b + −120 b ≈ 16.34

OR

Furthermore, if we assume b is is a positive real number, then there is only one choice: namely, b ≈ 16.34, and we have solved for one of the variables, leaving two more to go, B and A. We now apply the law of cosines from the A angle, aware of the number of unknown variables from such, in this case, one.

72 = 132 + (16.34)2 −2(13)(16.34) cos A A

13 B

c

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16. 34 50◦

7

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pg. 10

math hands Trigonometry 04 exercises

72 ≈ 132 + (16.34)2 − 2(13)(16.34) cos A 49 ≈ 169 + 267 + −424.84 cos A −387 ≈ −424.84 cos A −387 ≈ cos A −424.84 0.911 ≈ cos A

(LoC) (bi) (algebra) (algebra) (algebra)

Solve cos (A) ≈ 0.911 Solution: Ak ≈ 24.356◦ + k360◦

for k ∈ Z

Ak ≈ 335.644◦ + k360◦

for k ∈ Z

OR

Now, if we assume A is an interior angle of a triangle, then it has to be more than zero but less than 180◦. Then of all the possible choices above, only A ≈ 24.356◦ works since it it the only one less than 180◦ and larger than 0. Then to solve for B, we use the fact that B + 50◦ + 24.356◦ = 180◦ thus... B ≈ 105.64◦ ... completely solving the triangle..

24.356◦

13 105.64◦

c

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16. 34 50◦

7

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pg. 11

math hands Trigonometry 04 exercises 4. Solve A

b

13

B

65◦

11

Solution:

(13)2 = (11)2 + (b)2 − 2(11)(b) cos(65◦ ) 2

169 = 121 + (b) − 22b cos(65 ) ◦

2

0 = (b) − [22 cos(65 )]b + −48 ◦

OR

(BI) (BI)

2

(BI)

b ≈ − 3.69

(Quadratic Formula)

0 ≈ (b) + −9.298b + −48 b ≈ 12.99

(THE LoC)

Furthermore, if we assume b is is a positive real number, then there is only one choice: namely, b ≈ 12.99, and we have solved for one of the variables, leaving two more to go, B and A. We now apply the law of cosines from the A angle, aware of the number of unknown variables from such, in this case, one.

112 = 132 + (12.99)2 −2(13)(12.99) cos A A

12.99

13

B

c

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11

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65◦

pg. 12

math hands Trigonometry 04 exercises

112 ≈ 132 + (12.99)2 − 2(13)(12.99) cos A 121 ≈ 169 + 168.74 + −337.74 cos A −216.74 ≈ −337.74 cos A −216.74 ≈ cos A −337.74 0.642 ≈ cos A

(LoC) (bi) (algebra) (algebra) (algebra)

Solve cos (A) ≈ 0.642 Solution: Ak ≈ 50.059◦ + k360◦

for k ∈ Z

Ak ≈ 309.941◦ + k360◦

for k ∈ Z

OR

Now, if we assume A is an interior angle of a triangle, then it has to be more than zero but less than 180◦. Then of all the possible choices above, only A ≈ 50.059◦ works since it it the only one less than 180◦ and larger than 0. Then to solve for B, we use the fact that B + 65◦ + 50.059◦ = 180◦ thus... B ≈ 64.94◦ ... completely solving the triangle..

50.059◦

12.99

13

64.94◦

c

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65◦

pg. 13

math hands Trigonometry 04 exercises 5. Solve

C 3

a

20◦

B

5

Solution: We first note that applying the law of cosines from the 20◦ angle involves 4 quantities, 3 of which we already know.... thus we can solve for the only unknown one, namely, a. This diagram shows the quantities involved in applying the law of cosines. C 3

a b

20

◦

B

5 2

2

a2 = (3) + (5) − 2 (3) (5) cos 20◦

Thus.... 2

2

a2 = (3) + (5) − 2 (3) (5) cos 20◦

(L of Cos applied 20◦ )

a2 ≈ 5.809

(use calc. to estimate) (use calc. to estimate) (just length is positive, sometimes we may interpret both..)

