CHAPTER Trigonometric Graphs and Identities 11x Solutions Key · Trigonometric Graphs and Identities Solutions Key Are you reAdy? 1. D 12.B 3. E 4. C 5. 3 5 5 2 = 3 5 · 2__

Trigonometric Graphs and IdentitiesSolutions Key

Are you reAdy?

1. D 2. B

3. E 4. C

5. 3 __ 5 __

5 __ 2 = 3 __

5 · 2 __

5 = 6 ___

25 6.

3 __ 4 __

1 __ 2 = 3 __

4 · 2 __

1 = 3 __

2

7. - 3 __

8 ___

1 __ 8 = - 3 __

8 · 8 __

1 = -3 8.

2 __ 3 ___

- 7 __ 4 = 2 __

3 · - 4 __

7 = - 8 ___

21

9. √ 6 · √ 2 = √ 12 = 2 √ 3 10. √ 100 - 64 = 36 = 6

11. √ 9

____ √ 36

= 3 __ 6 = 1 __

2 12. √ 4 ___

25 = 2 __

5

13. (x + 11)(x + 7) = x 2 + 18x + 77

14. (y - 4)(y - 9) = y 2 - 13y + 36

15. (2x - 3)(x + 5) = 2 x 2 + 7x - 15

16. (k + 3)(3k - 3) = 3 k 2 + 6k - 9

17. (4z - 4)(z + 1) = 4 z 2 - 4

18. (y + 0.5)(y - 1) = y 2 - 0.5y - 0.5

19. (2x + 5) 2 = 4 x 2 + 20x + 25

20. (3y - 2) 2 = 9 y 2 - 12y + 4

21. (4x - 6)(4x + 6) = 16 x 2 - 36

22. (2m + 1)(2m - 1) = 4 m 2 - 1

23. (s + 7) 2 = s 2 + 14s + 49

24. (-p + 4)(-p - 4) = p 2 - 16

GrAphs of sIne And cosIne

CheCk it out!

1a. The function is not periodic.

b. The function is periodic with period 3.

2. 0.5

0

y

x

-0.5 - π π

Amplitude: ⎜a⎟ = ⎜ 1 __ 3 ⎟ = 1 __

3

Period: 2π ___ ⎜b⎟

= 2π ___ ⎜2⎟

= π

The curve is vertically compressed by a factor of 1 __ 3

and horizontally compressed by a factor of 1 __ 2 .

The maximum of h is 1 __ 3 , and the minimum is - 1 __

3 .

3.

2

0

y

x

-2 .001 .003 .005

frequency = 1 ______ period

= 1 _____ 0.004

= 250 Hz

4. The first x-intercept is at π __ 2 . Therefore, the intercepts

occur at π __ 2 + πn, where n is an integer.

Because h = π, the phase shift is π radians to the right.

0.5

0

y

x

- π π -0.5

5a. amplitude: ⎜a⎟ = ⎜-16⎟ = 16

period: 2π ___ ⎜b⎟

= 2π ____ ⎜ π ___ 45

⎟ = 90

vertical shift: 24 maxima: 24 + 16 = 40

at 45 and 135 minima: 24 - 16 = 8

at 0, 90, and 180 There are no x-intercepts. The curve is reflected over

the x-axis.

ft

Time (s)

Hei

ght

(ft)

s 0

10

20

30

40

160 120 80 40

b. The maximum height is 24 + 16 = 40 feet above ground.

think anD DisCuss

1. Possible answer: The frequency of a periodic function is the reciprocal of the period. The period of

f(x) = cos x is 2π, and the frequency is 1 ___ 2π

.

2. Possible answer: The period of sine and cosine functions tells how often the maxima and minima occur. The amplitude affects the value of the maxima and the minima.

3. Verticalcompression:

Horizontal stretch:

Reflection:y = -cos xperiod: 2π

Phase shift: y = sin(x + π)

period: 2π

CosineGraphs

( ) 1 _ 2 x y = sin

period: 4π

1 _ 2 y = cos x

period: 2π

315 Holt McDougal Algebra 2

xCHAPTER

x-1


11CHAPTER

11-1

CS10_A2_MESK710389_C11.indd 315 4/11/11 1:17:20 PM

exerCisesguided practice

1. cycles

2. The function is periodic with period 5.

3. The function is not periodic.

4. 1

0

y

x

-1-3π 3π

amplitude: ⎜a⎟ = ⎜2⎟ = 2; period:

2π ___ ⎜b⎟

= 2π ___ ⎜ 1 __ 2 ⎟ = 4π

5. 0.5

0

y

x

-0.5

-π π

amplitude:

⎜a⎟ = ⎜ 1 __ 4 ⎟ = 1 __

4 ;

period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π

6.

1

-1.5

x

y

0 2 -2


2π ___ ⎜b⎟

= 2π ___ ⎜π⎟

= 2

7. 642

-6-4-2

x

y

0.01 .03.02

frequency = 1 ______ period

= 1 ____ 0.01

= 100 Hz

8. The first x-intercept is at 3π ___ 2 . Therefore, the

intercepts occur at 3π ___ 2 + nπ, where n is an integer.

The phase shift is 3π ___ 2 radians to the left.

1

1

0

y

x

-π π

9. The first x-intercept is at 0. Therefore, the intercepts occur at 0 + nπ = πn, where n is an integer. The phase shift is π __

2 radians to the right.

.5

0

y

x

-π π



The phase shift is π __ 4 radians to the right.

.5

0

y

x

-.5-π π

11. amplitude: ⎜a⎟ = ⎜-4⎟ = 4


= 2π ___ ⎜1⎟

= 2π

vertical shift: 6 maxima: 6 + 4 = 10 at 180°, not in 0° ≤ θ ≤ 90° minima: 6 - 2 = 8 at 0 There are no x-intercepts. The curve is reflected over the x-axis.

0

2

4

6

8

80 60 40 20

Angle measure (°)

Hei

ght

(ft)

When θ = 60°, the height of the swing is H(60°) = -4 cos 60° + 6 = 4 ft off the ground.

practice and problem Solving

12. The function is not periodic.

13. The function is periodic with period 2π.

14.

2

0

y

x

-2 - π π


2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π

15.

1

0

y

x

- π π

amplitude:

⎜a⎟ = ⎜ 3 __ 2 ⎟ = 3 __

2 ;

period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π

16. 1

0

y

x

-1

- π _ 2 π _

2 π _

2 π _

2

amplitude: ⎜a⎟ = ⎜-1⎟ = 1; period:

2π ___ ⎜b⎟

= 2π ___ ⎜4⎟

= π __ 2

316 Holt McDougal Algebra 2 316 Holt McDougal Algebra 2

CS10_A2_MESK710389_C11.indd 316 4/11/11 1:17:22 PM

17. 4

0

y

x

-4 - 3π 3π

amplitude: ⎜a⎟ = 6 = 6; period:

2π ___ ⎜b⎟

= 2π ___ ⎜ 1 __ 3 ⎟ = 6π

18.

4

-6

x

y

0

0.0125 -0.0125 frequency = 1 ______

period

= 1 _____ 0.025

= 40 Hz

19. The first x-intercept is at π. Therefore, the intercepts occur at πn, where n is an integer. The phase shift is π radians to the left.

.5

0

y

x

-.5 - π π



The phase shift is 3π radians to the right.

.5

0

y

x

- π π



The phase shift is 3π ___ 4 radians to the left.

.5

0

y

x

-.5 - π π

22. The first x-intercept is at π __ 4 . Therefore, the

intercepts occur at π __ 4 + πn, where n is an integer.

The phase shift is π __ 4 radians to the left.

.5

0

y

x

-.5 - π π

23. amplitude: ⎜a⎟ = ⎜ 3 __ 2 ⎟ = 3 __

2 = 1.5


= 2π _____ ⎜ 5π ___ 31

⎟ = 62 ___

5 = 12.4

vertical shift: 23 maxima: 23 + 1.5 = 24.5 at 3.1 + 12.5n minima: 23 - 1.5 = 21.5 at 9.3 + 12.5n There are no x-intercepts.

h

Depth (ft)

Tim

e (h

)

0

21

22

23

24

2418126

ft

The maximum depth is 23 + 1.5 = 24.5 ft, and the minimum depth is 23 - 1.5 = 21.5 ft.

24a. ≈ 0.8 s

b. Let x represent the pulse rate.

x ___ 60

= 1 ___ 0.8

x = 75 beats/min

c. frequency = 1 ______ period

= 1 ___ 0.8

= 1.25 Hz

d. The pulse rate is measured in beats per minute and the frequency is measured in cycles, or beats, per second. They both measure the same quantity.

25. amplitude: ⎜a⎟ = ⎜1⎟ = 1; period: 2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

phase shift π __ 4 left and vertical shift 1 down

26. amplitude: ⎜a⎟ = ⎜ 3 __ 4 ⎟ = 3 __

4 ; period: 2π ___

⎜b⎟ = 2π ____

⎜ π __ 4

⎟ = 8;

vertical compression and horizontal stretch

27. amplitude: ⎜a⎟ = ⎜1⎟ = 1; period: 2π ___ ⎜b⎟

= 2π ____ ⎜2π⎟

= 1;

horizontal compression and vertical shift 2 down

28. amplitude: ⎜a⎟ = ⎜-3⎟ = 3; period: 2π ___ ⎜b⎟

= 2π ___ ⎜3⎟

= 2π ___ 3

;

horizontal compression, vertical stretch, and reflection across the x-axis

29. sin 160° = sin 8π ___ 9 30. cos 50° = cos 5π ___

18

≈ 0.3 ≈ 0.6

31. sin 15° = sin π ___ 12

32. cos 95° = cos 19π ____ 36

≈ 0.25 ≈ -0.1

33. f(x) = 6 sin 2x; f(x) = 6 cos 2x

34. f(x) = 1 __ 4

sin (x + 2 __ 3

π) ;

g(x) = 1 __ 4

cos (x + 1 __ 6

π)


CS10_A2_MESK710389_C11.indd 317 4/11/11 1:17:22 PM

35. f(x) = -4 sin 2x;

g(x) = 4 cos 2 (x + π __ 4 )

36. f(x) = 1 __ 4 sin 1 __

2 x + 1;

g(x) = 1 __ 4 cos 1 __

2 (x - π) + 1

37a. period: 6.1 + 6.1 = 12.2;

amplitude: 3 - 0 _____ 2 = 1.5;

max.: 3; min.: 0

b. h(0) = 3; h(6.1) = 0

c. h(t) = 1.5 cos 2π ____ 12.2

t + 1.5

38. No; possible answer: you would also need to know whether there is a phase shift or a vertical shift. The phase shift and period give the location of the maxima and minima. The amplitude and vertical shift give the values of the maxima and minima.

39. The period decreases for b > 1 and increases for b < 1 because the period is given by 2π ___

b .

teSt prep

40. D 41. Hamplitude: 2; amplitude: ⎜a⎟ = ⎜-4⎟ period: 4π = 4

42. B

43. 2

0

y

x

-2 π

phase shift π right, horizontal compression, vertical stretch, and reflection across the x-axis;

amplitude: ⎜a⎟ = ⎜-4⎟ = 4; period: 2π ___ ⎜b⎟

= 2π ___ ⎜2⎟

= π;

x-intercepts: 0, π __ 2 , π, 3π ___

2 , and 2π;

max.: 4, min.: -4

challenge and extend

44. y = sin -1 xsin y = x

0 -.5 .5

y

x

π - _ 4

π _ 4

y = cos -1 xcos y = x

0 -0.5 0.5

y

x

π - _ 2

π _ 2

45. amplitude of f : 1 __ 2 ; amplitude of g : 2

period of f : 360°; period of g : 360°

2

-2

x

y

0 45 180

46. (256°, -0.485) and (76°, 0.485)

47. f(θ) > g(θ) when 76° < θ < 256°

GrAphs of oTher TrIGonomeTrIc funcTIons

CheCk it out!

1.

2

y

x

3π _ 2 - 3π _

2

period: π ___ ⎜b⎟

= π ___ ⎜ 1 __ 2 ⎟ = 2π;

x-intercepts: first intercept occurs at x = 0. Because the period is 2π, the x-intercepts occur at 2πn;

asymptotes: π ____ 2 ⎜b⎟

+ πn ___ ⎜b⎟

= π ____ 2 ⎜ 1 __

2 ⎟ + πn ___

⎜ 1 __ 2 ⎟ = π + 2πn

2.

2

0

y

x

-2 - π_

2 - π_

2 π_

2 π_

2


= π ___ ⎜2⎟

= π __ 2 ;

x-intercepts: first intercept occurs at x = π __ 4 .

Because

the period is π __ 2 , the x-intercepts occur

at π __ 4 + π __

2 n;

asymptotes: πn ___ ⎜b⎟

= πn ___ ⎜2⎟

= π __ 2 n

3.

2

0

y

x

-2 - π π


= 2π ___ ⎜1⎟

= 2π;

asymptotes: first asymptote occurs at x = 0. Therefore, the asymptotes occur at 0 + πn = πn.

think anD DisCuss

1. Possible answer: Cosecant is the reciprocal of the sine function, so the graphs are related.


x-2


11-2

CS10_A2_MESK710389_C11.indd 318 4/11/11 1:17:23 PM

2. Possible answer: By using the unit circle, you can see that tangent is undefined at the same values where cosine is equal to 0. An undefined value corresponds to an asymptote, so the zeros of cosine correspont to the asymptotes of tangent.

3. Function Zeros Asymptotes Period

y = sec x

none

y = csc x

y = cot x

y = tan x

none

π _ 2

+ πn

πn

πn 2π

2π

πn π

π

π _ 2

+ πn

π _ 2

+ πn


1.

2

0

y

x

-2 - π _

2 - π _

2 π _

2 π _

2


= π ___ ⎜3⎟

= π __ 3 ; x-intercepts: π __

3 n;


+ πn ___ ⎜b⎟

= π ____ 2 ⎜3⎟

+ πn ___ ⎜3⎟

= π __ 6 + π __

3 n

2.

0

y

x

- 2π 2π

2

-2


= π ___ ⎜ 1 __ 4 ⎟ = 4π; x-intercepts: 4πn;


+ πn ___ ⎜b⎟

= π ____ 2 ⎜ 1 __

4 ⎟ + πn ___

⎜ 1 __ 4 ⎟ = 2π + 4πn

3.

x

y

0

2

-2

4 π -- 4

π -


= π ____ ⎜2π⎟

= 1 __ 2 ; x-intercepts: 1 __

2 n;


+ πn ___ ⎜b⎟

= π _____ 2 ⎜2π⎟

+ πn ____ ⎜2π⎟

= 1 __ 4 + n __

2

4.

2

y

x

-2 -π π


= π ___ ⎜1⎟

= π; x-intercepts: π __ 2 + πn;


= πn ___ ⎜1⎟

= πn

5.

4

y

x

-4 -π π


= π ___ ⎜2⎟


4 + π __

2 n;


= πn ___ ⎜2⎟

= π __ 2 n

6.

2

0

y

x

-π π


= π ___ ⎜1⎟

= π; x-intercepts: π __ 2

+ πn;


= πn ___ ⎜1⎟

= πn

7.

1

0 x

-1

- π π

y period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

asymptotes: π ____

2 ⎜1⎟ + π ___

⎜1⎟ n

= π __ 2

+ πn

8. 1

0

y

x

-1 - π _

2 - π _

2 π _

2 π _

2

period:

2π ___ ⎜b⎟

= 2π ___ ⎜4⎟

= π __ 2

;

asymptotes: π ____

2 ⎜4⎟ + π ___

⎜4⎟ n

= π __ 8

+ π __ 4

n

9.

4

0

y

x

-π π

period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

asymptotes: π ___

⎜1⎟ n = πn


10.

2

y

x

-2 - π _

2 - π _

2 3 π _

4


= π ___ ⎜ 3 __ 2 ⎟ = 2π ___

3 ; x-intercepts: 2π ___

3 n;


+ πn ___ ⎜b⎟

= π ____ 2 ⎜ 3 __

2 ⎟ + πn ___

⎜ 3 __ 2

⎟ = π __

3 + 2π ___

3 n


CS10_A2_MESK710389_C11.indd 319 4/11/11 1:17:25 PM

11. 2

0

y

x

-2 - π π


= π ___ ⎜1⎟

= π; x-intercepts: 3π ___ 4 + πn;


+ πn ___ ⎜b⎟

= π ____ 2 ⎜1⎟

- π __ 4 + πn ___

⎜1⎟ = π __

4 + πn

12.

x

y

0

2

-2

4 π -- 4

π -


= π ___ ⎜4⎟


4 n;


+ πn ___ ⎜b⎟

= π ____ 2 ⎜4⎟

+ πn ___ ⎜4⎟

= π __ 8 + π __

4 n

13.

-2

-1

x

y

0 1 -1


= π ____ ⎜ π __ 2 ⎟ = 2; x-intercepts: 2n;


+ πn ___ ⎜b⎟

= π _____ 2 ⎜ π __

2 ⎟ + πn ____

⎜ π __ 2 ⎟ = 1 + 2n

14.

2

0

y

x

-2 -π π


= π ___ ⎜1⎟



= πn ___ ⎜1⎟

= πn

15.

4

0

y

x

-4 - 2π 2π


= π ___ ⎜ 1 __ 4 ⎟ = 4π; x-intercepts: 2π + 4πn;


= πn ___ ⎜ 1 __ 4 ⎟ = 4πn

16.

2

0

y

x

-2 - π π


= π ___ ⎜1⎟



= πn ___ ⎜1⎟

= πn

17.

1

0

y

x

-1 - π π

period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

asymptotes: π ____

2 ⎜1⎟ + π ___

⎜1⎟ n

= π __ 2 + πn

18.

1

y

x

-1 - π π

period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

asymptotes: π ___

⎜1⎟ n = πn

19.

1

0

y

x

-1 - π π

period:

2π ___ ⎜b⎟

= 2π ____ ⎜-1⎟

= 2π;

asymptotes: π ____

⎜-1⎟ n = πn

20a. h

Time

Hei

ght

(m)

t0

1

2

3

4

54321

period:

2π ___ ⎜b⎟

= 2π _____ ⎜ 5π ___ 31

⎟ = 62 ___

5 ;

asymptotes:

π _____ ⎜ 5π ___ 31

⎟ n = 31 ___

5 n

not in 1 ≤ t ≤ 6

b. minimum: 31 ___ 5 ÷ 2 = 3.1

The low tide occurs at 3:06 P.M t =3.1

c. h(3.1) = 0.4 csc 5π ___ 31

(3.1) = 0.4 m

d. h(6) = 0.4 csc 5π ___ 31

(6) ≈ 3.95

The maximum height is about 3.95 m, and it occurs at 6:00 P.M.

21. asymptotes: π __ 2 + πn

possible answer: π __ 2 ; 3π ___

2 ; - π __

2 ; 5π ___

2

22. asymptotes: πn possible answer: -π; 0; π; 2π

23. asymptotes: π __ 2 + πn

possible answer: π __ 2 ; 3π ___

2 ; - π __

2 ; 5π ___

2


CS10_A2_MESK710389_C11.indd 320 4/11/11 1:17:26 PM

24. asymptotes: πn possible answer: -π; 0; π; 2π

25a. period: 2π ___ ⎜b⎟

= 2π _____ ⎜ 2 __ 3 π⎟

= 3 s

b. 20

-20

x

y

0.5

c. asymptotes: π ______ 2 ⎜ 2 __

3 π⎟

+ πn _____ ⎜ 2 __ 3 π⎟

= 3 __ 4 + 3 __

2 n

in 0 ≤ t ≤ 3 the asmptotes are at t = 3 __ 4 and t = 9 __

4 ;

Possible answer: The asymptotes at 0.75 + 1.5n represent the time when the light shines parallel to the wall.