a ≈ ±2.410 a ≈ 2.410

Now, we can apply the law of cosines from the perspective of angle B. Note the law of cosines from this side now gives an equation with 4 quantities 3 of which we now know. C 3

2.4 10

20◦

B

5 2

2

2

(3) ≈ (5) + (2.410) − 2 (5) (2.410) cos B

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pg. 14

math hands Trigonometry 04 exercises

Thus,

(3)2 ≈ (5)2 + (2.410)2 − 2 (5) (2.410) cos B

(L of cos applied angle B)

9.000 ≈ 30.809 − 24.100 cos B −21.809 ≈ −24.100 cos B 0.905 ≈ cos B cos B ≈ 0.905 B ≈ . . . , −25.18◦,

(now.. need to solve correctly, honestly, and completely) 25.18◦ , 334.82◦, 205.18◦, . . . etc. (solved correctly, honestly, and completely)

This is as far as the law of cosines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B is B ≈ 25.18◦ Finally, to solve for C we use that the sum of the interior angles of a triangle is 180◦ to get C ≈ 180◦ − 25.18◦ − 20◦ = 134.82◦ thus the final solution... 134.82◦ 3

2.4 10

20◦

25.18◦

5

6. Solve

C 5

a

35◦ 4

B

Solution: We first note that applying the law of cosines from the 35◦ angle involves 4 quantities, 3 of which we already know.... thus we can solve for the only unknown one, namely, a. This diagram shows the quantities involved in applying the law of cosines.

c

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pg. 15

math hands Trigonometry 04 exercises

C 5

a b

35

◦

B

4 2

2

a2 = (5) + (4) − 2 (5) (4) cos 35◦

Thus....

2

2

a2 = (5) + (4) − 2 (5) (4) cos 35◦

(L of Cos applied 35◦ )

a2 ≈ 8.234

(use calc. to estimate) (use calc. to estimate) (just length is positive, sometimes we may interpret both..)

a ≈ ±2.869 a ≈ 2.869

Now, we can apply the law of cosines from the perspective of angle B. Note the law of cosines from this side now gives an equation with 4 quantities 3 of which we now know. C 5

2.8 69

35◦

B

4 2

2

2

(5) ≈ (4) + (2.869) − 2 (4) (2.869) cos B

Thus,

2

2

2

(5) ≈ (4) + (2.869) − 2 (4) (2.869) cos B 25.000 ≈ 24.234 − 22.952 cos B

(L of cos applied angle B)

0.766 ≈ −22.952 cos B −0.033 ≈ cos B cos B ≈ −0.033

(now.. need to solve correctly, honestly, and completely) ◦

B ≈ . . . , −91.89 ,

c

2007-2009 MathHands.com

91.89◦ ,

268.11◦, 271.89◦, . . . etc. (solved correctly, honestly, and completely)

math hands

pg. 16

math hands Trigonometry 04 exercises

This is as far as the law of cosines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B is B ≈ 91.89◦ Finally, to solve for C we use that the sum of the interior angles of a triangle is 180◦ to get C ≈ 180◦ − 91.89◦ − 35◦ = 53.11◦ thus the final solution... 53.11◦ 5

2.8 69

35◦

91.89◦

4

7. Solve C b

5

A

35◦

4

Solution: We first note that applying the law of cosines from the 35◦ angle involves 4 quantities, 3 of which we already know.... thus we can solve for the only unknown one, namely, b. This diagram shows the quantities involved in applying the law of cosines. C b

5

A

35◦

4 2

2

b2 = (5) + (4) − 2 (5) (4) cos 35◦

Thus....

c

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pg. 17

math hands Trigonometry 04 exercises

2

2

b2 = (5) + (4) − 2 (5) (4) cos 35◦

(L of Cos applied 35◦ )

b2 ≈ 8.234 b ≈ ±2.869

(use calc. to estimate) (use calc. to estimate)

b ≈ 2.869

(just length is positive, sometimes we may interpret both..)

Now, we can apply the law of cosines from the perspective of angle A. Note the law of cosines from this side now gives an equation with 4 quantities 3 of which we now know. The only unknown quantity is A, thus we solve for it...

2.8 69

C 5

A

35◦

4 2

2

2

(5) ≈ (4) + (2.869) − 2 (4) (2.869) cos A

Thus,

2

2

2

(5) ≈ (4) + (2.869) − 2 (4) (2.869) cos A

(L of cos applied angle A)

25.000 ≈ 24.234 − 22.952 cos A 0.766 ≈ −22.952 cos A −0.033 ≈ cos A cos A ≈ −0.033

(now.. need to solve correctly, honestly, and completely) ◦

A ≈ . . . , −91.89 ,

91.89◦ ,

268.11◦, 271.89◦, . . . etc. (solved correctly, honestly, and completely)

This is as far as the law of cosines can help up.. but we can take it from here. Assuming A is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for A is A ≈ 91.89◦ Finally, to solve for C we use that the sum of the interior angles of a triangle is 180◦ to get C ≈ 180◦ − 91.89◦ − 35◦ = 53.11◦ 8. Solve

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pg. 18

math hands Trigonometry 04 exercises C b

2

A

67◦

4

Solution: We first note that applying the law of cosines from the 67◦ angle involves 4 quantities, 3 of which we already know.... thus we can solve for the only unknown one, namely, b. This diagram shows the quantities involved in applying the law of cosines. C b

2

A

67◦

4

b2 = (2)2 + (4)2 − 2 (2) (4) cos 67◦

Thus....

2

2

b2 = (2) + (4) − 2 (2) (4) cos 67◦

(L of Cos applied 67◦ )

b2 ≈ 13.748

(use calc. to estimate)

b ≈ ±3.708 b ≈ 3.708

(use calc. to estimate) (just length is positive, sometimes we may interpret both..)

Now, we can apply the law of cosines from the perspective of angle A. Note the law of cosines from this side now gives an equation with 4 quantities 3 of which we now know. The only unknown quantity is A, thus we solve for it...

3.7 08

C 2

A

67◦

4

(2)2 ≈ (4)2 + (3.708)2 − 2 (4) (3.708) cos A

Thus,

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pg. 19

math hands Trigonometry 04 exercises

2

2

2

(2) ≈ (4) + (3.708) − 2 (4) (3.708) cos A 4.000 ≈ 29.748 − 29.664 cos A

(L of cos applied angle A)

−25.748 ≈ −29.664 cos A 0.868 ≈ cos A cos A ≈ 0.868

(now.. need to solve correctly, honestly, and completely) ◦

A ≈ . . . , −29.77 ,

29.77◦ ,

330.23◦, 209.77◦, . . . etc. (solved correctly, honestly, and completely)

This is as far as the law of cosines can help up.. but we can take it from here. Assuming A is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for A is A ≈ 29.77◦ Finally, to solve for C we use that the sum of the interior angles of a triangle is 180◦ to get C ≈ 180◦ − 29.77◦ − 67◦ = 83.23◦ 9. Solve

C 3

a

40◦

B

4.5

Solution: We first note that applying the law of cosines from the 40◦ angle involves 4 quantities, 3 of which we already know.... thus we can solve for the only unknown one, namely, a. This diagram shows the quantities involved in applying the law of cosines. C 3

a b

40

◦

B

4.5

a2 = (3)2 + (4.5)2 − 2 (3) (4.5) cos 40◦

Thus....

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pg. 20

math hands Trigonometry 04 exercises

2

2

a2 = (3) + (4.5) − 2 (3) (4.5) cos 40◦

(L of Cos applied 40◦ )

a2 ≈ 8.567 a ≈ ±2.927

(use calc. to estimate) (use calc. to estimate)

a ≈ 2.927

(just length is positive, sometimes we may interpret both..)

Now, we can apply the law of cosines from the perspective of angle B. Note the law of cosines from this side now gives an equation with 4 quantities 3 of which we now know.