26a. cot θ = s __ h

s = h cot θ

b. s = 6 cot θ period: π ___

⎜b⎟ = π ___

⎜1⎟ = π

x-intercepts: π __ 2 + πn


= πn ___ ⎜1⎟

= πn

20

40

60

0

ft

Angle measure

Leng

th (f

t)

x

π_4

π_ π_8

π_ 3π _ 8

0 < x < π

__

2

π

__

2 < x < π π < x <

3π

___

2

3π

___

2 < x < 2π

27. sin x inc. dec. dec. inc.

28. csc x dec. inc. inc. dec.

29. cos x dec. dec. inc. inc.

30. sec x inc. inc. dec. dec.

31. tan x inc. inc. inc. inc.

32. cot x dec. dec. dec. dec.

33. Possible answer: For reciprocal pairs of trigonometric functions, when one increases, the other decreases and vice versa.

34. Possible answer: For reciprocal pairs of trigonometric functions, the signs are always the same.

35. Possible answer: First, graph g(x) = 3 cos 4x and draw vertical asymptotes through the zeros of g. Then start at the maximum value of g and draw upward toward the asymptotes on either side. Repeat this step for the minimum value, drawing downward.

teSt prep

36. B The domain of cotangent is {x | x ≠ πn}

37. G Range of y = 3 sin 2θ: {y | -3 ≤ y ≤ 3} So the range of y = 3 csc 2θ is: {y | y ≥ -3 or y ≤ 3}

38. B The period of the graph is π __

2 , and the asymptotes

are π __ 2 n.

39. H period: π ___

⎜b⎟ = π ___

⎜ 1 __ 2 ⎟ = 2π

40. B


41. period: 2π ___ ⎜b⎟

= 2π ___ ⎜π⎟

= 2;

local maximum: 4 - 3 = 1; local minimum: 4 + 3 = 7; phase shift: 1 right

42. period: π ___ ⎜b⎟

= π ___ ⎜ 1 __ 2 ⎟ = 2π;

local maximum: none; local minimum: none; phase shift: π __

2 right

43. period: 2π ___ ⎜b⎟

= 2π ___ ⎜2⎟

= π;

local maximum: 0 - 0.5 = -0.5; local minimum: 0 + 0.5 = 0.5; phase shift: π __

4 left

44. period: π ___ ⎜b⎟

= π ___ ⎜3⎟

= π __ 3 ;

local maximum: none; local minimum: none; phase shift: π left

45. period: 2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

local maximum: 0.62 - 0.76 = -0.14; local minimum: 0.62 + 0.76 = 1.38; phase shift: none

46. period: 2π ___ ⎜b⎟

= 2π ____ ⎜ π __ 2 ⎟ = 4;

local maximum: 0 - 1 = -1; local minimum: 0 + 1 = 1;

phase shift: 5 __ 7 left

47.

x

y

0

- π _ 2 - π _

- π _ 2 - π _

π _

2 π _

π _ 2 π _

D: {x | x ≤ -1 or x ≥ 1};

R: ⎧ ⎨

⎩ y | 0 ≤ y ≤ π and y ≠ π __

2 ⎫ ⎬

⎭


CS10_A2_MESK710389_C11.indd 321 4/11/11 1:17:27 PM

48.

x

y

- π _ 2

- π _

π _ 2 π _

π _ 2 π _

D: {x | x ∈ 핉};

R: ⎧ ⎨

⎩ y | - π __

2 < y < π __

2 ⎫ ⎬

⎭

49.

x

y

0

π _ 2 π _

π _ 2 π _

D: {x | x ≤ -1 or x ≥ 1};

R: ⎧ ⎨

⎩ y | - π __

2 ≤ y ≤ π __

2 and y ≠ 0 ⎫

⎬

⎭

50.

x

y

- π _ 2

- π _

- π _ 2

- π _

π _ 2 π _

0

D: {x | x ∈ 핉}; R: {y | 0 < y < π}

reAdy To Go on?

1. This function is not periodic.

2. This function is periodic with period 2π.

3. This function is periodic with period 4.

4. This function is not periodic.

5.

.5

0

y

x

-.5 - π _

2 - π _

2 π _

2 π _

2


2π ___ ⎜b⎟

= 2π ___ ⎜4⎟

= π __ 2

6.

1.5

0

y

x

-1.5

- π π

amplitude: ⎜a⎟ = ⎜-3⎟ = 3; period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π

7. .5

0

y

x

-.5 -2 2

amplitude: ⎜a⎟ = ⎜0.25⎟ = 1 __

4 ;

period:

2π ___ ⎜b⎟

= 2π ___ ⎜π⎟

= 2

8.

.25

0

y

x

-.5 - π π

x-intercepts: πn; phase shift:

3π ___ 2 right

9. .5

0

y

x

-.5

- π π

x-intercepts:

3π ___ 4 + πn;

phase shift:

3π ___ 4 right

10. .5

0

y

x

-.5 -π π

x-intercepts: π __

4 + πn;

phase shift:

5π ___ 4 left

11. τ(x) = 500(0.5) sin x = 250 sin x amplitude: ⎜a⎟ = 250


= 2π

maximum: 0 + 250 = 250 at π __ 2

minimum: 0 - 250 = -250 at 3π ___ 2 , not in 0 ≤ x ≤ π __

2

x-intercepts: πn

100

200

y

x

π _ 6 π _

3

Angle Measure

Torq

ue (N

·m)

τ ( π __ 3 ) = 250 sin ( π __

3 ) ≈ 216.5

The torque for an angle of π __ 3 is ≈ 216.5 nm.

12.

-1 - π _

2 π _

2


= π ___ ⎜4⎟


4 n;


+ πn ___ ⎜b⎟

= π ____ 2 ⎜4⎟

+ πn ___ ⎜4⎟

= π __ 8 + π __

4 n

13.

0

y

x

-2 -2π 2π


= π ___ ⎜ 1 __ 2 ⎟ = 2π; x-intercepts: 2πn;


+ πn ___ ⎜b⎟

= π ____ 2 ⎜ 1 __

2 ⎟ + πn ___

⎜ 1 __ 2 ⎟ = π + 2πn


CS10_A2_MESK710389_C11.indd 322 4/11/11 1:17:28 PM

14.

1

-1

x

y

0 2 -2


= π _____ ⎜ 1 __ 2 π⎟

= 2; x-intercepts: 2n;


+ πn ___ ⎜b⎟

= π ______ 2 ⎜ 1 __

2 π⎟

+ πn _____ ⎜ 1 __ 2 π⎟

= 1 + 2n

15.

1

0

y

x

-3

-π π


= π ___ ⎜1⎟



= πn ___ ⎜1⎟

= πn

16.

2

0

y

x

-2 -π π


= π _____ ⎜0.5⎟

= 2π; x-intercepts: π + 2πn;


= πn _____ ⎜0.5⎟

= 2πn

17.

- π _ 4

π π _ 4

2

0

y

x

-3


= π ___ ⎜4⎟


8 + π __

4 n;


= πn ___ ⎜4⎟

= π __ 4 n

18.

2

0

x

-2 -π π

y period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

asymptotes: π ____

2 ⎜1⎟ + π ___

⎜1⎟ n

= π __ 2 + πn

19.

.5

y

x

-.5 -π π

period:

2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

asymptotes: π ___

⎜1⎟ n = πn

20.

2

x

y

0 2 -2

period:

2π ___ ⎜b⎟

= 2π ___ ⎜π⎟

= 2;

asymptotes: π ____

2 ⎜π⎟ + π ___

⎜π⎟ n

= 1 __ 2

+ n

fundAmenTAl TrIGonomeTrIc IdenTITIes

CheCk it out!

1a. sin θ cot θ

= sin θ cos θ _____ sin θ

= cos θ

b. 1 - sec (-θ)

= 1 - 1 _______ cos (-θ)

= 1 - 1 _____ cos θ

= 1 - sec θ

2a. cos 2 θ ________ 1 - sin θ

= 1 - sin 2 θ ________ 1 - sin θ

= (1 - sin θ) (1 + sin θ)

_________________ 1 - sin θ

= 1 + sin θ

b. cot 2 θ = csc 2 θ - 1

= 1 _____ sin 2 θ

- 1

3. mg sin θ = µmg cos θ sin θ = µ cos θ sin θ = 0.4 cos θ

sin θ _____ cos θ

= 0.4

tan θ = 0.4 θ ≈ 22° The wood block will start to move when the wood

incline is raised to an angle of about 22°.

think anD DisCuss

1. Possible answer: Use identities to modify one side of the equation until it is written in the same form as the other side of the equation.

2. Possible answer: After multiplying the left side of the equation and simplifying by combining like terms, use the Pythagorean identity sin 2 θ + cos 2 θ = 1.

3.

1 + tan2 θ = sec2 θ

Pythagorean Identities

1 + cot2 θ = csc2 θ sin2 θ + cos2 θ = 1


x-3


11-3

CS10_A2_MESK710389_C11.indd 323 4/11/11 1:17:29 PM


1. sin θ sec θ

= sin θ ( 1 _____ cos θ

)

= sin θ _____ cos θ

= tan θ

2. cot(-θ)

= cos(-θ)

_______ sin(-θ)

= cos θ ______ -sin θ

= - ( cos θ _____ sin θ

)

= -cot θ

3. cos 2 θ ( sec 2 θ - 1)

= cos 2 θ ( tan 2 θ)

= cos 2 θ ( sinθ ____ cosθ

) 2

= cos 2 θ ( sin 2 θ _____ cos 2 θ

)

= sin 2 θ

4. csc θ tan θ

= sin θ ( sin θ _____ cos θ

)

= 1 _____ cos θ

5. (1 + sec 2 θ) (1 - sin 2 θ)

= (1 + 1 _____ cos 2 θ

) ( cos 2 θ)

= ( 1 + cos 2 θ _________ cos 2 θ

) ( cos 2 θ)

= 1 + cos 2 θ

6. sin 2 θ + cos 2 θ + tan 2 θ

= (1 - cos 2 θ) + cos 2 θ + ( sin θ _____ cos θ

) 2

= 1 + sin 2 θ _____ cos 2 θ

= cos 2 θ + sin 2 θ ____________ cos 2 θ

= 1 _____ cos 2 θ

7. mg sin θ = µmg cos θ sin θ = µ cos θ sin θ = 0.94 cos θ

sin θ _____ cos θ

= 0.94

tan θ = 0.94 θ ≈ 43° The glass-top table can be tilted to an angle of

about 43° before a glass plate on the table begins to slide.


8. sec θ cot θ

= ( 1 _____ cos θ

) ( cos θ _____ sin θ

)

= 1 ____ sin θ

= csc θ

9. sin θ - cos θ ___________ sin θ

= sin θ ____ sin θ

- cos θ _____ sin θ

= 1 - cot θ

10. tan θ sin θ

= sin 2 θ _____ cos θ

= 1 - cos 2 θ _________ cos θ

= sec θ - cos θ

11. sec 2 θ (1 - cos 2 θ)

= ( 1 _____ cos 2 θ

) sin 2 θ

= sin 2 θ _____ cos 2 θ

= tan 2 θ

12. cos 2 θ ________ 1 + sin θ

= 1 - sin 2 θ ________ 1 + sin θ

= (1 + sin θ) (1 - sin θ)

_________________ 1 + sin θ

= 1 - sin θ

13. tan θ ____ cot θ

= ( sin θ _____ cos θ

) ( sin θ _____ cos θ

)

= sin 2 θ _____ cos 2 θ

= sin 2 θ ________ 1 - sin 2 θ

14. cos θ cot θ + sin θ

= cos θ ( cos θ _____ sin θ

) + sin θ

= cos 2 θ _____ sin θ

+ sin 2 θ _____ sin θ

= cos 2 θ + sin 2 θ ____________ sin θ

= 1 ____ sin θ

15. sec 2 θ - 1 _________ 1 + tan 2 θ

=

1 _____ cos 2 θ

- 1

_________ 1 + sin 2 θ _____

cos 2 θ

= ( 1 - cos 2 θ _________ cos

2 θ

) ( cos 2 θ ____________ cos 2 θ + sin 2 θ

)

= 1 - cos 2 θ = sin 2 θ 16. mg sin θ = µmg cos θ sin θ = µ cos θ sin θ = 0.4 cos θ

sin θ _____ cos θ

= 0.9

tan θ = 0.9 θ ≈ 42° 17. tan θ cot θ

= ( sin θ _____ cos θ

) ( cos θ _____ sin θ

)

= 1

18. sin θ cot θ tan θ

= sin θ ( cos θ _____ sin θ

) ( sin θ _____ cos θ

)

= sin θ

19. cos θ + sin θ tan θ

= cos θ + sin θ ( sin θ _____ cos θ

)

= cos θ + sin 2 θ _____ cos θ

= cos 2 θ + sin 2 θ ____________ cos θ

= 1 _____ cos θ

= sec θ

20. sin θ csc θ - cos 2 θ

= sin θ ( 1 ____ sin θ

) - cos 2 θ

= 1 - cos 2 θ = sin 2 θ

21. cos 2 θ sec θ csc θ

= cos 2 θ ( 1 _____ cos θ

) ( 1 ____ sin θ

)

= cos θ _____ sin θ

= cot θ

22. cos θ ( tan 2 θ + 1)

= cos θ ( sec 2 θ)

= cos θ ( 1 _____ cos 2 θ

)

= 1 _____ cos θ

= sec θ

23. csc θ (1 - cos 2 θ)

= ( 1 ____ sin θ

) ( sin 2 θ)

= sin θ

24. csc θ cos θ tan θ

= ( 1 ____ sin θ

) (cos θ) ( sin θ _____ cos θ

)

= 1


CS10_A2_MESK710389_C11.indd 324 4/11/11 1:17:30 PM

25. sin θ _________ 1 - cos 2 θ

= sin θ _____ sin 2 θ

= 1 ____ sin θ

= csc θ

26. sin 2 θ _________ 1 - cos 2 θ

= sin 2 θ _____ sin 2 θ

= 1

27. tan θ ________ sin θ sec θ

= ( sin θ ____

cos θ ) _________

sin θ ( 1 ____ cos θ

)

= ( sin θ _____ cos θ

) ( cos θ _____ sin θ

)

= 1

28. cos θ ________ sin θ cot θ

= cos θ __________ sin θ ( cos θ _____

sin θ )

= cos θ _____ cos θ

= 1

29. tan θ (tan θ + cot θ)

= ( sin θ _____ cos θ

) ( sin θ _____ cos θ

+ cos θ _____ sin θ

)

= ( sin θ _____ cos θ

) ( sin 2 θ + cos 2 θ ____________ cos θ sin θ

)

= ( sin θ _____ cos θ

) ( 1 ________ cos θ sin θ

)

= 1 _____ cos 2 θ

= sec 2 θ

30. sin 2 θ + cos 2 θ + cot 2 θ

= 1 + cos 2 θ _____ sin 2 θ

= sin 2 θ + cos 2 θ ____________ sin 2 θ

= 1 _____ sin 2 θ

= csc 2 θ

31. sin 2 θ sec θ csc θ

= sin 2 θ ( 1 _____ cos θ

) ( 1 ____ sin θ

)


= tan θ

32. cos θ - 1 ________ cos 2 θ

= cos θ _____ cos 2 θ

- 1 _____ cos 2 θ

= 1 _____ cos θ

- 1 _____ cos 2 θ

= sec θ - sec 2 θ

33. sin 2 θ ( csc 2 θ - 1) = sin 2 θ cot 2 θ

= sin 2 θ ( cos 2 θ _____ sin 2 θ

)

= cos 2 θ

34. tan θ + cot θ



= sin 2 θ + cos 2 θ ____________ sin θ cos θ

= 1 ________ sin θ cos θ

= ( 1 ____ sin θ

) ( 1 _____ cos θ

)

= sec θ csc θ

35. cos θ ________ 1 - sin

2 θ

= cos θ _____ cos

2 θ

= 1 _____ cos θ

= sec θ

36. 1 - cos 2 θ _________

tan θ

= sin 2 θ ______ ( sin θ _____ cos θ

)

= sin 2 θ ( cos θ _____

sin θ )

= sin θ cos θ

37. csc 2 θ ________

1 + tan 2 θ

= csc 2 θ _____

sec 2 θ

= ( 1 _____ sin 2 θ

)

_____ ( 1 _____ cos

2 θ

)

= cos 2 θ _____

sin 2 θ

= cot 2 θ

38. tan θ = y __ x

= r sin θ _____ r cos θ


39. cot θ = x __ y

= r cos θ _____ r sin θ

= cos θ _____ sin θ

40. x 2 + y 2 = r 2

x 2 __ y 2

+ y 2

__ y 2

= r 2 __

y 2

( x __ y ) 2 + 1 = ( r __ y )

2

cot 2 θ + 1 = csc 2 θ 1 + cot 2 θ = csc 2 θ

41. csc θ = r __ y

= r _____ r sin θ

= 1 ____ sin θ

42. sec θ = r __ x

= r _____ r cos θ

= 1 _____ cos θ

43. x 2 + y 2 = r 2

x 2 __ x 2

+ y 2

__ x 2

= r 2 __

x 2

1 + ( y __ x )

2 = ( r __ x )

2

1 + tan 2 θ = sec 2 θ

44a. y(t) = z(t) 5 sin t = 2.6 cos t

sin t ____ cos t

= 2.6 ___ 5

tan t = 0.52 t ≈ 0.48 s

b. y(0.48) = 5 sin 0.48 ≈ 2.31

c. Because the masses have the same displacement at ≈ 0.48 s, they will also have the same displacement at times ≈ 0.48 + πn, where n is an integer.

45. (csc θ - 1) (csc θ + 1) = tan 2 θ

( 1 ____ sin θ

- 1) ( 1 ____ sin θ

+ 1) = tan 2 θ

not an identity

46. sec θ - cos θ = sin θ

1 _____ cos θ

- cos θ = sin θ

not an identity

47. cos θ (sec θ + cos θ csc 2 θ) = csc 2 θ

cos θ ( 1 _____ cos θ

+ cos θ _____ sin 2 θ

) = 1 _____ sin 2 θ

identity

48. cot θ (cos θ + sin θ tan θ) = csc θ

( cos θ _____ sin θ

) (cos θ + sin 2 θ _____ cos θ

) = 1 ____ sin θ

identity


CS10_A2_MESK710389_C11.indd 325 4/11/11 1:17:30 PM

49. cos θ = 0.99 cos θ not an identity

50. sin θ cos θ = tan θ - tan θ sin 2 θ

sin θ cos θ = sin θ _____ cos θ

- ( sin θ _____ cos θ

) ( sin 2 θ)

identity

51a. r = g tan θ

______ ω 2

= g tan θ

__________

( √

g ______

ℓ cos θ )

2

= g tan θ

______

g ______

ℓ cos θ

= (g tan θ) ( ℓ cos θ ______ g )

= ℓ tan θ cos θ

= ℓ ( sin θ _____ cos θ

) cos θ

= ℓ sin θ

b. ω = √

g ______

ℓ cos θ

ω 2 = g ______

ℓ cos θ

ℓ = g _______

ω 2 cos θ

ℓ = g ___

ω 2 sec θ

52. odd: sine, tangent, cotangent, cosecant; even: cosine, secant

53. The graphs of even functions show reflection symmetry across the y-axis. Odd functions show 180° rotational symmetry about the origin, or both a reflection across the x-axis and the y-axis.

54a. odd b. even 55. an infinite number of equivalent forms;

tan θ = sin θ _____ cos θ

, cos θ = sin θ ____ tan θ

, sin θ = tan θ cos θ

56. Because tan θ = sin θ _____ cos θ

, tan(-θ) = sin(-θ)

_______ cos(-θ)

.