C 3

2.9 27

40◦

B

4.5 2

2

2

(3) ≈ (4.5) + (2.927) − 2 (4.5) (2.927) cos B

Thus,

(3)2 ≈ (4.5)2 + (2.927)2 − 2 (4.5) (2.927) cos B

(L of cos applied angle B)

9.000 ≈ 28.817 − 26.343 cos B −19.817 ≈ −26.343 cos B 0.752 ≈ cos B cos B ≈ 0.752 B ≈ . . . , −41.24◦,

(now.. need to solve correctly, honestly, and completely) 41.24◦ , 318.76◦, 221.24◦, . . . etc. (solved correctly, honestly, and completely)

This is as far as the law of cosines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B is B ≈ 41.24◦ Finally, to solve for C we use that the sum of the interior angles of a triangle is 180◦ to get C ≈ 180◦ − 41.24◦ − 40◦ = 98.76◦ thus the final solution...

c

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pg. 21

math hands Trigonometry 04 exercises

98.76◦ 3

2.9 27

40◦

41.24◦

4.5

10. Solve

C 3

a

15◦

B

1

Solution: We first note that applying the law of cosines from the 15◦ angle involves 4 quantities, 3 of which we already know.... thus we can solve for the only unknown one, namely, a. This diagram shows the quantities involved in applying the law of cosines. C 3

a b

15

◦

B

1 2

2

a2 = (3) + (1) − 2 (3) (1) cos 15◦

Thus....

2

2

a2 = (3) + (1) − 2 (3) (1) cos 15◦

(L of Cos applied 15◦ )

2

(use calc. to estimate)

a ≈ ±2.050

(use calc. to estimate)

a ≈ 4.204 a ≈ 2.050

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2007-2009 MathHands.com

(just length is positive, sometimes we may interpret both..)

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pg. 22

math hands Trigonometry 04 exercises

Now, we can apply the law of cosines from the perspective of angle B. Note the law of cosines from this side now gives an equation with 4 quantities 3 of which we now know. C 3

2.0 50

15◦

B

1

(3)2 ≈ (1)2 + (2.050)2 − 2 (1) (2.050) cos B

Thus,

(3)2 ≈ (1)2 + (2.050)2 − 2 (1) (2.050) cos B

(L of cos applied angle B)

9.000 ≈ 5.204 − 4.100 cos B 3.796 ≈ −4.100 cos B −0.926 ≈ cos B cos B ≈ −0.926 B ≈ . . . , −157.82◦,

(now.. need to solve correctly, honestly, and completely) 157.82◦,

202.18◦, 337.82◦, . . . etc. (solved correctly, honestly, and completely)

This is as far as the law of cosines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B is B ≈ 157.82◦ Finally, to solve for C we use that the sum of the interior angles of a triangle is 180◦ to get C ≈ 180◦ − 157.82◦ − 15◦ = 7.18◦ thus the final solution... 7.18◦ 3

2.0 50

15◦ 1

157.82◦

11. Solve

c

2007-2009 MathHands.com

math hands

pg. 23

math hands Trigonometry 04 exercises C b

2

A

50◦

3

Solution: We first note that applying the law of cosines from the 50◦ angle involves 4 quantities, 3 of which we already know.... thus we can solve for the only unknown one, namely, b. This diagram shows the quantities involved in applying the law of cosines. C b

2

A

50◦

3

b2 = (2)2 + (3)2 − 2 (2) (3) cos 50◦

Thus....

2

2

b2 = (2) + (3) − 2 (2) (3) cos 50◦

(L of Cos applied 50◦ )

b2 ≈ 5.287

(use calc. to estimate) (use calc. to estimate) (just length is positive, sometimes we may interpret both..)

b ≈ ±2.299 b ≈ 2.299

Now, we can apply the law of cosines from the perspective of angle A. Note the law of cosines from this side now gives an equation with 4 quantities 3 of which we now know. The only unknown quantity is A, thus we solve for it...

2.2 99

C 2

A

50◦

3

(2)2 ≈ (3)2 + (2.299)2 − 2 (3) (2.299) cos A

Thus,

c

2007-2009 MathHands.com

math hands

pg. 24

math hands Trigonometry 04 exercises

2

2

2

(2) ≈ (3) + (2.299) − 2 (3) (2.299) cos A 4.000 ≈ 14.287 − 13.794 cos A

(L of cos applied angle A)

−10.287 ≈ −13.794 cos A 0.746 ≈ cos A cos A ≈ 0.746

(now.. need to solve correctly, honestly, and completely) 41.75◦ ,

◦

A ≈ . . . , −41.75 ,

318.25◦, 221.75◦, . . . etc. (solved correctly, honestly, and completely)

This is as far as the law of cosines can help up.. but we can take it from here. Assuming A is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for A is A ≈ 41.75◦ Finally, to solve for C we use that the sum of the interior angles of a triangle is 180◦ to get C ≈ 180◦ − 41.75◦ − 50◦ = 88.25◦ 12. PROVE the Law of COSINES

A b

c

B

C

a

c2 = a2 + b2 −2ab cos C

c

2007-2009 MathHands.com

math hands

pg. 25

Trigonometry Sec. 05 notes

MathHands.com M´ arquez

The LAW of SINES Main Idea The Law of Sines is similar to the law of cosines in that it is often helpful in solving non-right triangles. Roughly speaking, the Law of Sines says that the ratios of a sides to the sine of corresponding angles are all equal. Number of Triangles We take a moment here to emphasize the importance of solving all equations correctly, and completely. By doing so, we are sure to find out the correct number of triangle solutions. In general, when 3 of the items are given and 3 are missing, there will always be 0, 1, 2, or infinite many real-triangle solutions. To determine the total number of real triangles solutions for each specific problem one needs only to solve the equations correctly. Observe the following examples, as some of these contain two triangle solutions. Law of Sines: sin A a

A b

c

B

sin A a

sin B b

C

a

=

sin B b

=

sin C c

sin C c

Example: Solve C 5

7

65◦

B

c ◦

We first note that we know the ratio sin765 , and we know the side b = 5. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 5

7

we

w k no

65◦ c

c

2007-2009 MathHands.com

B

math hands

pg. 1

Trigonometry Sec. 05 notes

MathHands.com M´ arquez sin B sin 65◦ = 7 5 sin 65◦ = sin B 5· 7 sin 65◦ sin B = 5 · 7 sin B ≈ 0.647

(applying the law of sines) (algebra) (now, we will use calc. to approximate) (now we need to solve correctly and completely)

B ≈ . . . , −220.3 , 40.3 , 139.7 , 400.3◦, . . . ◦

◦

◦

This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B are B ≈ 40.3◦ Note that 139.7◦ + 65 is more than 180◦ Thus not a viable option for an interior angle.

74.7◦ 5

7

65◦

40.3◦

c

Then we just solve for c... again by using the law of sines. c 7 = ◦ sin 74.7 sin 65◦ 7 c= · sin 74.7◦ sin 65◦ c ≈ 7.45 Finally, the solution,

74.7◦ 5

7

65◦ 7.45

40.3◦

Example: Solve C 7

6

20◦

B

c

c

2007-2009 MathHands.com

math hands

pg. 2

Trigonometry Sec. 05 notes

MathHands.com M´ arquez ◦

We first note that we know the ratio sin620 , and we know the side b = 7. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 7

6

now we k

20◦

B

c

sin B sin 20◦ = 6 7 sin 20◦ 7· = sin B 6 sin 20◦ sin B = 7 · 6 sin B ≈ 0.399

(applying the law of sines) (algebra) (now, we will use calc. to approximate) (now we need to solve correctly and completely)

B ≈ . . . , −203.5◦, 23.5◦, 156.5◦ , 383.5◦, . . .

This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choices in the above list of candidates for B are B ≈ 23.5◦ or B ≈ 156.5◦ Thus we have two possible triangle solutions: 3.5◦

136.5◦

20◦

OR

23.5◦

c

7

20◦

6

7

6

156.5◦

c

then we just solve for c in each case... For the first triangle: 6 c = ◦ sin 136.5 sin 20◦ 6 c= · sin 136.5◦ sin 20◦ c ≈ 12.08

c 6 = ◦ sin 3.5 sin 20◦ 6 c= · sin 3.5◦ sin 20◦ c ≈ 1.07

or

3.5◦

136.5◦

20◦ 12.08

c

2007-2009 MathHands.com

OR

23.5◦

7

20◦

math hands

1.07

6

7

6

156.5◦

pg. 3

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez

The LAW of SINES 1. Solve C 5

7

55◦

B

c

◦

Solution: We first note that we know the ratio sin755 , and we know the side b = 5. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 5