Use sin(-θ) = - sin θ and cos(-θ) = cos θ to get

tan(-θ) = -sin θ ______ cos θ

= -tan θ.

teSt prep

57. D sec θ sin θ

= 1 ________ cos θ sin θ

= tan θ 58. J

F sec θ csc θ = 1 ________ cos θ sin θ

G 1 ________ sin θ cos θ

H tan θ _____ sin 2 θ

= sin θ _________ cos θ sin 2 θ

= 1 ________ cos θ sin θ

J cos 2 θ _____ cot θ

= cos 2 θ ( sin θ _____ cos θ

) = cos θ sin θ

59. A 1 + cos 2 θ = sin 2 θ

1 + x 2 __ r 2

= y 2

__ r 2

r 2 + x 2 ______

r 2

= y 2

__ r 2

-y 2

____ r 2

≠

y 2 __

r 2

60. G 1 + tan 2 θ = sec 2 θ so, 1 - sec 2 θ = - tan 2 θ

61. sin θ + cot θ cos θ

= sin θ + ( cos θ _____ sin θ

) cos θ Given, ratio identity

= sin 2 θ _____ sin θ

+ cos 2 θ _____ sin θ

Common denominators

= sin 2 θ + cos 2 θ ____________ sin θ

Add fractions.

= 1 ____ sin θ

Pythagorean identity

= csc θ Ratio identity


62. 1 _____ cos θ

+ 1 _____ cos 2 θ

= cos θ _____ cos 2 θ

+ 1 _____ cos 2 θ

= cos θ + 1 ________ cos 2 θ

63. cos θ _____ sin θ

+ sin θ _____ cos θ

= cos 2 θ + sin 2 θ ____________ sin θ cos θ

= 1 ________ sin θ cos θ

64. 1- cos θ _____ sin θ

= sin θ ____ sin θ


= sin θ - cos θ ___________ sin θ

65. 1 ________ 1 - cos θ

- cos θ _________ 1 - cos 2 θ

= 1 + cos θ _________ 1 - cos 2 θ

- cos θ _________ 1 - cos 2 θ

= 1 _________ 1 - cos 2 θ

66.

1 _____ sin 2 θ

- 1

________ cos 2 θ _____ sin 2 θ

= ( 1 - sin 2 θ ________ sin 2 θ

) ( sin 2 θ _____ cos 2 θ

)

= ( cos 2 θ _____

sin 2 θ

) ( sin 2 θ _____

cos 2 θ )

= 1

67. 1 ____ sin θ

+ 1 _____ cos θ

___________

1 ________ sin θ cos θ

= ( cos θ + sin θ ___________ sin θ cos θ

) (sin θ cos θ)

= sin θ + cos θ


CS10_A2_MESK710389_C11.indd 326 4/11/11 1:17:31 PM

68. 1 ____ sin θ

- 1 _____ cos θ

_________

sin θ _____ cos θ


= ( cos θ - sin θ ___________ sin θ cos θ

) ( sin θ cos θ ____________ sin 2 θ - cos 2 θ

)

= cos θ - sin θ ____________ sin 2 θ - cos 2 θ

= - (sin θ + cos θ)

_______________________ (sin θ + cos θ)(sin θ - cos θ)

= - 1 __________ sin θ + cos θ

69. 1 - 1 ____

sin θ ________

1 - 1 _____ sin 2 θ

= ( sin θ - 1 ________ sin θ

) ( sin 2 θ ________ sin 2 θ - 1

)

= (sin θ - 1) sin θ

_____________ sin 2 θ - 1

= (sin θ - 1)sin θ

_________________ (sin θ - 1) (sin θ + 1)

= sin θ ________ sin θ + 1

sum And dIfference IdenTITIes

CheCk it out!

1a. tan 105° = tan (45° + 60°)

= tan 45° + tan 60° _______________ 1 - tan 45° tan 60°

= 1 + √ 3

________ 1 - 1 √ 3

= 1 + √ 3

_______ 1 - √ 3

· 1 + √ 3

_______ 1 + √ 3

= 4 + 2 √ 3

________ -2

= -2 - √ 3

b. sin (- 11π ____ 12

) = sin ( π __ 3 - 5π ___

4 )

= sin π __ 3 cos 5π ___

4 - cos π __

3 sin 5π ___

4

= √ 3

___ 2 · (- 1 ___

√ 2 ) - 1 __

2 · (- 1 ___

√ 2 )

= - √ 3

____ 2 √ 2

+ 1 ____ 2 √ 2

= 1 - √ 3

_______ 2 √ 2

· 2 √ 2 ____ 2 √ 2

= 2 √ 2 - 2 √ 6

__________ 4 · 2

= √ 2 - √ 6

________ 4

2. cos (x + π __ 2 )

= cos ( π __ 2 ) cos x - sin ( π __

2 ) sin x

= (0)cos x - (1)sin x = -sin x

3. 90° < A < 180°

sin A = y __ r = 4 __

5

x 2 + 4 2 = 5 2 x = - √ 25 - 16 x = -3

Thus, cos A = x __ r = - 3 __ 5 .

0° < B < 90°

cos B = x __ r = 3 __ 5

3 2 + y 2 = 5 2 y = √ 25 - 9 y = 4

Thus, sin B = y __ r = 4 __

5 .

sin (A - B) = sin A cos B - cos A sin B

= ( 4 __ 5 ) ( 3 __

5 ) - (- 3 __

5 ) ( 4 __

5 )

= 12 ___ 25

+ 12 ___ 25

= 24 ___ 25

4. R 60° = ⎡

⎢

⎣ cos 60°

sin 60°

-sin 60°

cos 60°

⎤

⎥

⎦ , S =

⎡

⎢

⎣ 0

2

0

4

√ 3

1

- √ 3

1

⎤

⎥

⎦

R 60° × S = ⎡

⎢

⎣ cos 60°

sin 60°

-sin 60°

cos 60°

⎤

⎥

⎦ ⎡

⎢

⎣ 0

2

0

4

√ 3

1

- √ 3

1

⎤ ⎥

⎦

= ⎡ ⎢

⎣ - √ 3

1

-2 √ 3

2 0

2 - √ 3

-1

⎤

⎥

⎦

The coordinates of the points after a 60° rotation are

A' (-2.00, 0.00) , B' (-4.00, 0.00) , C' (-1.00, 1.73) , D' (-1.00, -1.73) .

think anD DisCuss

1. Possible answer: Evaluate sin (60° - 45°) , sin (45° - 30°) , or sin (135° - 120°) .

2. Possible answer: Both the sine and cosine identities are in the form of a sum or difference of 2 products. The products each involve the 2 different angles given in the same order (A then B) .

In sine identities, a cosine is multiplied by a sine, and the sign of the second term matches the sign between A and B. In cosine identities, cosines are multiplied by each other, sines are multiplied by each other, and the signs are opposites.

3. tan ( A + B ) =

sin ( A + B )

Sum and Difference Identities

Sine: = sin A cos B + cos A sin B = sin A cos B - cos A sin B

Tangent: tan A + tan B tan A - tan B

Cosine: = cos A cos B - sin A sin B = cos A cos B + sin A sin B

1 - tan A tan B

1 + tan A tan B

sin ( A - B )

tan ( A - B ) =

cos ( A + B ) cos ( A - B )


1. A rotation matrix assumes a counterclockwise rotation about the origin.


x-4


11-4

CS10_A2_MESK710389_C11.indd 327 4/11/11 1:17:31 PM

2. cos105° = cos (135° - 30°) = cos 135° cos30 + sin135°sin30°

= - 1 ___ √ 2

· √ 3

___ 2 + 1 ___

√ 2 · 1 __

2

= - √ 6

_____ 4 +

√ 2 ___ 4 =

√ 2 - √ 6 ________

4

3. sin 11π ____ 12

= sin ( π __ 4 + 2π ___

3 )

= sin π __ 4 cos 2π ___

3 + cos π __

4 sin 2π ___

3

= 1 ___ √ 2

· - 1 __ 2 + 1 ___

√ 2 ·

√ 3 ___

2

= - 1 ____ 2 √ 2

+ √ 3

____ 2 √ 2

= -1 + √ 3

________ 2 √ 2

· 2 √ 2 ____ 2 √ 2

= -2 √ 2 + 2 √ 6

____________ 4 · 2

= √ 6 - √ 2

________ 4

4. tan π ___ 12

= tan ( 4π ___ 12

- 3π ___ 12

)

= tan 4π ___

12 - tan 3π ___

12 _______________

1 + tan 4π ___ 12

tan 3π ___ 12

= tan π __

3 - tan π __

4 _____________

1 + tan π __ 3 tan π __

4

= √ 3 - 1

_______ 1 + √ 3

= √ 3 - 1

_______ 1 + √ 3

· 1 - √ 3

_______ 1 - √ 3

= -1 + 2 √ 3 - 3

_____________ -2

= -4 + 2 √ 3

_________ -2

= 2 - √ 3

5. cos(-75°) = cos(60° - 135°) = cos 60° cos 135° + sin 60° sin 135°

= 1 __ 2 · (- 1 ___

√ 2 ) +

√ 3 ___

2 · 1 ___

√ 2

= - 1 ____ 2 √ 2

+ √ 3

____ 2 √ 2

= √ 3 - 1

_______ 2 √ 2

= √ 3 - 1

_______ 2 √ 2

· √ 2

___ √ 2

= √ 6 - √ 2

________ 2 · 2

= √ 6 - √ 2

________ 4

6. sin ( π __ 2 + x) = cos x

sin π __ 2 cos x + cos π __

2 sin x =

1 · cos x + 0 · sin x = cos x = cos x

7. tan (π + x) = tan x

tan π + tan x ____________ 1 - tan π tan x

=

0 + tan x __________ 1 - 0 · tan x

=

tan x = tan x

8. cos ( 3π ___ 2 - x) = -sin x

cos 3π ___ 2 cos x + sin 3π ___

2 sin x =

0 · cos x + (-1) · sin x = -sin x = -sin x

9. step 1 Find cos A and cos B.

A is in QIII, and sinA = - 12 ___ 13

.

x 2 + (-12) 2 = 13 2 x = - √ 169 - 144 x = -5

Thus, cos A = - 5 ___ 13

.

B is in QII, and sin B = 4 __ 5 .

x 2 + 4 2 = 5 2 x = - √ 25 - 16 x = -3

Thus, cos B = - 3 __ 5 .

step 2 Use the angle-sum identity to find sin (A + B).sin (A + B) = sin A cos B + cos A sin B

= - 12 ___ 13

· (- 3 __ 5 ) + (- 5 ___

13 ) · 4 __

5

= 36 ___ 65

- 20 ___ 65

= 16 ___ 65

10. step 1 Use cos A and cos B obtained in question 9.step 2 Use the angle-sum identity to find cos (A - B).cos (A - B) = cos A cos B + sin A sin B

= - 5 ___ 13

· (- 3 __ 5 ) + (- 12 ___

13 ) · 4 __

5

= 15 ___ 65

- 48 ___ 65

= - 33 ___ 65

11. step 1 Use cos A and cos B obtained in question 9 to find tan A and tan B.

tan A = sin A _____ cos A

tan B = sin B _____ cos B

tan A = - 12 ___

13 ____

- 5 ___ 13

= 12 ___

5 tan B =

4 __ 5 ___

- 3 __ 5 = - 4 __

3

step 2 Use the angle-sum identity to find tan (A + B).

tan (A + B) = tan A + tan B ____________ 1 - tan A tan B

= 12 ___ 5 + (- 4 __

3 ) ____________

1 - 12 ___ 5 · (- 4 __

3 )

= 16 ___ 15

______

1 + 48 ___ 15

=

16 ___ 15

___

63 ___ 15

= 16 ___

63


CS10_A2_MESK710389_C11.indd 328 4/11/11 1:17:32 PM

12. step 1 Use tan A and tan B obtained in question 11.step 2 Use the angle-sum identity to find tan (A - B).

tan (A - B) = tan A - tan B ____________ 1 + tan A tan B

= 12 ___ 5 - (- 4 __

3 ) ____________

1 + 12 ___ 5 · (- 4 __

3 )

= 56 ___ 15

______

1 - 48 ___ 15

=

56 ___ 15

____

- 33 ___ 15

= - 56 ___

33

13. step 1 Write matrices for a 120° rotation and for the points in the figure.

R 120° = ⎡ ⎢ ⎣ cos 120°

sin 120°

-sin 120°

cos 120°

⎤ ⎥ ⎦

S = ⎡ ⎢ ⎣ 0

2

0

-1

3

0 ⎤ ⎥ ⎦

step 2 Find the matrix product.

R 120° × S = ⎡ ⎢ ⎣ cos 120°

sin 120°

-sin 120°

cos 120°

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

2

0

-1

3

0 ⎤ ⎥ ⎦

= ⎡ ⎢ ⎣ -1.73

-1

0.87

0.5

-1.5

2.6

⎤ ⎥ ⎦

step 3 The approximate coordinates of the points after a 120° rotation are A′ (-1.73, -1) , B′ (0.87, 0.5) , and C′ (-1.5, 2.6) .

14. sin 7π ___ 12

= sin ( 3π ___ 12

+ 4π ___ 12

)

= sin ( π __ 4 + π __

3 )

= sin π __ 4 cos π __

3 + cos π __

4 sin π __

3

= 1 ___ √ 2

· 1 __ 2 + 1 ___

√ 2 ·

√ 3 ___

2

= 1 ____ 2 √ 2

+ √ 3

____ 2 √ 2

= 1 + √ 3

_______ 2 √ 2

· √ 2 ___ √ 2

= √ 2 + √ 6

________ 4

15. tan 165° = tan (30° + 135°)

= tan 30° + tan 135° ________________ 1 - tan 30° tan 135°

=

1 ___ √ 3

- 1

_______ 1 + 1 ___

√ 3

=

1 - √ 3

_______ √ 3

_______ 1 + √ 3

_______ √ 3

= 1 - √ 3

_______ 1 + √ 3

· 1 - √ 3

_______ 1 - √ 3

= 4 - 2 √ 3

________ -2

= √ 3 - 2

16. sin 195° = sin (135° + 60°) = sin 135° cos 60° + cos 135° sin 60°

= 1 ___ √ 2

· 1 __ 2 - 1 ___

√ 2 ·

√ 3 ___

2

= 1 ____ 2 √ 2

- √ 3

____ 2 √ 2

= 1 - √ 3

_______ 2 √ 2

· √ 2 ___ √ 2

= √ 2 - √ 6

________ 4

17. cos 11π ____ 12

= cos ( 8π ___ 12

+ 3π ___ 12

)

= cos ( 2π ___ 3 + π __

4 )

= cos 2π ___ 3 cos π __

4 - sin 2π ___

3 sin π __

4

= - 1 __ 2 · 1 ___

√ 2 -

√ 3 ___

2 · 1 ___

√ 2

= - 1 ____ 2 √ 2

- √ 3

____ 2 √ 2

= - 1 - √ 3

_______ 2 √ 2

· √ 2 ___ √ 2

= - √ 2 - √ 6

__________ 4

18. cos ( 3π ___ 2 + x) = sin x

cos 3π ___ 2 cos x - sin 3π ___

2 sin x =

0 · cos x + 1 · sin x = sin x = sin x

19. sin ( 3π ___ 2 + x) = -cos x

sin 3π ___ 2 cos x + cos 3π ___

2 sin x =

-1 · cos x + 0 · sin x = -cos x = -cos x

20. tan (x - 2π) = tan x

tan x - tan 2π _____________ 1 + tan x tan 2π

=

tan x - 0 ________ 1 =

tan x = tan x 21. step 1 Find sin A and cos B.

A is in QII, and cosA = - 12 ___ 13

.

(-12) 2 + y 2 = 13 2 y = √ 169 - 144 y = 5

Thus, sin A = 5 ___ 13

.

B is in QIV, and sin B = - 4 __ 5 .

x 2 + (-4) 2 = 5 2 x = √ 25 - 16 x = 3

Thus, cos B = 3 __ 5 .


CS10_A2_MESK710389_C11.indd 329 4/11/11 1:17:32 PM

step 2 Use the angle-sum identity to find sin (A + B).sin (A + B) = sin A cos B + cos A sin B

= 5 ___ 13

· 3 __ 5 + (- 12 ___

13 ) · (- 4 __

5 )

= 15 ___ 65

+ 48 ___ 65

= 63 ___ 65

22. step 1 Use cos A and cos B obtained in question 21 to find tan A and tan B.

tan A = sin A _____ cos A

tan B = sin B _____ cos B

tan A = 5 ___ 13

____

- 12 ___ 13

= - 5 ___

12 tan B =

- 4 __ 5 ___

3 __ 5 = - 4 __

3

step 2 Use the angle-sum identity to find tan (A - B).


= - 5 ___

12 - (- 4 __

3 ) ____________

1 + 5 ___ 12

· (- 4 __ 3 )

= 11 ___ 12

______

1 - 20 ___ 36

=

11 ___ 12

___

16 ___ 36

= 33 ___

63

23. step 1 Use cos A and cos B obtained in question 21.step 2 Use the angle-sum identity to find cos (A + B).cos (A + B) = cos A cos B - sin A sin B

= - 12 ___ 13

· 3 __ 5 - 5 ___

13 · (- 4 __

5 )

= - 36 ___ 65

+ 20 ___ 65

= - 16 ___ 65

24. step 1 Use cos A and cos B obtained in question 21.step 2 Use the angle-sum identity to find cos (A - B).cos (A - B) = cos A cos B + sin A sin B

= - 12 ___ 13

· 3 __ 5 + 5 ___

13 · (- 4 __

5 )

= - 36 ___ 65

- 20 ___ 65

= - 56 ___ 65

25. step 1 Write matrices for a 45° rotation and for the points in the figure.

R 45° = ⎡ ⎢ ⎣ cos 45°

sin 45°

-sin 45°

cos 45°

⎤ ⎥ ⎦

S = ⎡ ⎢ ⎣ 0

2 1

2

0

1 ⎤ ⎥ ⎦

step 2 Find the matrix product.

R 45° × S = ⎡ ⎢ ⎣ cos 45°

sin 45°

-sin 45°

cos 45°

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

2 1

2

0

1 ⎤ ⎥ ⎦

= ⎡ ⎢ ⎣ -1.41

1.41

-0.71

2.12

-0.71

0.71 ⎤ ⎥ ⎦

step 3 The approximate coordinates of the points after a 45° rotation are A′ (-1.41, 1.41) , B′ (-0.71, 2.12) , and C′ (-0.71, 0.71) .