7

now we k

55◦

B

c

sin 55◦ sin B = (applying the law of sines) 7 5 ◦ sin 55 = sin B (algebra) 5· 7 ◦ sin 55 (now, we will use calc. to approximate) sin B = 5 · 7 sin B ≈ 0.585 (now we need to solve correctly and completely) B ≈ . . . , −215.8◦, 35.8◦, 144.2◦ , 395.8◦, . . . This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B are B ≈ 35.8◦ Note that 144.2◦ + 55 is more than 180◦ Thus not a viable option for an interior angle.

89.2◦ 5

7

55◦ c

c

2007-2009 MathHands.com

35.8◦

math hands

pg. 4

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez

Then we just solve for c... again by using the law of sines. 7 c = sin 89.2◦ sin 55◦ 7 c= · sin 89.2◦ sin 55◦ c ≈ 8.54 Finally, the solution,

89.2◦ 5

7

55◦

35.8◦

8.54

2. Solve C 5

4

20◦

B

c

◦

Solution: We first note that we know the ratio sin420 , and we know the side b = 5. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 5

4

we

w k no

20◦

B

c

sin 20◦ sin B = 4 5 sin 20◦ = sin B 5· 4 sin 20◦ sin B = 5 · 4 sin B ≈ 0.428

(applying the law of sines) (algebra) (now, we will use calc. to approximate) (now we need to solve correctly and completely)

B ≈ . . . , −205.3 , 25.3 , 154.7 , 385.3◦, . . .

c

2007-2009 MathHands.com

◦

◦

math hands

◦

pg. 5

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez

This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choices in the above list of candidates for B are B ≈ 25.3◦ or B ≈ 154.7◦ Thus we have two possible triangle solutions: 5.3◦

134.7◦

20◦

OR

25.3◦

c

5

20◦

4

5

4

154.7◦

c

then we just solve for c in each case... For the first triangle: c 4 = ◦ sin 134.7 sin 20◦ 4 · sin 134.7◦ c= sin 20◦ c ≈ 8.31

c 4 = ◦ sin 5.3 sin 20◦ 4 c= · sin 5.3◦ sin 20◦ c ≈ 1.08

or

5.3◦

134.7◦

20◦

OR

25.3◦

8.31

5

20◦

1.08

4

5

4

154.7◦

3. Solve C 5

8

40◦

B

c

◦

Solution: We first note that we know the ratio sin840 , and we know the side b = 5. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 5

8

now we k

40◦ c

c

2007-2009 MathHands.com

B

math hands

pg. 6

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez sin B sin 40◦ = (applying the law of sines) 8 5 ◦ sin 40 5· = sin B (algebra) 8 ◦ sin 40 (now, we will use calc. to approximate) sin B = 5 · 8 sin B ≈ 0.402 (now we need to solve correctly and completely) B ≈ . . . , −203.7◦, 23.7◦, 156.3◦ , 383.7◦, . . .

This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B are B ≈ 23.7◦ Note that 156.3◦ + 40 is more than 180◦ Thus not a viable option for an interior angle.

116.3◦ 5

8

40◦

23.7◦

c

Then we just solve for c... again by using the law of sines. 8 c = ◦ sin 116.3 sin 40◦ 8 c= · sin 116.3◦ sin 40◦ c ≈ 11.16 Finally, the solution,

116.3◦ 5

8

40◦ 11.16

23.7◦

4. Solve

c

2007-2009 MathHands.com

math hands

pg. 7

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez C 5

7

40◦

B

c

◦

Solution: We first note that we know the ratio sin740 , and we know the side b = 5. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 5

7

now we k

40◦

B

c

sin B sin 40◦ = (applying the law of sines) 7 5 ◦ sin 40 = sin B (algebra) 5· 7 ◦ sin 40 (now, we will use calc. to approximate) sin B = 5 · 7 sin B ≈ 0.459 (now we need to solve correctly and completely) B ≈ . . . , −207.3◦, 27.3◦, 152.7◦ , 387.3◦, . . . This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B are B ≈ 27.3◦ Note that 152.7◦ + 40 is more than 180◦ Thus not a viable option for an interior angle.

112.7◦ 5

7

40◦ c

27.3◦

Then we just solve for c... again by using the law of sines. c 7 = ◦ sin 112.7 sin 40◦ 7 c= · sin 112.7◦ sin 40◦ c ≈ 10.05

c

2007-2009 MathHands.com

math hands

pg. 8

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez

Finally, the solution,

112.7◦ 5

7

40◦

27.3◦

10.05

5. Solve C 5

3

70◦

B

c

◦

Solution: We first note that we know the ratio sin370 , and we know the side b = 5. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 5

3

we

w k no

70◦

B

c

sin 70◦ sin B = 3 5 sin 70◦ = sin B 5· 3 sin 70◦ sin B = 5 · 3 sin B ≈ 1.566

(applying the law of sines) (algebra) (now, we will use calc. to approximate) (now we need to solve correctly and completely)

There are NO real solutions to sin B = 1.566 Thus no real triangle solutions.

6. Solve

c

2007-2009 MathHands.com

math hands

pg. 9

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez C 8

7

51◦

B

c

◦

Solution: We first note that we know the ratio sin751 , and we know the side b = 8. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 8

7

now we k

51◦

B

c

sin B sin 51◦ = (applying the law of sines) 7 8 ◦ sin 51 = sin B (algebra) 8· 7 ◦ sin 51 (now, we will use calc. to approximate) sin B = 8 · 7 sin B ≈ 0.888 (now we need to solve correctly and completely) B ≈ . . . , −242.6◦, 62.6◦, 117.4◦ , 422.6◦, . . . This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choices in the above list of candidates for B are B ≈ 62.6◦ or B ≈ 117.4◦ Thus we have two possible triangle solutions: 11.6◦

66.4◦

51◦ c

OR

62.6◦

8

51◦

c

7

8

7

117.4◦

then we just solve for c in each case... For the first triangle: c 7 = ◦ sin 66.4 sin 51◦ 7 · sin 66.4◦ c= sin 51◦ c ≈ 8.25

c

2007-2009 MathHands.com

or

math hands

c 7 = ◦ sin 11.6 sin 51◦ 7 c= · sin 11.6◦ sin 51◦ c ≈ 1.81

pg. 10

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez 11.6◦

66.4◦

51◦

OR

62.6◦

8.25

8

51◦

1.81

7

8

7

117.4◦

7. Solve C 8

16

51◦

B

c

◦

Solution: We first note that we know the ratio sin1651 , and we know the side b = 8. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 8

16

now we k

51◦

B

c

sin 51◦ sin B = (applying the law of sines) 16 8 ◦ sin 51 = sin B (algebra) 8· 16 ◦ sin 51 sin B = 8 · (now, we will use calc. to approximate) 16 sin B ≈ 0.389 (now we need to solve correctly and completely) ◦ ◦ ◦ B ≈ . . . , −202.9 , 22.9 , 157.1 , 382.9◦, . . . This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choice in the above list of candidates for B are B ≈ 22.9◦ Note that 157.1◦ + 51 is more than 180◦ Thus not a viable option for an interior angle.

c

2007-2009 MathHands.com

math hands

pg. 11

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez

106.1◦ 8

16

51◦

22.9◦

c

Then we just solve for c... again by using the law of sines. 16 c = sin 106.1◦ sin 51◦ 16 c= · sin 106.1◦ sin 51◦ c ≈ 19.78 Finally, the solution,

106.1◦ 8

16

51◦

22.9◦

19.78

8. Solve C 8

4

51◦

B

c

◦

Solution: We first note that we know the ratio sin451 , and we know the side b = 8. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 8

4

we

w k no

51◦ c

c

2007-2009 MathHands.com

B

math hands

pg. 12

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez sin B sin 51◦ = 4 8 sin 51◦ 8· = sin B 4 sin 51◦ sin B = 8 · 4 sin B ≈ 1.554

(applying the law of sines) (algebra) (now, we will use calc. to approximate) (now we need to solve correctly and completely)

There are NO real solutions to sin B = 1.554 Thus no real triangle solutions.