= 1 ___ √ 2

· √ 3

___ 2 - 1 ___

√ 2 · 1 __

2

= √ 3

____ 2 √ 2

- 1 ____ 2 √ 2

= √ 3 - 1

_______ 2 √ 2

· √ 2 ___ √ 2

= √ 6 - √ 2

________ 4

27. tan (-105°) = tan (30° - 135°)

= tan 30° - tan 135° _______________ 1 + tan30°tan135°

=

1 ___ √ 3

+ 1

_______ 1 - 1 ___

√ 3

=

1 + √ 3

_______ √ 3

_______ √ 3 - 1

_______ √ 3

= √ 3 + 1

_______ √ 3 - 1

· √ 3 + 1

_______ √ 3 + 1

= 2 √ 3 + 4

________ 2 = 2 + √ 3

28. cos 195° = cos (135° + 60°) = cos 135° cos 60° - sin 135° sin 60°

= - 1 ___ √ 2

· √ 3

___ 2 - 1 ___

√ 2 · 1 __

2

= - √ 3

_____ 2 √ 2

- 1 _____ 2 √ 2

= - √ 3 - 1

________ 2 √ 2

· √ 2 ___ √ 2

= - √ 6 - √ 2

__________ 4

29. sin (-15°) = sin (30° - 45°) = sin 30° cos 45° - cos 30° sin 45°

= 1 __ 2 · 1 ___

√ 2 -

√ 3 ___

2 · 1 ___

√ 2

= 1 ____ 2 √ 2

- √ 3

____ 2 √ 2

= 1 - √ 3

_______ 2 √ 2

· √ 2 ___ √ 2

= √ 2 - √ 6

________ 4


CS10_A2_MESK710389_C11.indd 330 4/11/11 1:17:33 PM

30. cos 19π ____ 12

= cos ( 3π ___ 12

+ 16π ____ 12

)

= cos ( π __ 4 + 4π ___

3 )

= cos π __ 4 cos 4π ___

3 - sin π __

4 sin 4π ___

3

= 1 ___ √ 2

· (- 1 __ 2 ) - 1 ___

√ 2 · (-

√ 3 ___

2 )

= -1 ____ 2 √ 2

+ √ 3

____ 2 √ 2

= √ 3 - 1

_______ 2 √ 2

· √ 2 ___ √ 2

= √ 6 - √ 2

________ 4

31. tan 5π ___ 12

= tan ( 9π ___ 12

- 4π ___ 12

)

= tan ( 3π ___ 4 - π __

3 )

= tan 3π ___

4 - tan π __

3 ______________

1 + tan 3π ___ 4 tan π __

3

= -1 - √ 3

____________ 1 + (-1) · √ 3

= -1 - √ 3

________ 1 - √ 3

· 1 + √ 3

_______ 1 + √ 3

= -4 - 2 √ 3

_________ -2

= 2 + √ 3


= 1 ___ √ 2

· (- 1 __ 2 ) + (- 1 ___

√ 2 ) ·

√ 3 ___

2

= ( -1 ____ 2 √ 2

) - √ 3

____ 2 √ 2

= -1 - √ 3

________ 2 √ 2

· √ 2 ___ √ 2

= - √ 2 - √ 6

__________ 4

33. tan 195° = tan (135° + 60°)

= tan 135° + tan 60° ________________ 1 - tan 135° tan 60°

= -1 + √ 3

____________ 1 - (-1) · √ 3

= -1 + √ 3

________ 1 + √ 3

· 1 - √ 3

_______ 1 - √ 3

= -4 + 2 √ 3

_________ -2

= 2 - √ 3

34. cos π ___ 12

= cos ( 4π ___ 12

- 3π ___ 12

)

= cos ( π __ 3 - π __

4 )

= cos π __ 3 cos π __

4 + sin π __

3 sin π __

4

= 1 __ 2 · 1 ___

√ 2 +

√ 3 ___

2 · 1 ___

√ 2

= 1 ____ 2 √ 2

+ √ 3

____ 2 √ 2

= 1 + √ 3

_______ 2 √ 2

· √ 2 ___ √ 2

= √ 2 + √ 6

________ 4

35. cos (θ - 30°) = 1 __ 2

θ is in QI or QII. 1 2 + y 2 = 2 2 y = √ 4 - 1 y = √ 3 θ - 30° = 60° θ = 90°

36. cos (20° + θ) = √ 2 ___ 2

θ is in QI or QII.

( √ 2 ) 2 + y 2 = 2 2

y = √ 4 - 2 y = √ 2 20° + θ = 45° θ = 25°

37. sin (180° - θ) = 1 __ 2

θ is in QI or QII. x 2 + 1 2 = 2 2 x = ± √ 4 - 1 x = ± √ 3 180° - θ = 30° 180° - θ = 150° θ = 150° θ = 30°

38a. △ = ( sin ( θ i - θ r )

__________ sin θ i cos θ r

) t

= ( sin θ i cos θ r - cos θ i sin θ r

____________________ sin θ i cos θ r

) t

= ( sin θ i cos θ r

_________ sin θ i cos θ r

- cos θ i sin θ r

_________ sin θ i cos θ r

) t

= (1 - cos θ i _____ sin θ i

sin θ r _____ cos θ r

) t

= (1 - cot θ i tan θ r ) t

b. sin ( θ i - θ r ) = ℓ __ h

39. step 1 Find cos A, sin B, tan A, and tan B.

A is in QIII, and sin A = - 7 ___ 25

.

x 2 + (-7) 2 = 2 5 2 x = - √ 625 - 49 x = -24

Thus, cos A = - 24 ___ 25

, tan A = 7 ___ 24

.

B is in QI, and cos B = 12 ___ 13

.

12 2 + y 2 = 13 2 y = √ 169 - 144 y = 5

Thus, sin B = 5 ___ 13

, tan B = 5 ___ 12

.

step 2 Use the angle-sum identity to find tan (A + B), cos (A + B), and sin (A - B).


CS10_A2_MESK710389_C11.indd 331 4/11/11 1:17:33 PM


= 7 ___ 24

+ 5 ___ 12

__________

1 - 7 ___ 24

· 5 ___ 12

= 204 ____ 288

_______

1 - 35 ____ 288

= 204 ____

253

cos (A + B) = cos A cos B - sin A sin B

= - 24 ___ 25

· 12 ___ 13

- (- 7 ___ 25

) · 5 ___ 13

= - 288 ____ 325

+ 35 ____ 325

= - 253 ____ 325


= - 7 ___ 25

· 12 ___ 13

- (- 24 ___ 25

) · 5 ___ 13

= - 84 ____ 325

+ 120 ____ 325

= 36 ____ 325

40. step 1 Find cos A, cos B, tan A, and tan B.

A is in QVI, and sin A = - 1 __ 3 .

x 2 + (-1) 2 = 3 2 x = - √ 9 - 1 x = 2 √ 2

Thus, cos A = 2 √ 2

____ 3 , tan A = - 1 ____

2 √ 2 .

B is in QI, and sin B = 4 __ 5 .

x 2 + 4 2 = 5 2 x = √ 25 - 16 x = 3

Thus, cos B = 3 __ 5 , tan B = 4 __

3 .

step 2 Use the angle-sum identity to find tan (A + B), cos (A + B), and sin (A - B).


=

- 1 ____ 2 √ 2

+ 4 __ 3

______________ 1 - 4 __

3 · (- 1 ____

2 √ 2 )

= 16 - 3 √ 2 _________

12 _________

12 - 4 √ 2 _________ 12

= 16 - 3 √ 2 _________ 12 - 4 √ 2

· 12 + 4 √ 2 _________ 12 + 4 √ 2

= 216 - 100 √ 2 ____________ 112

= 54 - 25 √ 2 __________ 28

cos (A + B) = cos A cos B - sin A sin B

= 2 √ 2 ____ 3 · 3 __

5 - (- 1 __

3 ) · 4 __

5

= 6 √ 2 ____ 15

+ 4 ___ 15

= 4 + 6 √ 2 ________ 15


= - 1 __ 3 · 3 __

5 - 2 √ 2 ____

3 · 4 __

5

= - 3 ___ 15

- 8 √ 2 ____ 15

= -3 - 8 √ 2 _________ 15

41a. R 90° = ⎡ ⎢ ⎣ cos 90°

sin 90°

-sin 90°

cos 90°

⎤ ⎥ ⎦ =

⎡ ⎢ ⎣ 0

1 -1

0 ⎤ ⎥ ⎦

R 180° = ⎡ ⎢ ⎣ cos 180°

sin 180°

-sin 180°

cos 180°

⎤ ⎥ ⎦ =

⎡ ⎢ ⎣ -1

0

0

-1 ⎤ ⎥ ⎦

R 270° = ⎡ ⎢ ⎣ cos 270°

sin 270°

-sin 270°

cos 270°

⎤ ⎥ ⎦ =

⎡ ⎢ ⎣

0

-1 1

0 ⎤ ⎥ ⎦

b. S = ⎡ ⎢ ⎣ 0

0 1

1 4

0 1

-1 ⎤ ⎥ ⎦

R 90° × S = ⎡ ⎢ ⎣ 0

1 -1

0 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0 1

1 4

0 1

-1 ⎤ ⎥ ⎦

= ⎡ ⎢ ⎣ 0

0 -1

1

0

4 1

1 ⎤ ⎥ ⎦

P ′(0, 0), Q ′(-1, 1), R ′(0, 4), S ′(1, 1)

R 180° × S = ⎡ ⎢ ⎣ -1

0

0

-1 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0 1

1 4

0 1

-1 ⎤ ⎥ ⎦

= ⎡ ⎢ ⎣ 0

0 -1

-1 -4

0 -1

1 ⎤ ⎥ ⎦

P ′′(0, 0), Q ′′(-1, -1), R ′′(-4, 0), S ′′(-1, 1)

R 270° × S = ⎡ ⎢ ⎣

0

-1 1

0 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0 1

1 4

0 1

-1 ⎤ ⎥ ⎦

= ⎡ ⎢ ⎣ 0

0 1

-1

0

-4 -1

-1 ⎤ ⎥ ⎦

P ′′′(0, 0), Q ′′′(1, -1), R ′′′(0, -4), S ′′′(-1, -1)

c. 4

-4

x

y

-4 4

42. Possible answer: No; 11π ___ 24

cannot be expressed as asum or difference of values from the unit circle.

43a. amplitude = 4.2; period = 2π · 3 ___ 2π

= 3

b. y(t) = 4.2sin ( 2π ___ 3 t - π __

2 )

= 4.2 (sin 2π ___ 3 t cos π __

2 - cos 2π ___

3 t sin π __

2 )

= 4.2 (sin 2π ___ 3 t · 0 - cos 2π ___

3 t · 1)

= -4.2 (cos 2π ___ 3 t)

c. y(t) = -4.2 (cos 2π ___ 3 t)

y(8) = -4.2 (cos 2π ___ 3 (8))

= -4.2cos 16π ____ 3

= -4.2 (- 1 __ 2 ) = 2.1


CS10_A2_MESK710389_C11.indd 332 4/11/11 1:17:34 PM

44. R 45° × S ≈ ⎡ ⎢ ⎣ 0.71

0.71

-0.71

0.71

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

3 1

4 2

3 2

0 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ -2.12

2.12 -2.12

3.54

-0.71

3.54 1.41

1.41 ⎤ ⎥ ⎦

A ′(-2.12, 2.12), B ′(-2.12, 3.54), C ′(-0.71, 3.54), D ′(1.41, 1.41)

45. R 60° × S ≈ ⎡ ⎢ ⎣ 0.5

0.87

-0.87

0.5 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

3 1

4 2

3 2

0 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ -2.60

1.5

-2.96

2.87

-1.60

3.23

1 1.73

⎤ ⎥ ⎦

A ′(-2.60, 1.5), B ′(-2.96, 2.87), C ′(-1.60, 3.23), D ′(1, 1.73)

46. R 120° × S ≈ ⎡ ⎢ ⎣ -0.5

0.87

-0.87

-0.5

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

3 1

4 2

3 2

0 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ -2.60

-1.5

-3.96

-1.13

-3.60

0.23

-1 1.73

⎤ ⎥ ⎦

A ′(-2.60, -1.5), B ′(-3.96, -1.13), C ′(-3.60, 0.23), D ′(-1, 1.73)

47. R -30° × S ≈ ⎡ ⎢ ⎣ 0.87

-0.5

0.5

0.87 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

3 1

4 2

3 2

0 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 1.50

2.60

2.87

2.96

3.23

1.60

1.73

-1

⎤ ⎥ ⎦

A ′(1.50, 2.60), B ′(2.87, 2.96), C ′(3.23, 1.60), D ′(1.73, -1)

48. In general, sin(A + B) ≠ sin A + sin B. sin(45° + 45°) ≠ sin45° + sin45°, which results in

sin 90° ≠ √ 2 ___ 2 · 1 __

2 +

√ 2 ___ 2 · 1 __

2 , or 1 ≠ √ 2 .

49. A; 50. H;

cos 60° = 1 __ 2 sin 5π ___

6 = 1 __

2

51. A;

cos A = √ 3

___ 2 , sin B = 4 __

5

cos(A - B) = cos A cosB + sin A sinB

= √ 3

___ 2 · 3 __

5 + 1 __

2 · 4 __

5

= 3 √ 3 + 4

________ 10

52. Possible answer: sin(-15°) = sin(45° - 60°) = sin 45° cos 60° - cos 45° sin 60°

= √ 2 ___ 2 · 1 __

2 -

√ 2 ___ 2 ·

√ 3 ___

2

= √ 2 - √ 6

________ 4


53. The rotation matrix for -θ is

⎡ ⎢ ⎣ cos (-θ)

sin (-θ)

-sin (-θ)

cos (-θ)

⎤ ⎥ ⎦ = ⎡

⎢ ⎣ cos θ

-sin θ

sin θ

cos θ

⎤ ⎥ ⎦ .

The inverse of this rotation matrix is

1 ____________ cos 2 θ + sin 2 θ

⎡ ⎢ ⎣ cos θ

sin θ

-sin θ

cos θ

⎤ ⎥ ⎦ = ⎡

⎢ ⎣ cos θ

sin θ

-sin θ

cos θ

⎤ ⎥ ⎦ ,

which is equivalent to the rotation matrix for θ.

54. tan(A + B)

= sin(A + B)

_________ cos(A + B)

= sin A cos B + cos A sin B ___________________ cos A cos B - sin A sin B

= sin A cos B + cos A sin B ___________________ cos A cos B - sin A sin B

· 1 _______ cos A cos B

_______

1 _______ cos A cos B

= sin A ____ cos A

+ sin B ____ cos B

___________

1 - sin A sin B _______ cos A cos B

= tan A + tan B ____________ 1 - tan A tan B

55. x ′ = r cos (α + θ) = r cos α(cos θ) - r sin α (sin θ) = x(cos θ) - y(sin θ)

y ′ = r sin (α + θ) = r sin α(cos θ) + r cos α (sin θ) = y(cos θ) + x(sin θ)

⎡ ⎢ ⎣ cos θ

sin θ

-sin θ

cos θ

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ x y ⎤ ⎥ ⎦ = ⎡

⎢ ⎣ x ′ y ′

⎤ ⎥ ⎦

56. Point A lies on the positive x-axis, it is easy to see the angle formed by A ′ and A is the angle of the rotation. By inspection, A(1, 0) and A ′(0, 1) forms a 90° angle. Similarly 90° angles are formed between B and B ′ and C and C ′. Students can check this by applying the 90° rotation matrix to the figure.

57. Similar reasoning as question 56. A ′ ( √ 2

___ 2

, √ 2 ___ 2

)

θ is in QI, tan θ = y __ x = 1, θ = 45°.

58. Similar reasoning as question 56. A ′(-1, 0)

θ is in QII, tan θ = y __ x = 0, θ = 180°.

59. Similar reasoning as question 56. A ′ ( √ 3

___ 2

, 1 __ 2

)

θ is in QI, tan θ = y __ x = 1 ___

√ 3 , θ = 30°.

double-AnGle And hAlf-AnGle IdenTITIes

CheCk it out!

1. step 1 Solve sin 2 θ = 1 - cos 2 θ. sin 2 θ = 1 - cos 2 θ

sin θ = - √

1 - ( 1 __ 3 )

2

= - √

8 __ 9 = -

2 √ 2 ____

3

step 2 Find tan 2θ.

tan 2θ = 2 tan θ ________ 1 - tan 2 θ

= 2(-2 √ 2 )

________ 1 - 8

= 4 √ 2 ____ 7

step 3 Find cos 2θ.cos 2θ = 2 cos 2 θ - 1

= 2 ( 1 __ 3 )

2 - 1 = - 7 __

9


x-5


11-5

CS10_A2_MESK710389_C11.indd 333 4/11/11 1:17:35 PM

2a. cos 4 θ - sin 2 θ = cos 2θ

( cos 2 θ + sin 2 θ)( cos 2 θ - sin 2 θ) = (1)(cos 2θ) = cos 2θ

b. sin 2θ = 2 tan θ ________ 1 + tan 2 θ

= 2 ( sin θ ____

cos θ ) ______

sec 2 θ

= 2 ( sin θ ____

cos θ ) ______

1 ____ cos 2 θ

·

( cos 2 θ ____ 1 ) ______

( cos 2 θ ____ 1 )

= 2 ( sin θ _____ cos θ

) ( cos 2 θ _____ 1 )

= 2 sinθ cos θ = sin 2θ

3a. tan 75° = tan 150° ____ 2

= √

1 - cos 150° ___________ 1 + cos 150°

= √

1 +

√ 3 ___ 2 ______

1 - √ 3 ___ 2

=

2 + √ 3

_______ 2 _______

2 - √ 3

_______ 2

=

2 + √ 3

_______ 2 - √ 3

=

2 + √ 3

_______ 2 - √ 3

• 2 + √ 3

________ 2 + √ 3

=

7 + 4 √ 3

________ 1

= 7 + 4 √ 3

b. cos 5π ___ 8 = cos 5π ___

4 ( 1 __

2 )

= - √

1 + cos 5π ___

4 __________

2

= - √

1 - 1 ___

√ 2 ______

2

= - √

2 - √ 2 _______ 4 = -

√ 2 - √ 2 ________

2

4. step 1 Find cos θ and sin θ

θ is in QI, and tan θ = 4 __ 3

x 2 + y 2 = r 2

r = √ 4 2 + 3 2 = √ 16 + 9 = √ 25 = 5

Thus, sin θ = 4 __ 5 and cos θ = 3 __

5 .

step 2 Evaluate sin θ __ 2 . step 3 Evaluate cos θ __

2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = √

1 + cos θ ________

2

= √

1 - 3 _

5 _____

2 = √

1 + 3 _

5 _____

2

= √ 1 __ 5 = √ 4 __

5

= 1 ___ √ 5

· √ 5

___ √ 5

= 2 ___ √ 5

· √ 5

___ √ 5

= √ 5

___ 5 =

2 √ 5 ____

5

think anD DisCuss

1. cos 2θ = cos 2 θ - sin 2 θ, because it can be factored into (cos θ + sin θ) (cos θ - sin θ) , and the factor (cos θ + sin θ) can be divided out to eliminate the denominator.

2. Determine the quadrant that θ _ 2 lies in based on θ

and identify the signs of sine and cosine in that quadrant.

3. Double Angle Identity

for Cosine

cos 2 θ = cos 2 θ - sin 2 θ cos 2 θ = 2 2 cos 2 θ - 1 cos 2 θ = 1 - sin 2 θ



sin θ = √

1 - (- 5 ___ 13

) 2

= √ 144 ____ 169

= 12 ___ 13

step 2 Find sin 2θ. step 3 Find cos 2θ.sin 2θ = 2sin θ cos θ cos 2θ = 2 cos 2 θ - 1

= 2 ( 12 ___ 13

) (- 5 ___ 13

) = 2 (- 5 ___ 13

) 2 - 1

= - 120 ____ 169

= - 119 ____ 169


tan 2θ = 2tan θ ________ 1 - tan 2 θ

= 2 (- 12 __

5 ) _______

1 - 144 ___ 25

= (- 120 ___

25 ) ______

(- 119 ___ 25

) = 120 ____

119


CS10_A2_MESK710389_C11.indd 334 4/11/11 1:17:35 PM

2. step 1 Solve cos 2 θ = 1 - sin 2 θ. cos 2 θ = 1 - sin 2 θ

cos θ = √

1 - ( 4 __ 5 )

2 = √

9 ___

25 = 3 __

5


= 2 ( 3 __ 5 ) ( 4 __

5 ) = 2 ( 3 __

5 )

2 - 1

= 24 ___ 25

= - 7 ___ 25


tan 2θ = 2tan θ ________ 1 - tan 2 θ

= 2 ( 4 _

3 ) ______

1 - 16 __ 9 =

24 __ 9 _____

(- 7 _ 9 ) = - 24 ___

7

3. 2 cos 2θ = 4 cos 2 θ - 22(2 cos 2 θ - 1) = 4 cos 2 θ - 2 = 4 cos 2 θ - 2

4. s in 2 θ = 1 - cos 2θ + 1 _________ 2

1 - cos 2 θ =

1 - 2 cos 2 θ ______ 2 =

1 - (2 cos 2 θ - 1) + 1

_______________ 2 =

1 - cos 2θ + 1 _________ 2

= 1 - cos 2θ + 1 _________ 2

5. 1 + cos 2θ _________ sin 2θ

= cot θ

1 + (2 cos 2 θ - 1)

_______________ 2 sin θ cosθ

=

2 cos 2 θ __________ 2 sin θ cos θ

=

cos θ _____ sin θ

=

cot θ = cot θ

6. sin 2θ = 2 tan θ ________ 1 + tan 2 θ

= 2 tan θ ______ sec 2 θ

= 2 ( sin θ _____ cos θ

) cos 2 θ

= 2sin θ cos θ sin 2θ = sin 2θ

7. cos 67.5° = cos 135° ____ 2

= √

1 + cos 135° ___________ 2

= √

1 -

√ 2 ___ 2 ______

2 =

√ 2 - √ 2 ________

2

8. cos π ___ 12

= cos π __ 6 ( 1 __

2 )

= √

1 + cos π __

6 _________

2

= √

1 +

√ 3 ___ 2 ______

2 =

√ 2 + √ 3 ________

2

9. tan 3π ___ 8 = cos 3π ___

4 ( 1 __

2 )

= √

1 - cos 3π __

4 _________

1 + cos 3π __ 4

= √

1 +

√ 2 ___ 2 ______

1 - √ 2 ___ 2

= √

2 + √ 2 _____

2 _____

2 - √ 2 _____ 2

= √

2 + √ 2 _______ 2 - √ 2

10. sin 112.5° = cos 225° ____ 2

= √

1 + cos 225° ___________ 2

= √

1 +

√ 2 ___ 2 ______

2 =

√ 2 + √ 2 ________

2

11. step 1 Find cos θ.