9. Solve C 5

4

41◦

B

c

◦

Solution: We first note that we know the ratio sin441 , and we know the side b = 5. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 5

4

we

w k no

41◦

B

c

sin B sin 41◦ = 4 5 sin 41◦ 5· = sin B 4 sin 41◦ sin B = 5 · 4 sin B ≈ 0.820

(applying the law of sines) (algebra) (now, we will use calc. to approximate) (now we need to solve correctly and completely)

B ≈ . . . , −235.1 , 55.1 , 124.9 , 415.1◦, . . . ◦

◦

◦

This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choices in the above list of candidates for B are B ≈ 55.1◦ or B ≈ 124.9◦ Thus we have two possible triangle solutions:

c

2007-2009 MathHands.com

math hands

pg. 13

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez 14.1◦

83.9◦

41◦

OR

55.1◦

c

5

41◦

4

5

4

124.9◦

c

then we just solve for c in each case... For the first triangle: c 4 = sin 83.9◦ sin 41◦ 4 c= · sin 83.9◦ sin 41◦ c ≈ 6.06

c 4 = sin 14.1◦ sin 41◦ 4 c= · sin 14.1◦ sin 41◦ c ≈ 1.49

or

14.1◦

83.9◦

41◦

OR

55.1◦

6.06

5

41◦

1.49

4

5

4

124.9◦

10. Solve C 7

4

24◦

B

c

◦

Solution: We first note that we know the ratio sin424 , and we know the side b = 7. Thus.. we can apply the law of sines to solve for the angle B. The following diagram shows the ration we know and the ratio we will compare it to. C 7

4

we

w k no

24◦ c

c

2007-2009 MathHands.com

B

math hands

pg. 14

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez sin B sin 24◦ = (applying the law of sines) 4 7 ◦ sin 24 7· = sin B (algebra) 4 ◦ sin 24 (now, we will use calc. to approximate) sin B = 7 · 4 sin B ≈ 0.712 (now we need to solve correctly and completely) B ≈ . . . , −225.4◦, 45.4◦, 134.6◦ , 405.4◦, . . .

This is as far as the law of sines can help us.. but we can take it from here. Assuming B is an interior angle of triangle [in flatland.. Euclidean space] we can assume it must measure somewhere between 0 and 180◦ . The only such choices in the above list of candidates for B are B ≈ 45.4◦ or B ≈ 134.6◦ Thus we have two possible triangle solutions: 21.4◦

110.6◦

24◦

OR

45.4◦

c

7

24◦

4

7

4

134.6◦

c

then we just solve for c in each case... For the first triangle: 4 c = ◦ sin 110.6 sin 24◦ 4 c= · sin 110.6◦ sin 24◦ c ≈ 9.21

c 4 = ◦ sin 21.4 sin 24◦ 4 c= · sin 21.4◦ sin 24◦ c ≈ 3.59

or

21.4◦

110.6◦

24◦ 9.21

OR

45.4◦

7

24◦

3.59

4

7

4

134.6◦

11. PROVE the Law of SINES 12. Give an example of a triangle where 3 of the quantities: a, b, c, A, B, C are given and there are infinite many possible real triangle solutions. Solution: A = 40◦ , B = 40◦ , C = 10o◦

13. Give an example of a triangle where 3 of the quantities: a, b, c, A, B, C are given and there is NO possible real triangle solution.

c

2007-2009 MathHands.com

math hands

pg. 15

Trigonometry Sec. 05 exercises

MathHands.com M´ arquez

Solution: a = 3, b = 1, c = 10

14. Give an example of a triangle where 3 of the quantities: a, b, c, A, B, C are given and there are exactly 5 possible real triangle solutions. Solution: not possible..

c

2007-2009 MathHands.com

math hands

pg. 16