θ is in QIII, and sin θ = - 24 ___ 25

.

x 2 + 24 2 = 25 2

x = - √ 25 2 - 24 2 = - √ 625 + 576 = - √ 49 = -7

Thus, cos θ = - 7 ___ 25

.


2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = - √

1 + cos θ ________

2

= √

1 + 7 __

25 ______

2 = - √

1 - 7 __ 25

______

2

= √

16 ___ 25

= - √

9 ___ 25

= 4 __ 5 = - 3 __

5


CS10_A2_MESK710389_C11.indd 335 4/11/11 1:17:35 PM

step 4 Evaluate tan θ __ 2 .

tan θ __ 2 =

sin θ __ 2

____

cos θ __ 2

= 4 __ 5

____

(- 3 __ 5

) = - 4 __

3

12. step 1 Given θ is in QIV, and cos θ = 1 __ 4 .


2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = - √

1 + cos θ ________

2

= √

1 - 1 _

4 _____

2 = - √

1 + 1 _ 4 _____

2

= √

3 __ 8 = - √

5 __

8

= √ 3

____ 2 √ 2

· √ 2 ___ √ 2

= - √ 5

____ 2 √ 2

· √ 2 ___ √ 2

= √ 6

___ 4 = -

√ 10 ____

4


tan θ __ 2 =

sin θ __ 2

____

cos θ __ 2

= √ 6

___ 4

______

(- √ 10

____ 4

)

= - √ 6

____ √ 10

· √ 5 · √ 2

________ √ 5 · √ 2

= - √ 15

____ 5



sin θ = √

1 - (- 7 ___ 25

) 2

= √

576 ____ 625

= 24 ___ 25


= 2 ( 24 ___ 25

) (- 7 ___ 25

) = 2 (- 7 ___ 25

) 2 - 1

= - 336 ____ 625

= - 527 ____ 625


tan 2θ = sin 2θ ______ cos 2θ

= (- 336 ____

625 )

_____ (- 527 ____

625 ) = 336 ____

527

14. step 1 Find cos θ and sin θ.

θ is in QI, and tan θ = 20 ___ 21

.

x 2 + y 2 = r 2

r = √ 20 2 + 21 2 = √ 400 + 441 = √ 841 = 29

Thus, sin θ = 20 ___ 29

and cos θ = 21 ___ 29

.


= 2 ( 20 ___ 29

) ( 21 ___ 29

) = 2 ( 21 ___ 29

) 2 - 1

= 840 ____ 841

= 41 ____ 841



= ( 840 ____ 841

)

____ ( 41 ____ 841

) = 840 ____

41

15. sin 2θ _____ sin θ

= 2 cos θ

2 sin θ cos θ __________ sin θ

=

2 cos θ = 2 cos θ

16. cos 2 θ = 1 __ 2 (1 + cos 2θ)

= 1 + cos 2θ _________ 2

= 1 + 2 cos 2 θ - 1 _____________ 2

= 2 cos 2 θ ______ 2

cos 2 θ = cos 2 θ

17. tan θ = 1 - cos 2θ _________ sin 2θ

= 1 - (1 - 2 sin 2 θ)

______________ 2 sin θ cos θ

= 2 sin 2 θ __________ 2 sin θ cos θ


tan θ = tan θ

18. tan θ = sin 2θ _________ 1 + cos 2θ

= 2 sinθ cos θ _____________ 1 + 2 cos 2 θ - 1

= 2 sinθ cos θ _________ 2 cos 2 θ


tan θ = tan θ

19. sin 7π ___ 12

= sin 7π ___ 6 ( 1 __

2 )

= √

1 - cos 7π ___

6 __________

2

= √

1 +

√ 3 ___ 2 ______

2 =

√ 2 + √ 3 ________

2


CS10_A2_MESK710389_C11.indd 336 4/11/11 1:17:36 PM

20. cos 5π ___ 12

= cos 5π ___ 6 ( 1 __

2 )

= √

1 + cos 5π ___

6 __________

2

= √

1 -

√ 3 ___ 2 ______

2 =

√ 2 - √ 3 ________

2

21. sin 22.5° = sin 45° ___ 2

= √

1 - cos 45° __________ 2

= √

1 -

√ 2 ___ 2 ______

2 =

√ 2 - √ 2 ________

2

22. t an 15° = tan 30° ___ 2

= √

1 - cos 30° __________ 1 + cos 30°

= √

1 -

√ 3 ___ 2 ______

1 + √ 3 ___ 2

= √

2 - √ 3 _____

2 _____

2 + √ 3 _____ 2

= √

2 - √ 3

_______ 2 + √ 3


θ is in QIV, and tan θ = - 12 ___ 35

.

12 2 + 35 2 = r 2

r = √ 12 2 + 35 2 = √ 144 + 1225 = √ 1369 = 37

Thus, cos θ = 35 ___ 39

.


2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = - √

1 + cos θ ________

2

= √

1 - 35 __

37 ______

2 = - √

1 + 35 __

37 ______

2

= √ 1 ___ 37

= √ 37

____ 37

= - √

36 ___ 37

= - 6 √ 37 _____

37


tan θ __ 2 =

sin θ __ 2

____

cos θ __ 2

= √ 37

____ 37

_______

(- 6 √ 37

_____ 37

)

= - 1 __ 6


θ is in QIII, and sin θ = - 3 __ 5 .

x 2 + 3 2 = 5 2

x = - √ 5 2 - 3 2 = - √ 25 - 9 = - √ 16 = -4

Thus, cos θ = - 4 __ 5 .


2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = - √

1 + cos θ ________

2

= √

1 + 4 _

5 _____

2 = - √

1 - 4 _ 5 _____

2

= √

9 ___ 10

= - √ 1 ___ 10

= 3 √ 10

_____ 10

= - √ 10

____ 10


tan θ __ 2 =

sin θ __ 2 ____

cos θ __ 2

= 3 √ 10

_____ 10

______

(- √ 10

____ 10

)

= -3

25. sin 3θ= sin (2θ + θ)= sin 2θ cos θ + cos 2θ sin θ= (2sin θ cos θ)cos θ + ( cos 2 θ - sin 2 θ)sinθ

= 2sin θ cos 2 θ + sin θ cos 2 θ - sin 3 θ= 3sin θ cos 2 θ - sin 3 θ

26. sin 4θ= sin 2(2θ)= 2(sin 2θ cos 2θ)= 2(2sin θ cos θ)( cos 2 θ - sin 2 θ)= 4sin θ cos 3 θ - 4 sin 3 θ cosθ

27. cos 3θ= cos (2θ + θ)= cos 2θ cos θ - sin 2θ sin θ= (1 - 2 sin 2 θ)cos θ - (2sin θ cos θ)sin θ= cos θ(1 - 4 sin 2 θ)

28. cos 4θ= cos (3θ + θ)= cos 3θ cos θ - sin 3θ sin θ= ( cos 3 θ - 3 sin 2 θ cos θ)cos θ - (3sin θ cos 2 θ - sin 3 θ)sinθ

= cos 4 θ - 3 sin 2 θ cos 2 θ - 3 sin 2 θ cos 2 θ - sin 4 θ= sin 4 θ + cos 4 θ - 6 sin 2 θ cos 2 θ

29. cos 2θ + 2 sin 2 = (1 - 2 sin 2 θ) + 2 sin 2 θ= 1

30. cos 2θ + 1= (2 cos 2 θ - 1) + 1= 2 cos 2 θ


CS10_A2_MESK710389_C11.indd 337 4/11/11 1:17:36 PM

31. tan 2θ(2 - sec 2 θ)

= 2 tanθ ________ 1 - tan 2 θ

(2 - sec 2 θ)

= 2 tanθ ________ ( 2 cos 2 - 1 _______

cos 2 θ )

(2 - sec 2 θ)

= 2 tanθ _________ 2 - sec 2 θ

(2 - sec 2 θ)

= 2 tanθ

32. cos 2θ ___________ cos θ + sin θ

= ( cos 2 θ - sin 2 θ)

_____________ cos θ + sin θ

= (cos θ + sin θ)(cos θ - sin θ)

_______________________ cos θ + sin θ

= cos θ - sin θ

33. cos θ sin 2θ _________ 1 + cos 2θ

= 2sin θ cos 2 θ ____________________________ ( sin 2 θ + cos 2 θ) + ( cos 2 θ - sin 2 θ)

= 2sin θ cos 2 θ __________ 2 cos 2 θ

= sin θ

34. cos 2θ - 1 _________ sin 2 θ

= 1 - 2 sin 2 θ - 1 _____________ sin 2 θ

= -2 sin 2 θ _______ sin 2 θ

= -2

35a. y(t) = 3.1sin2t = 3.1(2sin t cos t) = 6.2sin t cos t

35b. 3.8cos t = 6.2sin t cos t sin t ≈ 0.6129 sin -1 t ≈ 0.66

35c. y(0.66) = 3.1sin2(0.66) ≈ 3.1 · 0.969 ≈ 3.00 m.


sin θ = √ 1 - (- 3 __

8 )

2

= √ 55 ___

64 =

√ 55 ____

8


= 2 ( √ 55

____ 8 ) (- 3 __

8 ) = 2 (- 3 __

8 )

2 - 1

= - 3 √ 55

_____ 32

= - 23 ___ 32



= (-

3 √ 55 _____

32 )

_______ (- 23 ___

32 ) =

3 √ 55 _____

23


2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = √

1 + cos θ ________

2

= √

1 + 3 _

8 _____

2 = √

1 - 3 _

8 _____

2

= √ 11 ___ 16

= √

5 ___ 16

= √ 11

____ 4 =

√ 5 ___

4


tan θ __ 2 =

sin θ __ 2 ____

cos θ __ 2

= √ 11 ____

4 ___

√ 5

___ 4

= √ 11 ___ 5 =

√ 55 ____

5


sin θ = - √ 1 - (-

√ 5 ___

3 )

2

= - √ 4 __ 9 = - 2 __

3


= 2 (- 2 __ 3 ) (

5 ____

3 ) = 2 (-

√ 5 ___

3 )

2

- 1

= 4 √ 5

____ 9 = 1 __

9



= 4 √ 5

____ 9 ____

1 __ 9 = 4 √ 5


2 .

sin θ __ 2 = - √

1 - cos θ ________ 2 cos θ __

2 = - √

1 + cos θ ________ 2

= - √

1 +

√ 5 ___

3 ______

2 = - √

1 -

√ 5 ___

3 ______

2

= - √

3 + √ 5 _______

6 = - √

3 - √ 5

_______ 6

= √ 18 + 6 √ 5

__________ 6 = -

√ 18 - 6 √ 5 __________

6


CS10_A2_MESK710389_C11.indd 338 4/11/11 1:17:37 PM


tan θ __ 2 =

sin θ __ 2

____

cos θ __ 2

= ( √ 18 + 6 √ 5

__________ 6

)

__________

(- √ 18 - 6 √ 5

__________ 6

)

= - √ 3 + √ 5

________ √ 3 - √ 5

38. step 1 Solve cos 2 θ = 1 - sin 2 θ. cos 2 θ = 1 - sin 2 θ

cos θ = √

1 - ( 2 __ 5 )

2

= √ 21 ___ 25

= √ 21

____ 5

step 2 Find sin 2θ. step 3 Find cos 2θ.sin 2θ = 2sin θ cos θ cos 2θ = 1 - 2 sin 2 θ

= 2 ( 2 __ 5 ) (

√ 21 ____ 5 ) = 1 - 2 ( 2 __

5 )

2

= 4 √ 21 _____ 25

= 17 ___ 25



= 4 √ 21

_____ 25

____

17 ___ 25

= 4 √ 21 _____

17


2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = - √

1 + cos θ ________

2

= √

1 -

√ 21 ___ 5 _______

2 = - √

1 +

√ 21 ___ 5 _______

2

= √

5 - √ 21 ________ 10

= - √

5 + √ 21 ________ 10

= √ 50 - 10 √ 21

____________ 10

= - √ 50 + 10 √ 21

____________ 10


tan θ __ 2 =

sin θ __ 2

____

cos θ __ 2

= √ 50 - 10 √ 21

____________ 10

__________

√ 50 + 10 √ 21

____________ 10

= - √ 5 - √ 21

_________ √ 5 + √ 21

39. step 1 Find cos θ and sin θ.

θ is in QIV, and tan θ = - 1 __ 2 .

x 2 + y 2 = r 2

r = √ (-1) 2 + 2 2 = √ 1 + 4 = √ 5

Thus, sin θ = - √ 5

___ 5 and cos θ =

2 √ 5 ____

5 .


= 2 (- √ 5

___ 5 ) (

2 √ 5 ____

5 ) = 2 (-

2 √ 5 ____

5 )

2

- 1

= - 20 ___ 25

= - 4 __ 5 = 15 ___

25 = 3 __

5



= - 4 __

5 ___

3 __ 5 = - 4 __

3


2 .

sin θ __ 2 = √

1 - cos θ ________

2 cos θ __

2 = - √

1 + cos θ ________

2

= √

1 - 2 √ 5 ___

5 _______

2 = - √

1 + 2 √ 5 ___

5 _______

2

= √

5 - 2 √ 5

________ 10

= - √

5 + 2 √ 5

________ 10

= √ 50 - 20 √ 5

___________ 10

= - √ 50 + 20 √ 5

___________ 10


tan θ __ 2 =

sin θ __ 2 ____

cos θ __ 2

= √ 50 - 20 √ 5

___________ 10

________ -

√ 50 + 20 √ 5 ___________

10

= - √ 5 - 2 √ 5

_________ √ 5 + 2 √ 5

40. cos 7π ___ 8 = cos 7π ___

4 ( 1 __

2 )

= - √

1 + cos 7π ___

4 __________

2

= - √

1 +

√ 2 ___ 2 ______

2 = -

√ 2 + √ 2 ________

2

41. sin 11π ____ 12

= sin 11π ____ 6 ( 1 __

2 )

= √

1 - cos 11π ____

6 __________

2

= √

1 -

√ 3 ___ 2 ______

2 =

√ 2 - √ 3 ________

2


CS10_A2_MESK710389_C11.indd 339 4/11/11 1:17:37 PM

42. cos 105° = cos 210° ____ 2

= - √

1 + cos 210° ___________ 2

= - √

1 -

√ 3 ___ 2 ______

2 = -

√ 2 - √ 3 ________

2

43. sin (-15°) = sin -30° _____ 2

= - √

1 - cos (-30°)

_____________ 2

= - √

1 -

√ 3 ___ 2 ______

2 = -

√ 2 - √ 3 ________

2

44a. △P = 2 P i sin θ __ 2 ; △P = 2 P f sin θ __

2

b. It does not change. c. △P = 2 P f √

1 - cos θ ________ 2

45. cos 2 θ __ 2 = (cos θ __

2 )

2

= ( √

1 + cos θ ________ 2 )

2

= ( 1 + cos θ ________ 2 ) ( 1 - cos θ ________

1 - cos θ )

= 1 - cos 2 θ __________ 2(1 - cos θ)

= sin 2 θ __________ 2(1 - cos θ)

46. 1 - tan 2 θ ________ 1 + tan 2 θ

= 1 - sin 2 θ ____

cos 2 θ ________

sec 2 θ

= 1 - ( sin 2 θ ____

cos 2 θ )

_______ ( 1 ____

cos 2 θ ) · cos 2 θ _____

cos 2 θ

= cos 2 θ - sin 2 θ = cos 2θ

47. tan θ + sin θ __________ 2 tan θ

= ( sin θ ____

cos θ ) + sin θ ( cos θ ____

cos θ ) ________________

2 ( sin θ ____ cos θ

)

= ( 1 ___

cosθ ) (sin θ + sin θ cos θ)

____________________ ( 1 ___

cosθ ) (2 sin θ)

= sin θ + sin θ cos θ ______________ 2 sin θ

= sin θ(1 + cos θ)

_____________ 2 sin θ

= 1 + cosθ ________ 2

= ( √

1 + cos θ ________ 2 )

2

= (cos θ __ 2 )

2 = cos 2 θ __

2

48.

-4

4

2π -2π

Graph appears to be equal to sin x.

(cos x)(1 - cos 2x)

_______________ sin 2x

= (cos x)[1 - (1 - 2 sin 2 x)]

____________________ 2 sin x cos x

= 2 sin 2 x ______ 2 sin x

= sin x

49a. d(θ) = v o 2 sin θ cos θ

___________ 16

= v o 2 2 sin θ cos θ

____________ 16 · 2

= v o 2 sin 2θ

________ 32

b. d(θ) = 80 2 sin θ cos θ ___________ 16

Using the calculator: d(15°) = 100 ft d(30°) ≈ 173 ft d(45°) = 200 ft d(60°) ≈ 173 ft d(75°) = 100 ft

c. 45° d. 31° < θ < 59°

50. First find sin 15° by using sin 30° __ 2 , and then find

sin 7.5° by using sin 15° __ 2 .

51. Possible answer: If θ is multiplied by 2, use a double-angle identity. If θ is divided by 2, use a half-angle identity.

52. D; 2 ( √ 2 ___ 2 ) (-

√ 2 ___

2 ) = -1 53. F; sin 2 θ - sin 2 θ = 0

54. C; √

1 - (- 12 ___

13 ) _________

2 = 5 ____

√ 26

= 5 √ 26

_____ 26

55. G; sin 157.5° = sin 315° ____ 2

= √ 2 - √ 2

________ 2

56. Possible answer:

cos 2θ ___________ sin θ + cos θ

= cos 2 θ - sin 2 θ ____________ sin θ + cos θ

= (cos θ + sin θ)(cos θ - sin θ)

_______________________ sin θ + cos θ

= cos θ - sin θ


CS10_A2_MESK710389_C11.indd 340 4/11/11 1:17:37 PM


57. Possible answer:


= 2 sin θ cos θ __________ 1 - 2 sin 2 θ

= ( 2 sin θ cos θ __________ 1 - 2 sin 2 θ

) ( 1 ___

cos 2 θ ) _____

( 1 ___ cos 2 θ

)

= 2 ( sin θ ____

cos θ ) _____________

1 ____ cos 2 θ

- 2 ( sin 2 θ ____ cos 2 θ

)

= 2 tan θ _____________ sec 2 θ - 2 tan 2 θ

= 2 tan θ ________ 1 - tan 2 θ

58. tan θ __ 2 =

sin θ __ 2

____

cos θ __ 2

= ± √

1 - cos θ ________

2 _________

± √

1 + cos θ ________ 2

= ± √

1 - cos θ ________ 2 ( √ 2 ________

1 + cos θ )

= ± √

1 - cos θ ________ 1 + cos θ

59. tan ⎡

⎢ ⎣ 1 __ 2 ( 30° ___

2 )

⎤

⎥ ⎦ = √

1 - cos 30° ___ 2

________

1 + cos 30° ___ 2

=

1 -

√ 2 + √ 3 ________

2 _________

1 + √ 2 + √ 3

________ 2

= √

2 - √ 2 + √ 3

____________ 2 + √ 2 + √ 3

60. tan ⎡

⎢ ⎣ 1 __ 2 ( 1 __

2 · π __

4 )

⎤

⎥ ⎦ = √

1 - cos ( 1 __

2 · π __

4 )

__________ 1 + cos ( 1 __

2 · π __

4 )

=

1 -

√ 2 + √ 2 ________

2 _________

1 + √ 2 + √ 2

________ 2

= √

2 - √ 2 + √ 2

____________ 2 + √ 2 + √ 2

61. sin ⎡

⎢ ⎣ 1 __ 2 ( 1 __

2 · π __

6 )

⎤

⎥ ⎦ = √

1 - cos ( 1 __

2 · π __

6 )

__________ 2

= √

1 -

√ 2 + √ 3 ________

2 _________

2 =

√

2 - √ 2 + √ 3 _____________

2

62. cos ⎡

⎢ ⎣ 1 __ 2 ( 45° ___

2 )

⎤

⎥ ⎦ = √

1 + cos 45° ___ 2

________

2

= √

1 +

√ 2 + √ 2 ________

2 _________

2 =

√

2 + √ 2 + √ 2 _____________

2

63. πn, where n is an integer; possible answer:

1

x

-ππ

0

y

Solve the equation: sin 2θ = 2 sin θ2 sin θ cosθ - 2 sin θ = 0 2 sin θ(cos θ - 1) = 0so cos θ = 1 or sin θ = 0, which are both true when θ = πn, where n is an integer.

64. cos(A - B) - cos(A + B) = (cosA cosB + sinA sinB) - (cosA cosB - sinA sinB) = 2 sinA sinB

sinA sinB = 1 __ 2 ⎡ ⎣ cos(A - B) - cos(A + B) ⎤ ⎦ ;

cos(A + B) + cos(A - B) = (cosA cosB - sinA sinB) + (cosA cosB + sinA sinB) = 2 cosA cosB

cosA cosB = 1 __ 2 ⎡ ⎣ cos(A + B) + cos(A - B) ⎤ ⎦

solvInG TrIGonomeTrIc equATIons

CheCk it out!

1. 2 cos θ + √ 3 = 0 2 cos θ = - √ 3

cos θ = - √ 3

___ 2

θ = cos -1 - √ 3

___ 2

θ = 150° + 360°n, 210 + 360°n

2a. cos 2 θ + 2 cos θ = 3 cos 2 θ + 2 cos θ - 3 = 0Use the Quadratic Formula.

cos θ = -2 ± √ 2 2 - 4(1)(-3)

__________________ 2(1)

= -2 ± √ 16

_________ 2

= -1 ± 2 = 1 or -3Only 1 works, so cos θ = 1 θ = 0

b. Use Quadratic Formula.

sin θ = 5 ± √ 5 2 - 4(1)(-2)

_________________ 2(1)

= 5 ± √ 32

________ 2

= 5 __ 2 + 2 √ 2 or 5 __

2 - 2 √ 2

Only 5 __ 2 - 2 √ 2 is between -1 and 1, so

θ = sin -1 ( 5 __ 2 - 2 √ 2 ) ≈ 21.9° or 158.1°


x-6


11-6

CS10_A2_MESK710389_C11.indd 341 4/11/11 1:17:38 PM

3a. 4 si n 2 θ + 4 cos θ = 54(1 - co s 2 θ ) + 4 cos θ - 5 = 0 4 - 4co s 2 θ + 4cos θ - 5 = 0 -4 cos 2 θ + 4cos θ - 1 = 0Use the Quadratic Formula.

cos θ = 4 ± √ 4 2 - 4(-4)(-1)

__________________ 2(4)

= 1 __ 2

θ = cos -1 ( 1 __ 2 ) = 60° or 300°

b. sin 2θ = - cos θ 2 sin θ cos θ = -cos θ Divide both sides by cos θ . However, note that this is not valid if cos θ = 0, i.e., θ = 90° or 270°. So, these two are solutions, since they make the equation 0 = 0. 2 sin θ = -1

sin θ = - 1 __ 2

θ = 210° or 330°So, the solutions are: 210°, 330°, 90° or 270°.

4. We want to let h(d) = 12 and solve for d. 12 = 3.31 sin π _____

182.5 (d - 85.25) + 12.22

-0.22 = 3.31 sin π _____ 182.5

(d - 85.25)

0.06646 = sin π _____ 182.5

(d - 85.25)

si n -1 0.06646 = π _____ 182.5

(d - 85.25)

182.5 _____ π ( sin -1 0.06646 ) + 85.25 = dd = 89 or 264These days correspond to late March and late September.

think anD DisCuss

1. Possible answer: First solve the equation for a restricted domain equal to the period of the given function. Then use an understanding of periodicity (i.e., cos θ = cos θ + 2π ) of the given function to find all solutions.

2. Method Most useful when... Example

Graphing approximate solutions are okay or only practical solution method

equation has one function in terms of θ

equation has degree of 2

equation has degree of 2

equation contains two different trigonometric functions

Solving linear equations

Factoring

Quadratic Formula

Identity substitution

2 sinθ = 1

sin2θ + sinθ - 2 = 0

sin2θ + sinθ - 5 = 0

cos (2.1θ) = 0.25

cos 2θ + 3 sinθ + 2 = 0


1. 6 cos θ -1 = 2 6 cos θ = 3

cos θ = 3 __ 6 = 1 __

2

θ = co s -1 ( 1 __ 2 )

θ = 60° + 360°n or 300° + 360°n

2. 2 sin θ - √ 3 = 0

sin θ = √ 3

___ 2

θ = si n -1 ( √ 3

___ 2 )

θ = 60° + 360°n or 120° + 360°n

3. cos θ = √ 3 - cos θcos θ + cos θ = √ 3 2 cos θ = √ 3

cos θ = √ 3

___ 2

θ = co s -1 ( √ 3

___ 2 )

θ = 30° + 360°n or 330° + 360°n

4. 2 si n 2 θ + 3 sin θ + 1 = 0(2 sin θ + 1)( sin θ + 1) = 0

sin θ = - 1 __ 2 or sin θ = -1.

θ = si n -1 (- 1 __ 2 ) or s in -1 (-1)

θ = 3π ___ 2 , 7π ___

6 , 11π ____

6

5. cos 2 θ - 4 cos θ + 1 = 0Use the Quadratic Formula.

cos θ = 4 ± √ (-4) 2 - 4(1)(1)

__________________ 2(1)

= 2 ± √ 12 ____

2

= 2 ± √ 3 Only 2 + √ 3 works, so cos θ = 2 - √ 3 θ = 74.5° or 285.5°

6. 2 si n 2 θ - cos 2θ = 02(1 - c os 2 θ) - (2 c os 2 θ - 1) = 0 2 - 2 co s 2 θ - 2 co s 2 θ + 1 = 0 -4 co s 2 θ + 3 = 0

co s 2 θ = 3 __ 4

cos θ = ± √ 3

___ 2

θ = 30°, 150°, 210°, 330°

7. sin 2 θ + cos θ = -1

1 - c os 2 θ + cos θ = -1

co s 2 θ - cos θ - 2 = 0( cos θ - 2)( cos θ + 1) = 0 cos θ cannot equal 2, so cos θ = -1 θ = π


CS10_A2_MESK710389_C11.indd 342 4/11/11 1:17:38 PM

8. We want to solve for m, when E(m) = 825. 825 = 350 sin π __

6 (m + 1.5) + 650

175 = 350 sin π __ 6 (m + 1.5)

1 __ 2 = sin π __

6 (m + 1.5)

s in -1 1 __ 2 = π __

6 (m + 1.5)

m = 6 __ π ( s in -1 ( 1 __ 2 ) ) - 1.5

In order to complete this, we need to know what

range of values for s in -1 1 __ 2 to use. π __

6 would result in

m being negative. So, we use 5π ___ 6 and 13π ____

6 .

Thus, m = 3.5 (mid April) or 11.5 (mid December).


9. 1 - 2 cos θ = 0 2 cos θ = 1

cos θ = 1 __ 2

θ = 60° + 360°n, 300° + 360°n

10. √ 3 tan θ - 3 = 0 √ 3 tan θ = 3

tan θ = 3 ___ √ 3

= √ 3 θ = 60° + 180°n

11. 2 cos θ + √ 3 = 0 2 cos θ = - √ 3

cos θ = - √ 3

___ 2

θ = 150° + 360°n, 210° + 360°n

12. 2 sin θ + 1 = 2 + sin θ2 sin θ - sin θ = 2 - 1 sin θ = 1θ = 90° + 360°n

13. 2 co s 2 θ + cos θ - 1 = 0(2 cos θ - 1)( cos θ + 1) = 0

cos θ = - 1 __ 2 or -1.

θ = π __ 3 , π, or 5π ___

3 .

14. s in 2 θ + 2 sin θ - 2 = 0Use the Quadratic Formula

sin θ = -2 ± √ (2) 2 - 4(1)(-2)

___________________ 2(1)

= -1 ± √ 12 ____

2

= -1 ± √ 3 Only -1 + √ 3 works, so sin θ = -1 + √ 3 θ ≈ 47.1° or 132.9°

15. cos 2θ + cos θ + 1 = 02 cos 2 θ - 1 + cos θ + 1 = 0 2 c os 2 θ + cos θ = 0 cos θ (2 cos θ + 1) = 0

cos θ = 0 or - 1 __ 2

θ = 90°, 270°, 120°, 240°

16. cos 2θ = sin θ 1 - 2 si n 2 θ - sin θ = 0(1 - 2 sin θ)(1 + sin θ) = 0

sin θ = 1 __ 2 or -1

θ = π __ 6 , 5π ___

6 , 3π ___

2

17a. Let E(t) = 850, and solve for t 850 = 100 sin π ___

12 (t - 8) + 800

50 = 100 sin π ___ 12

(t - 8)

1 __ 2 = sin π ___

12 (t - 8)

s in -1 ( 1 __ 2 ) = π ___

12 (t - 8)

t = 12 ___ π si n -1 ( 1 __ 2 ) + 8

t = 10 or 18Thus, it occurs at 10:00 A.M. or 6:00 P.M.

b. E(t) will be maximized when sin π ___ 12

(t - 8) = 1. π ___ 12

(t - 8) = sin -1 1

t = 12 ___ π si n -1 (1) + 8 t = 14 = 2:00 P.M. This occurs when the building is full, and early afternoon is often the hottest time of the day, so electricity usage would be high due to air conditioning.E(t) will be minimized when sin π ___

12 (t - 8) = -1.

π ___ 12

(t - 8) = sin -1 -1

t = 12 ___ π si n -1 (-1) + 8 t = 26 = 2:00 A.M. This makes sense since nobody is in the building.

18. 2 si n 2 θ = sin θ 2 s in 2 θ - sin θ = 0 sin θ(2 sin θ - 1) = 0

sin θ = 0 or 1 __ 2

θ = 0°, 30°, 150°, 180°

19. 2 c os 2 θ - sin θ = 12(1 - s in 2 θ) - sin θ - 1 = 0 2 - 2 si n 2 θ - sin θ - 1 = 0 2 si n 2 θ + sin θ - 1 = 0 (2 sin θ - 1)( sin θ + 1) = 0

sin θ = 1 __ 2 or -1

θ = 30°, 150°, 270°


CS10_A2_MESK710389_C11.indd 343 4/11/11 1:17:39 PM

20. cos 2θ - 2 sin θ + 2 = 01 - 2 s in 2 θ - 2 sin θ + 2 = 0. 2 si n 2 θ + 2 sin θ - 3 = 0.Use the Quadratic Formula.

sin θ = -2 ± √ (2) 2 - 4(2)(-3)

___________________ 2(2)

= - 1 __ 2 ±

√ 28 ____

4

= -1 ± √ 7 ________ 2

Only -1 + √ 7 ________ 2 works, so

sin θ = -1 + √ 7 ________ 2

θ ≈ 55.4° or 124.6°

21. 2 c os 2 θ + 3 sin θ = 32(1 - si n 2 θ) + 3 sin θ = 32 - 2 si n 2 θ + 3 sin θ - 3 = 0.2 si n 2 θ - 3 sin θ + 1 = 0 (2 sin θ - 1)( sin θ - 1) = 0

sin θ = 1 __ 2 or 1

θ = 30°, 90°, 150°

22. co s 2 θ + sin θ - 1 = 0. 1 - s in 2 θ + sin θ - 1 = 0 sin θ - sin 2 θ = 0 sin θ(1 - sin θ) = 0 sin θ = 0 or 1 θ = 0°, 90°, 180°

23. 2 si n 2 θ + sin θ = 0 sin θ(2 sin θ + 1) = 0

sin θ = 0 or - 1 __ 2

θ = 0°, 180°, 210°, 330°

24. si n 2 θ - sin θ = 0 sin θ( sin θ - 1) = 0 sin θ = 0 or 1 θ = 0, π __

2 , π

25. co s 2 θ - 3 cos θ - 4 = 0( cos θ - 4)( cos θ + 1) = 0cos θ = -1 or 4 θ = π

26. cos θ(0.5 + cos θ) = 0 cos θ = 0 or -0.5

θ = π __ 2 , 3π ___

2 , 2π ___

3 , 4π ___

3

27. 2 s in 2 θ - 3 sin θ - 2 = 0(2 sin θ + 1)( sin θ - 2) = 0

sin θ = - 1 __ 2 or 2

θ = 7π ___ 6 , 11π ____

6

28. co s 2 θ + 0.5 cos θ = 5 2 co s 2 θ + cos θ - 10 = 0(2 cos θ + 5)( cos θ - 2) = 0 cos θ = 2 or -5.So, no solutions exist.

29. si n 2 θ + 3sin θ + 3 = 0Use the Quadratic Formula.

sin θ = -3 ± √ (3) 2 - 4(1)(3)

__________________ 2(1)

= -3 ± -3 _____

2

Since this is not a real number, no solutions exist.

30. c os 2 θ + 4cos θ - 3 = 0Use the Quadratic Formula.

cos θ = -4 ± √ (4) 2 - 4(1)(-3)

___________________ 2(1)

= -4 ± √ 28

____ 2 1

=-2 ± √ 7

Only -2 + √ 7 works, so cos θ = -2 + √ 7

θ ≈ 0.869, 5.414

31. ta n 2 θ = √ 3 tan θ tan θ (tan θ - √ 3 ) = 0tan θ = 0 or √ 3

θ = 0, π, π __ 3 , 4π ___

3 .

32a. Let R(θ) = 250 and v = 96 then solve for θ.

250 = (96 ) 2 sin 2θ _____ 32

32(250)

_______ 9 6 2

= sin 2θ

2000 _____ 2303

= sin 2θ

2θ ≈ 60° or 120° θ ≈ 30° or 60°

b. Let H max (θ) = 50 and v = 96 then solve for θ.

50 = (9 6) 2 si n 2 θ

_________ 64

(50)(64)

_______ 9 6 2

= si n 2 θ

25 ___ 72

= s in 2 θ

5 _____ 6 √ 2

= 5 ____ 6 √ 2

= sin θ

θ ≈ 36°. θ could also be 144°, but this implies that the the ball was thrown backwards.

33a. Let A = 92, r = 18 and solve for θ.

92 = (18 ) 2

_____ 2 (θ - sin θ )

2(92)

_____ 18 2

= θ - sin θ

46 ___ 81

= θ - sin θ

To solve this, graph the functions sin θ and θ - 46 ___ 81

and note that they intersect at θ = π __ 2 . Thus this is

the solution. This would allow 4 sets since 2π ___ π __ 2 = 4.


CS10_A2_MESK710389_C11.indd 344 4/11/11 1:17:39 PM

b. Let A = 50, r = 18 and solve for θ.

50 = (18 ) 2

_____ 2 (θ - sin θ )

2(50)

_____ 18 2

= θ - sin θ

25 ___ 81

= θ - sin θ

To solve this, graph the functions sin θ and θ - 25 ___ 81

and note that they intersect at θ = 2π ___ 5 . Thus this is

the solution. This would allow 5 sets as 2π ___ 2π __ 5 = 5.

34a. Let h = 5 and solve for t. 5 = 4.5sin π ____

6.25 (t + 4) + 7.5

5 - 7.5 _______ 4.5

= sin π ____ 6.25

(t + 4)

π ____ 6.25

(t + 4) = si n -1 (- 2.5 ___ 4.5

)

t ≈ 3.717, 5.6968, 10.011, 12 t ≈ 3:25 A.M., 7:20 A.M., 3:55 P.M., 7:50 P.M.

b. Max occurs when sin π ____ 6.25

(t + 4) = 1.

sin π ____ 6.25

(t + 4) = 1

π ____ 6.25

(t + 4) = 5π ___ 2

t = 11.625 Max occurs when sin π ____

6.25 (t + 4) = -1.

sin π ____ 6.25

(t + 4) = -1

π ____ 6.25

(t + 4) = 3π ___ 2

t = 5.375 Difference = 11.625 - 5.375 = 6.25 h c. The period is twice the difference between max

and min, so it 2 × 6.25 = 12.5 h.

d. No; 12.5 does not divide evenly into 24.

35. B is incorrect. Dividing by sin θ eliminates solutions when sin θ = 0.

36. Possible answer: An equation that includes a trigonometric function is a trigonometric equation, such as sin 2 θ - sin θ = 0, and may only be true for some values. A trigonometric identity is true for all values, such as tan θ = sin θ ___

cos θ .

37. This can be simplified to 8 cosx = x. Graph both of these functions, and note that they intersect at x ≈ -4.165, -1.797, 1.395, 5.464, 6.831.

38a. This will reach maximum when the cos term is 1. So, it would be 2.9(1) + 3 = 5.9 cm. The minimum will occur when the cos term is -1. 2.9(-1) + 3 = 0.1 cm.

b. Let y = 1 and solve for t.

1 = 2.9cos ( 2π ___ 3 t + π __

4 ) + 3

- 2 ___ 2.9

= cos ( 2π ___ 3 t + π __

4 )

2.332 = 2π ___ 3 t + π __

4

t = 0.74 s c. 0.74 + 3n and 1.51 + 3n

This is because the period is 2π __ 3 ___

2π = 3.

39. ≈ 84.8°, ≈ 264.8°

40. ≈ 197.1°, ≈ 252.9°

41. 60°, 150°, 240°, 330°

42. ≈ 52.5°, ≈ 113.5°, ≈ 232.5°, ≈ 293.5

43. ≈ 38.5°, ≈ 141.5°

44. no solution

45. A trigonometric equation may have no solution (sin x = 2), an infinite number of solutions (sin x = 1), or a finite number of solutions if a restricted domain is given (sin x = 1, 0° ≤ x ≤ 180°).

teSt prep

46. B2cos θ + √ 3 = 2 √ 3 2cos θ = 2 √ 3 - √ 3

cos θ = √ 3

___ 2

θ = 30°, 330°

47. J 5tan θ - √ 3 - tan θ = 0 4tan θ = √ 3

tan θ = ⎜ √ 3 ___

4 ⎟

θ ≈ 23.4°

48. Dsi n 2 (270°) = sin (270°) (- 1) 2 = (1) This is false.

49. J

cos θ - 1 = - 1 __ 2

cos θ = 1 __ 2

θ = π __ 3 , 5π ___

3

50. B si n 2 θ - sin θ - 2 = 0(sin θ - 2)(sin θ + 1) = 0 sin θ = 2 or -1 θ = 270°

51. cos θ = -1 ± √ (1 ) 2 - 4(2)(-2)

___________________ 2(2)

= -1 ± √ 17

_________ 4

θ = co s -1 ( -1 + 17 _________ 4 ) or c os -1 ( -1 - √ 17 _________

4 )

θ = 38.7° or 321.3°


CS10_A2_MESK710389_C11.indd 345 4/11/11 1:17:40 PM


52. cos θ (9co s 2 θ - 1) = 0 cos θ (3cos θ + 1)(3cos θ - 1) = 0

cos θ = 0, ± 1 __ 3

θ = 90°, 270°, 109.5°, 250.5°, 70.5°, 289.5°

53. cos θ (4c os 2 θ - 1) = 0 cos θ (2cos θ + 1)(2cos θ - 1) = 0

cos θ = 0, ± 1 __ 2

θ = 90°, 270°, 120°, 240°, 60°, 300°

54. 16si n 4 θ - 16si n 2 θ + 3 = 0(4si n 2 θ - 3)(4si n 2 θ - 1) = 0.

s in 2 θ = 3 __ 4 or 1 __

4

sin θ = ± √ 3

___ 2 or ± 1 __

2

θ = 60°, 150°, 240°, 330°, 30°, 120°, 210°, 300°

55. si n 2 θ - 4.5sin θ - 2.5 = 0 2si n 2 θ - 9sin θ - 5 = 0 (sin θ - 5)(2sin θ + 1) = 0

sin θ = 5 or - 1 __ 2

θ = 210°, 330°

56. ⎜sin θ ⎟ = 1 __ 2 means sin θ = ± 1 __

2 .

Thus, θ = 30°, 150°, 210°, 330°.

57. ⎜cos θ ⎟ = √ 3

___ 2 means cos θ = ±

√ 3 ___

2 .

Thus, θ = 30°, 150°, 210°, 330°.

reAdy To Go on?

1. si n 2 θ sec θ csc θ

= si n 2 θ ( 1 _____ cos θ

) ( 1 ____ sin θ

)

= sin θ ( 1 _____ cos θ

) = tan θ

2. sin(-θ) sec θ cot θ

= -sin θ ( 1 _____ cos θ

) ( 1 _____ tan θ

)

= (- sin θ _____ cos θ

) ( 1 _____ tan θ

)

= -tan θ ( 1 _____ tan θ

) = -1

3. c ot 2 θ - 1 _________ cot θ + 1

= c ot 2 θ - 1 _________ cs c 2 θ

= si n 2 θ (c ot 2 θ - 1) = si n 2 θ c ot 2 θ - sin 2 θ

= si n 2 θ ( c os 2 θ ______ s in 2 θ

) - s in 2 θ

= co s 2 θ - si n 2 θ = cos 2θ = 1 - 2 si n 2 θ

4. cot θ sec θ

= ( cos θ _____ sin θ

) ( 1 _____ cos θ

)

= 1 _____ sin θ

= csc θ

5. 1 _______ cos (-θ)

= 1 _____ cos θ

= sec θ

6. c sc 2 θ 1 ___________ tan θ + cot θ

= ( 1 _____ si n 2 θ

) ( 1 ___________ tan θ + cot θ

)

= ( 1 _____ si n 2 θ

) ( 1 _______________ ( sin θ _____ cos θ

) + ( cos θ _____ sin θ

) )

= 1 ________________________ si n 2 θ ( sin θ _____

cos θ ) + s in 2 θ ( cos θ _____

sin θ )

= 1 _________________

( si n 3 θ _____ cos θ

) + sin θ cos θ

= 1 _________________

s in 3 θ + sin θ c os 2 θ ________________ cos θ

= 1 __________________

sin θ ( si n 2 θ + co s 2 θ ____________ cos θ

)

= 1 ____ sin θ

( 1 _____ cos θ

)

= 1 _____ tan θ

= cot θ

7. cos 5π ___ 12

= cos ( 2π ___ 12

+ 3π ___ 12

) = cos ( π __ 6 + π __

4 )

= cos π __ 6 cos π __

4 - sin π __

6 sin π __

4

= √ 3

___ 2 · 1 ___

√ 2 - 1 __

2 · 1 ___

√ 2

= 1 ___ √ 2

( √ 3

___ 2 - 1 __

2 )

= √ 3 - 1

_______ 2 √ 2

= √ 2 √ 3 - √ 2

___________ 4 =

√ 6 - √ 2 ________

4

8. sin(-75°) = sin(60° - 135°) = sin 60° cos 135° - cos 60° sin 135°

= √ 3

___ 2 · (- 1 ___

√ 2 ) - 1 __

2 · 1 ___

√ 2

= - 1 ___ √ 2

( √ 3

___ 2 + 1 __

2 )

= - √ 3 - 1

________ 2 √ 2

= - √ 2 √ 3 - √ 3

____________ 4 =

- √ 6 - √ 2 __________

4


CS10_A2_MESK710389_C11.indd 346 4/11/11 1:17:40 PM

9. tan 75° = tan (45° + 30°)

= tan 45° + tan 30° ________________ 1 - tan 45° tan 30°

=

1 + 1 ___ √ 3

___________ 1 - (1) ( 1 ___

√ 3 )

(

1 + 1 ___ √ 3

_______ 1 + 1 ___

√ 3 )

=

1 + 2 ___ √ 3

+ 1 __ 3

___________ 2 __ 3

= 3 __ 2 ( 4 __

3 + 2 ___

√ 3 ) = 2 + √ 3

10. sin (A + B) = sin A cos B + cos A sin B . We are given sin A and cos B and need to find cos A and sin B .

Recall, sin A = a __ c = 1 __ 4 .

Also, a 2 + b 2 = c 2 , so 1 2 + b 2 = 4 2 . Thus, b = √ 15 .

So, cos A = b __ c = √ 15

____ 4 . However, since A is in QII,

the cos must be negative, so cos A = - √ 15

______ 4 .

Likewise, cos B = b __ c = 12 ___ 13

, and a 2 + 12 2 = 13 2 , so a = 5.

Thus, sin B = a __ c = 5 ___ 13

. However, since B is in QIV,

the sin must be negative, so sin B = -5 ___ 13

.

So, sin (A + B) = sin A cos B + cos A sin B

= 1 __ 4 · 12 ___

13 +

- √ 15 ______

4 - 5 ___

13

= 12 + 5 √ 15

__________ 52

11. cos (A + B) = cos A cos B - sin A sin B

= - √ 15

______ 4 · 12 ___

13 - 1 __

4 · (- 5 ___

13 )

= -12 √ 15 + 5

___________ 52

12. cos (A - B) = cos A cos B + sin A sin B

= - √ 15

______ 4 · 12 ___

13 + 1 __

4 · (- 5 ___

13 )

= -12 √ 15 - 5

___________ 52

13. R 120° = ⎡

⎢

⎣ cos 120°

sin 120°

-sin 120°

cos 120°

⎤

⎥

⎦

S = ⎡

⎢

⎣ 0

0 4

1 0

2 -1

1 ⎤

⎥

⎦

R 120° ×S = ⎡

⎢

⎣ cos 120°

sin 120°

-sin 120°

cos 120°

⎤

⎥

⎦ ⎡

⎢

⎣ 0

0 4

1 0

2 -1

1 ⎤

⎥

⎦

= ⎡

⎢

⎣ 0

0

-2.87

2.96

-1.73

-1.00

-0.37

-1.37 ⎤

⎥

⎦

Thus, the new points are (0, 0), (-2.87, 2.96), (-1.73, -1), (-0.37, -1.37).

14. If cos θ = - 4 __ 5 = b __ c , then a 2 = 5 2 - ( -4) 2 .

Thus, a = 3.

sin θ = a __ c = 3 __ 5 . Since θ is in QIII, sin is < 0.

So sin θ = - 3 __ 5 .

sin 2θ = 2sin θ cos θ

= 2 ( -3 ___ 5 ) ( -4 ___

5 ) = 24 ___

25

15. cos 2θ = 2co s 2 θ - 1 = 2 (- 4 __ 5

) 2 - 1

= 32 ___ 25

- 1 = 7 ___ 25

16. tan 2θ = sin 2θ ______ cos 2θ

= 24 ___ 25

___

7 ___ 25

= 24 ___

7

17. sin θ __ 2 = ± √

1 - cos θ ________

2 = ± √

1 + 4 __

5 _____

2

= ± √

9 ___ 10

= ±3 √ 10

____ 10

.

Since 180° < θ < 270°, 90° < θ __ 2

< 135°. This is in QII,

so the sin should be positive. So it is 3 √ 10

_____ 10

.

18. cos θ __ 2 = ± √

1 + cos θ ________

2 = ± √

1 - 4 __

5 _____

2 = √ 1 ___

10 = ±

√ 10 ____

10

Since 180° < θ < 270°, 90° < θ __ 2

< 135°. This is in

QII, so the cos should be negative. So, it is - √ 10

____ 10

.

19. tan θ __ 2 =

sin θ __ 2 _____

cos θ __ 2 =

3 √ 10

_____ 10

______

- √ 10

____ 10

= -3.

20. cos 22.5° = cos 45° ___ 2 = ± √

1 + cos 45° __________

2

= ± √

1 + 1 _____ √ 2 /2

= ± √

√ 2 + 1 _______

√ 2

_______ 2 = ± √

2 + √ 2 _______

2

Since 22.5° is in QI, cos > 0. Thus, it is

2 + 2 _______ 2

.

21. 1 + 2sin θ = 0 2sin θ = -1 sin θ = - 1 _

2

θ = 7π ___ 6 , 11π ____

6 for 0 ≤ θ ≤ 2π

Since sin has a period of 2π, all solutions are

7π ___ 6 + 2πn , 11π ____

6 + 2πn, for n ∈ 핑.

22. cos 2θ + 2cos θ = 32co s 2 θ - 1 + 2 cos θ - 3 = 0 2(co s 2 θ + cos θ - 2) = 0 2(cos θ + 2)(cos θ - 1) = 0cos θ = -2 or 1 θ = 0°


CS10_A2_MESK710389_C11.indd 347 4/11/11 1:17:41 PM

23. 8si n 2 θ - 2sin θ - 1 = 0 (2sin θ - 1)(4sin θ + 1) = 0

sin θ = 1 __ 2 ,- 1 __

4

θ = 30°, 150°, ≈ 194.5°, ≈ 345.5°

24. cos 2θ = 3cos θ + 1 2c os 2 θ - 1 = 3cos θ + 1 2c os 2 θ - 3cos θ - 2 = 0(cos θ - 2)(2cos θ + 1) = 0

cos θ = 2, -1 ___ 2

θ = 2π ___ 3 , 4π ___

3

25. si n 2 θ + cos θ + 1 = 01 - co s 2 θ + cos θ + 1 = 0 co s 2 θ - cos θ - 2 = 0 (cos θ + 1)(cos θ - 2) = 0cos θ = -1, 2 θ = π

26. Let T(x) = 65°F. 65 = 15.85cos π __

6 (x - 1) + 76.85

-11.85 = -15.85cos π __ 6 (x - 1)

237 ____ 317

= cos π __ 6 (x - 1)

x = 6 __ π co s -1 ( 237 ____ 317

) + 1

= 2.39 or 11.61

This is mid-March and mid-December.Let T(x) = 85°F. 85 = -15.85cos π __

6 (x - 1) + 76.85

8.15 = -15.85cos π __ 6 (x - 1)

- 163 ____ 317

= cos π __ 6 (x - 1)

x = 6 __ π c os -1 (- 163 ____ 317

) + 1

= 5.03 or 8.97

This is early June and late September.

sTudy GuIde: revIew

1. cycle 2. frequency

3. period 4. phase shift

Graph of sine anD Cosine

5. because a = 1, amplitude: 1;because b = 3,

period: 2π ___ 3

1.5

0

y

x

-1- π _

2 π_2

6. because a = 1, amplitude: 1;

because b = 1 __ 2 ,

period: 4π

1.5

0

y

x

-1-2π 2π

7. because a = 1 __ 3 ,

amplitude: 1 __ 3 ;

because b = 1,period: 2π

.5

0

y

x

-.5-π π

8. because a = 2, amplitude: 2; because b = π,period: 2

x

y

1

-1

0 1.5-1.5

9. because a = 2, amplitude: 1 __

2 ;

because b = 2,period: π

.75

0

y

x

-.75

-π π

10. because a = π __ 2 ,

amplitude: π __ 2 ;

because b = π,period: 2

π- _2

π _

2

x

y

0-2 2

11. π indicates a shift π units left. The first intercept occurs at π __

2 .

x-intercepts: π __ 2 + πn;

phase shift: π left

.5

0

y

x

-.5

-π π

12. π __ 4 indicates a shift

π __ 4 units left. The first

intercept occurs at 3π ___ 2

.

x-intercepts: 3π ___ 4

+ πn;

phase shift: π __ 4 left

.5

0

y

x

-.5

-π π

13. - 3π ___ 4 indicates a shift

3π ___ 4 units right. The first

intercept occurs at π __ 2 .

x-intercepts: π __ 2 + πn;

phase shift: 3π ___ 2 right

.5

0

y

x

-.5

-π π

14. 3π ___ 2 indicates a shift

3π ___ 2 units left. The first

intercept occurs at π.x-intercepts: πn;phase shift: 3π ___

2 left

π_2

.5

0x

-.5

y

- 3π _ 2


x-1


11-1

CS10_A2_MESK710389_C11.indd 348 4/11/11 1:17:42 PM

15. amplitute: 1.2;period: 24phase shift: 6 right

y

x

0

6

4

8

362412

16. Because b = π ___ 12

,

period is 2π ____ ⎜ π ___ 12

⎟ = 24 h.

17. From the graph, max is 8.2 at noon.

Graphs of other triGonometriC funCtions

18. period: π ___ ⎜b⎟

= π ___ ⎜1⎟

= π;

x-intercepts: first intercept occurs at x = 0. Because the period is π, the x-intercepts occur at πn.


+ πn ___ ⎜b⎟

= π __ 2 + πn

0

-1

1

-π π

y

x

19. period: π ___ ⎜b⎟

= π ___ ⎜π⎟

= 1;

x-intercepts: first intercept occurs at x = 0. Because the period is 1, the x-intercepts occur at n.


+ πn ___ ⎜b⎟

= π ____ 2 ⎜π⎟

+ πn ___ ⎜π⎟

= 1 __ 2 + n

x

y

0 1-1

20. period: π ___ ⎜b⎟

= π ____ ⎜ π __ 2 ⎟ = 2;

x-intercepts: first intercept occurs at x = 0. Because the period is 2, the x-intercepts occur at 2n.asymptotes: π ____

2 ⎜b⎟ + πn ___

⎜b⎟ = π _____

2 ⎜ π __ 2 ⎟ + πn ____

⎜ π __ 2 ⎟ = 1 + 2n

x

y

2

-2

0 2-2

21. period: π ___ ⎜b⎟

= π ___ ⎜1⎟

= π;

x-intercepts: first intercept occurs at x = π __ 2

. Because

the period is π, the x-intercepts occur at π __ 2

+ πn.


= πn ___ ⎜1⎟

= πn

6

y

x

-3-π π

22. period: π ___ ⎜b⎟

= π ___ ⎜1⎟

= π;

x-intercepts: first intercept occurs at x = π __ 2

. Because

the period is π, the x-intercepts occur at π __ 2

+ πn.


= πn ___ ⎜1⎟

= πn

2

x

-2-1 π _

2 π_

2 0

y

23. period: π ___ ⎜b⎟

= π ___ ⎜π⎟

= 1;

x-intercepts: first intercept occurs at x = 1 __ 2

. Because

the period is 1, the x-intercepts occur at 1 __ 2

+ n.


= πn ___ ⎜π⎟

= n

2

-2

x

y

1-1

24. period: 2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;

asymptotes: π ____ 2 ⎜1⎟

+ πn ___ ⎜1⎟

= π __ 2 + πn

1 x

-2

-π π

y

0


x-2


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25. period: 2π ___ ⎜b⎟

= 2π ___ ⎜2⎟

= π;

asymptotes: πn ___ ⎜2⎟

= π __ 2 n

1

x0-π π

y

26. period: 2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;


= πn

4

x

-4

y

π _2

- 3π _ 2

27. period: 2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;


+ πn ___ ⎜1⎟

= π __ 2 + πn

.5x

-.5

-π π

y

0

28. period: 2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;


+ πn ___ ⎜1⎟

= π __ 2 + πn

x

-1

1

-π π

y

0

29. period: 2π ___ ⎜b⎟

= 2π ___ ⎜1⎟

= 2π;


= πn

2

x

-2

-π π0

y

funDamental triGonometriC iDentities

30. sec θ sin θ cot θ = ( 1 _____ cos θ

) sin θ ( cos θ _____ sin θ

)

= ( cos θ _____ cos θ

) ( sin θ ____ sin θ

) = 1

31. sin 2 (- θ)

________ tan θ

= (-sin θ)(-sin θ)

_____________ sin θ ____ cos θ

= (sin θ)(sin θ) ( cos θ _____ sin θ

)

= sin θ cos θ

32. (sec θ + 1)(sec θ - 1) = sec 2 θ - 1 = tan 2 θ

33. cos θ sec θ + cos 2 θ csc 2 θ = 1 + cos 2 θ ( 1 _____ sin 2 θ

)

= 1 + cot 2 θ = csc 2 θ

34. (tan θ + cot θ)2 = tan 2 θ + 2 tan θ cot θ + cot 2 θ = tan 2 θ + 2 + cot 2 θ = ( tan 2 θ + 1) + (1 + cot 2 θ) = sec 2 θ + csc 2 θ

35. tan θ + cot θ = sin θ _____ cos θ


= sin 2 θ + cos 2 θ ____________ sin θ cos θ

= 1 ________ sin θ cos θ

= sec θ csc θ

36. sin 2 θ tan θ = (1 - cos 2 θ)tan θ = tan θ - cos 2 θ tan θ

= tan θ - cos 2 θ ( sin θ _____ cos θ

)

= tan θ - sin θ cos θ

37. tan θ _________ 1 - cos 2 θ

= ( sin θ ____

cos θ ) _____

sin 2 θ

= ( sin θ _____ cos θ

) ( 1 _____ sin 2 θ

)

= ( 1 _____ cos θ

) ( 1 ____ sin θ

)

= sec θ csc θ

38. cot θ sec θ = ( cos θ _____ sin θ

) ( 1 _____ cos θ

)

= csc θ

39. sec θ sin θ ________ cot θ

= ( 1 _____ cos θ

) sin θ ( sin θ _____ cos θ

)

= tan 2 θ

40. tan(-θ)

_______ cot θ

= -tan θ tan θ

= - tan 2 θ

41. cos θ cot θ _________ csc 2 θ - 1

= cos θ ( cos θ ____

sin θ ) _________

( cos 2 θ ____ sin 2 θ

)

= cos 2 θ sin 2 θ __________ cos 2 θ sin θ

= sin θ


x-3


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CS10_A2_MESK710389_C11.indd 350 4/11/11 1:17:45 PM

sum anD DifferenCe iDentities

42. sin 19π ____ 12

= sin ( 15π ____ 12

+ 4π ___ 12

)

= sin ( 5π ___ 4 + π __

3 )

= sin 5π ___ 4 cos π __

3 + cos 5π ___

4 sin π __

3

= - 1 ___ √ 2

· 1 __ 2 - 1 ___

√ 2 ·

√ 3 ___

2

= - 1 ____ 2 √ 2

- √ 3

____ 2 √ 2

= -1 - √ 3

________ 2 √ 2

· √ 2 ___ √ 2

= - √ 2 - √ 6

__________ 4

43. cos 165° = cos (135° + 30°) = cos 135° cos 30° - sin 135° sin 30°

= - 1 ___ √ 2

· √ 3

___ 2 - 1 ___

√ 2 · 1 __

2

= - √ 3

_____ 2 √ 2

- 1 _____ 2 √ 2

= - √ 3 - 1

________ 2 √ 2

· √ 2 ___ √ 2

= - √ 6 - √ 2

__________ 4

44. cos15° = cos (45° - 30°) = cos 45° cos30 + sin45°sin30°

= 1 ___ √ 2

· √ 3

___ 2 + 1 ___

√ 2 · 1 __

2

= √ 6

___ 4 +

√ 2 ___ 4

= √ 6 + √ 2

________ 4

45. tan π ___ 12

= tan ( 4π ___ 12

- 3π ___ 12

)

= tan 4π ___

12 - tan 3π ___

12 _______________

1 + tan 4π ___ 12

tan 3π ___ 12

= tan π __

3 - tan π __

4 _____________

1 + tan π __ 3 tan π __

4

= √ 3 - 1

_______ 1 + √ 3

= √ 3 - 1

_______ 1 + √ 3

· 1 - √ 3

_______ 1 - √ 3

= -1 + 2 √ 3 - 3

_____________ -2

= -4 + 2 √ 3

_________ -2

= 2 - √ 3

46. step 1 Find sin A , cos A, sin B , and cos B.

A is in QI, and tan A = 3 __ 4 .

x 2 + y 2 = r 2

r = √ 3 2 + 4 2 r = √ 25 = 5

Thus, sin A = 3 __ 5 , cos A = 4 __

5 .

B is in QII, and tan B = - 5 ___ 12

.

x 2 + y 2 = r 2

r = √ 5 2 + 12 2 r = √ 169 = 13

Thus, sin B = 5 ___ 13

, cos B = - 12 ___ 13

.

step 2 Find sin A cos B, cos A sin B.

sin A cos B = 3 __ 5 · (- 12 ___

13 ) = - 36 ___

65

cos A sin B = 4 __ 5 · 5 ___

13 = 20 ___

65 .

step 3 Use the angle-sum identity.sin (A + B) = sin A cos B + cos A sin B

= - 36 ___ 65

+ 20 ___ 65

= - 16 ___ 65

47. step 1 Use cos A, sin B , and cos B found in question 46 to find cos A cos B, sin A sin B.

cos A cos B = 4 __ 5 · (- 12 ___

13 ) = - 48 ___

65

sin A sin B = 3 __ 5 · 5 ___

13 = 15 ___

65

step 2 Use the angle-sum identity.cos (A + B) = cos A cos B - sin A sin B

= - 48 ___ 65

- 15 ___ 65

= - 63 ___ 65

48. step 1 Use tan A, tan B to find tan A tan B.

tan A tan B = 3 __ 4 · (- 5 ___

12 ) = - 15 ___

48

step 2 Use the angle-sum identity.


= 3 __ 4 + 5 ___

12 _____

1 - 15 ___ 48

=

56 ___ 48

__

33 ___ 48

= 56 ___

33

49. step 1 Use tan A tan B found in question 48.step 2 Use the angle-sum identity.


= 3 __ 4 - 5 ___

12 _____

1 + 15 ___ 48

=

16 ___ 48

__

63 ___ 48

= 16 ___

63

50. step 1 Use sin A cos B and cos A sinB found in question 46.step 2 Use the angle-sum identity.sin (A - B) = sin A cos B - cos A sin B

= - 36 ___ 65

- 20 ___ 65

= - 56 ___ 65


x-4


11-4

CS10_A2_MESK710389_C11.indd 351 4/11/11 1:17:45 PM

51. step 1 Use cos A cos B and sin A sinB found in question 47.step 2 Use the angle-sum identity.cos (A - B) = cos A cos B + sin A sin B

= - 48 ___ 65

+ 15 ___ 65

= - 33 ___ 65

52. step 1 Find tan A , cos A, tan B , and sin B.

A is in QI, and sin A = √ 7

___ 4 .

x 2 + y 2 = r 2

x = √ 4 2 - √ 7 2 x = √ 9 = 3

Thus, tan A = √ 7 ___ 3 , cos A = 3 __

4 .

B is in QII, and cos B = - 5 ___ 13

.

x 2 + y 2 = r 2

y = √ 13 2 - 5 2 y = √ 144 = 12

Thus, tan B = - 12 ___ 5 , sin B = 12 ___

13 .

step 2 Find sin A cos B, cos A sin B.

sin A cos B = √ 7 ___ 4 · (- 5 ___

13 ) = -

5 √ 7 ____

52

cos A sin B = 3 __ 4 · 12 ___

13 = 36 ___

52

step 3 Use the angle-sum identity.sin (A + B) = sin A cos B + cos A sin B

= - 5 √ 7

____ 52

+ 36 ___ 52

= 36 - 5 √ 7 _________ 52

53. step 1 Use cos A, sin B , and cos B found in question 46 to find cos A cos B, sin A sin B.

cos A cos B = 3 __ 4 · (- 5 ___

13 ) = - 15 ___

52

sin A sin B = √ 7 ___ 5 · 12 ___

13 = 12 √ 7 _____

52

step 2 Use the angle-sum identity.cos (A + B) = cos A cos B - sin A sin B

= - 15 ___ 52

- 12 √ 7 _____ 52

= -15 - 12 7 ____________ 52

54. step 1 Use tan A, tan B to find tan A tan B.

tan A tan B = √ 7 ___ 3 · (- 12 ___

5 ) = -

12 √ 7 _____

15

step 2 Use the angle-sum identity.


= √ 7 ___ 3

+ 12 ___ 5

_______

1 - 12 √ 7 _____ 15

= 5 √ 7 + 36 __________ 15 - 12 √ 7

55. step 1 Use tan A tan B found in question 48.step 2 Use the angle-sum identity.


= √ 7 ___ 3

- 12 ___ 5

_______

1 + 12 √ 7 _____ 15

= 5 √ 7 - 36 __________ 15 + 12 √ 7

56. step 1 Use sin A cos B and cos A sinB found in question 46.step 2 Use the angle-sum identity.sin (A - B) = sin A cos B - cos A sin B

= - 5 √ 7

____ 52

- 36 ___ 52

= -36 - 5 √ 7 __________ 52

57. step 1 Use cos A cos B and sin A sinB found in question 47.step 2 Use the angle-sum identity.cos (A - B) = cos A cos B + sin A sin B

= - 15 ___ 52

+ 12 √ 7 _____ 52

= -15 + 12 7 ____________ 52

58. R 30° × S ≈ ⎡ ⎢ ⎣ 0.87

0.50

-0.50

0.87

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

3

0 4

2 1

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0

2.60

1.50

2.46

3.73

-0.13

2.23 ⎤ ⎥ ⎦

A ′(0, 0), B ′(2.60, 1.50), C ′(2.46, 3.73), D ′(-0.13, 2.23)

59. R 45° × S ≈ ⎡ ⎢ ⎣ 0.71

0.71

-0.71

0.71

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

3

0 4

2 1

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0 2.12

2.12 1.41

4.24

-0.71

2.12 ⎤ ⎥ ⎦

A ′(0, 0), B ′(2.12, 2.12), C ′(1.41, 4.24), D ′(-0.71, 2.12)

60. R 60° × S ≈ ⎡ ⎢ ⎣ 0.5

0.87

-0.87

0.5 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

3

0 4

2 1

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0

1.50

2.60

0.27

4.46

-1.23

1.87 ⎤ ⎥ ⎦

A ′(0, 0), B ′(1.50, 2.60), C ′(0.27, 4.46), D ′(-1.23, 1.87)

61. R 90° × S ≈ ⎡ ⎢ ⎣ 0

1 -1

0 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

3

0 4

2 1

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0

0

3 -2

4 -2

1 ⎤ ⎥ ⎦

A ′(0, 0), B ′(0, 3), C ′(-2, 4), D ′(-2, 1)

62. R 120° × S ≈ ⎡ ⎢ ⎣ -0.5

0.87

-0.87

-0.5

⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

5

2

0

4

-5

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0

-4.23

3.33

-3.46

-2.00

0.77

-5.33 ⎤ ⎥ ⎦

A ′(0, 0), B ′(-4.23, 3.33), C ′(-3.46, -2.00), D ′(0.77, -5.33)

63. R 180° × S ≈ ⎡ ⎢ ⎣ -1

0

0

-1 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

5

2

0

4

-5

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0

-5

-2

0

-4

5

-2 ⎤ ⎥ ⎦

A ′(0, 0), B ′(-5, -2), C ′(0, -4), D ′(5, -2)

64. R 240° × S ≈ ⎡ ⎢ ⎣

-0.5

-0.87

0.87

-0.5 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

5

2

0

4

-5

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0

-0.77

-5.33

3.46

-2.00

4.23

3.33 ⎤ ⎥ ⎦

A ′(0, 0), B ′(-0.77, -5.33), C ′(3.46, -2.00), D ′(4.23, 3.33)

65. R 270° × S ≈ ⎡ ⎢ ⎣

0

-1 1

0 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ 0

0

5

2

0

4

-5

2 ⎤ ⎥ ⎦

≈ ⎡ ⎢ ⎣ 0

0 2

-5 4

0 2

5 ⎤ ⎥ ⎦

A ′(0, 0), B ′(2, -5), C ′(4, 0), D ′(2, 5)


CS10_A2_MESK710389_C11.indd 352 4/11/11 1:17:46 PM

Double-anGle anD half-anGle iDentities

66. For tan θ = 4 __ 3 in QI, sin θ = 4 __

5 , cos θ = 3 __

5 .

sin 2θ = 2 sin θ cos θ

= 2 ( 4 __ 5 ) ( 3 __

5 ) = 24 ___

25

67. For tan θ = 4 __ 3 in QI, cos θ = 3 __

5 .

cos 2θ = 2 cos 2 θ - 1

= 2 ( 3 __ 5 )

2 - 1 = - 7 ___

25


5 .

tan θ __ 2 = √

1 - cos θ ________

1 + cos θ

= √

1 - 3 __

5 _____

1 + 3 __ 5 = √ 2 __

8 = 1 __

2


5 .

sin θ __ 2 = √

1 - cosθ ________

2

= √

1 - 3 __

5 _____

2 = 1 ___

√ 5 =

√ 5 ___

5

70. For cos θ = 3 __ 4 in QIV, tan θ = -

√ 7 ___

3 .

tan 2θ = 2 tan θ ________ 1 - tan 2 θ

= -

2 √ 7 ____

3 ______

1 - 7 __ 9 = -

6 √ 7 ____

2 = -3 √ 7

71. cos 2θ = 2 cos 2 θ - 1

= 2 ( 3 __ 4 )

2 - 1 = 1 __

8

72. cos θ __ 2 = - √ 1 + cos θ __

2

= - √

1 + 3 __

4 _____

2

= - √ 7 __ 8 ·

√ 2 ___ √ 2

= - √ 14

____ 4

73. sin θ __ 2 = √ 1 - cos θ __

2

= √

1 - 3 __

4 _____

2

= 1 ___ √ 8

· √ 2 ___ √ 2

= √ 2 ___ 4

74. sin π ___ 12

= sin π __ 6 ( 1 __

2 )

= √

1 - cos π __

6 _________

2

= √

1 -

√ 3 ___ 2 ______

2 =

√ 2 - √ 3 ________

2

75. cos 75° = cos 150° ____ 2

= √

1 + cos 150° ___________ 2

= √

1 -

√ 3 ___ 2 ______

2 =

√ 2 - √ 3 ________

2

solvinG triGonometriC equations

76. √ 2 cos θ + 1 = 0 √ 2 cos θ = -1

cosθ + 1 = - 1 ___ √ 2

θ = cos -1 (- 1 ___ √ 2

)

θ = 135°for all real number values of θ, where n is an integer.θ = 135° + 260°n, 225° + 360°n

77. cosθ = 2 + 3 cosθ 2 cos θ = -2 cos θ = -1 θ = cos -1 (-1) θ = 180°for all real number values of θ, where n is an integer.θ = 180° + 360°n

78. tan 2 θ + tan θ = 0tan θ(tan θ + 1) = 0tan θ = 0 or tan θ = -1 θ = 0° θ = 135°for all real number values of θ, where n is an integer.θ = 180°n, 135° + 180°n

79. sin 2 θ - cos 2 θ = 1 __ 2

cos 2 θ - sin 2 θ = - 1 __ 2

cos 2θ = - 1 __ 2

2θ = cos -1 (- 1 __ 2 )

2θ = 120° θ = 60°for all real number values of θ, where n is an integer.θ = 60° + 180°n, 120° + 180°n

80. 2 cos 2 θ - 3 cos θ = 2 2 cos 2 θ - 3 cos θ - 2 = 0(2 cos θ + 1)(cos θ - 2) = 0

cos θ = - 1 __ 2 or cos θ = 2

θ = 2π ___ 3 , 4π ___

3


x-5

x-6


11-5

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CS10_A2_MESK710389_C11.indd 353 4/11/11 1:17:46 PM

81. cos 2 θ + 5 cos θ - 6 = 0(cos θ + 6)(cos θ - 1) = 0cos θ = -6 or cos θ = 1θ = 0

82. sin 2 θ - 1 = 0(sin θ - 1)(sin θ + 1) = 0sin θ = 1 or sin θ = -1

θ = π __ 2 , 3π ___

2

83. 2 sin 2 θ - sin θ = 3 2 sin 2 θ - sin θ - 3 = 0(2 sin θ - 3)(sin θ + 1) = 0

sin θ = -1 or sin θ = 3 __ 2

θ = 3π ___ 2

84. cos 2θ = cos θ 2 cos 2 θ - cos θ - 1 = 0(2 cos θ + 1)(cos θ - 1) = 0

cos θ = - 1 __ 2 or cos θ = 1

θ = 0, 2π ___ 3

, 4π ___ 3

85. sin 2θ + cos θ = 02 sin θ cos θ + cos θ = 0 cos θ(2 sin θ + 1) = 0

cos θ = 0 or sin θ = - 1 __ 2

θ = π __ 2 , 7π ___

6 , 3π ___

2 , 11π ____

6

86a. 900 min; late June

b. 540 min; late December

chApTer TesT

1. amplitude = 1 __ 2 ; period = 2π ___

2 = π

.75

x

-.5

y

0

- π _ 2

π_2

2. phase shift: π __ 3 left

x-intercepts: sin (x + π __ 3 ) = 0. So, x + π __

2 = π. x = 2π ___

3

Since sin hits zero every π, the x-intercepts are

2π ___ 3 + πn, n ∈ 핑.

1

x

-1

-π π

y

0

3. Let r = 0.5 m and F = 500 N. Thus,τ(θ) = (0.5)(500)cos θ = 250 cos θ .The max will occur at cos θ = 1. So θ = 0°, and τ = 250 N · m.The min will occur at cos θ = 0 (since this is the minimum of cos on 0° ≤ θ ≤ 90°). So θ = 90°, and τ = 0 N · m.

4. period: 1;intercepts: tan θ = 0 for all mulitiples of π, so tan πx has intercepts at n. tan θ has asymptotes

at π __ 2 + πn. So, tan πx has them at 1 __

2 + n.

-1.5 1.5

x

-1

1

y

0

5. period = π __ 4 ;

Intercepts: cot θ = 0 for π __ 2 + πn;

so cot 4x has intercepts at π __ 8 + π __

4 n. cot θ has

aysmptotes at multiples of π. So, cot 4x has themat π __

4 n.

1

x

-1

y

- π _ 2 π_

2

6. The period is the same as sin x , which is 2π. sin x = 0 for πn. Thus, this is where csc x is undefined, so asymptotes occur at x = πn.

0.5

x

-π π

y

0

7. cot θ = cos θ _____ sin θ

= ( cos θ _____ sin θ

) ( cos θ _____ cos θ

)

= co s 2 θ ( 1 ____ sin θ

) ( 1 _____ cos θ

)

= co s 2 θ sec θ csc θ 8. (sec θ + 1)(sec θ - 1) = se c 2 θ - 1 = t an 2 θ (since

ta n 2 θ + 1 = se c 2 θ )


CS10_A2_MESK710389_C11.indd 354 4/11/11 1:17:47 PM

9. sin (-θ) = -sin θ and cos (-θ) = cos θ

Thus, sin (-θ)

_______ cos (-θ)

= -sin θ ______ cos θ

= -tan θ

10. tan A = b __ a = 3 __ 4 , and a 2 + b 2 = c 2 . Thus,

c =

3 2 + 4 2 = 5. So, sin A = b __ c = 3 __ 5 .

cos A = a __ c = 4 __ 5 .

All values are positive since it is in QI.

If sin B = -12 ____ 13

= b __ c , then a = √ 13 2 - (- 12) 2 = 5.

So, cos B = a __ c = -5 ___ 13

since it is in QIII.

sin (A + B) = sin A cos B + sin B cos A

= 3 __ 5 · (- 5 ___

13 ) + (- 12 ___

13 ) · 4 __

5 = -15 - 48 _________

63 = - 63 ___

65

11. Using what we determined in question 10:cos(A - B) = cos A cos B + sin A sin B

= 4 __ 5 · (- 5 ___

12 ) + 3 __

5 · (- 12 ___

13 ) = -20 - 36 _________

65 = - 56 ___

65

12. R 30° = ⎡

⎢

⎣ cos 30°

sin 30°

-sin 30°

cos 30°

⎤

⎥

⎦

S = ⎡

⎢

⎣ 0

1 2 1 3

3 -1

3 ⎤

⎥

⎦

R 30° ×S = ⎡

⎢

⎣ cos 30°

sin 30°

-sin 30°

cos 30°

⎤

⎥

⎦ ⎡

⎢

⎣ 0

1 2

1

3

3 -1

3 ⎤

⎥

⎦

= ⎡

⎢

⎣ -0.50

0.87

1.23

1.87

1.10

4.10

-2.37

2.10

⎤

⎥

⎦

Thus, the new points are (-0.50, 0.87), (1.23, 1.87), (1.10, 4.10), (-2.37, 2.10).

13. If tan θ = -12 ____ 5 = b __ a , then c = √ (-1 2) 2 + 5 2 = 13.

So, cos θ = a __ c = 5 ___ 13

. However, since it is in QII, it is

negative: -5 ___ 13

. sin θ = b __ c = 12 ___ 13

, and is positive.

sin 2θ = 2 sin θ cos θ = 2 ( 12 ___ 13

) ( -5 ___ 13

) = -120 _____ 169

14. Using the information from question 13.

cos 2θ = 2 co s 2 θ - 1 = 2 ( 5 ___ 13

) 2 - 1 = 50 ____

169 - 1

= - 119 ____ 169

15. Using the information from question 13.

cos θ __ 2 = ± √

1 + cos θ ________

2 = ± √

1 - 5 __

13 ______

2 =

2 √ 13 _____

13

It is positive since 45° < θ __ 2 < 90°, which is QI.

16. sin 3π ___ 8 = sin 1 __

2 ( 3π ___

4 ) = ± √

1 - sin 3π __ 4 ________

2 = ± √

1 - 1 ___

√ 2 ______

2

= √

2 + √ 2 _______ 2

It is positive since 3π ___ 8 is in QI.

17. tan θ = - √ 3 θ = ta n -1 (- √ 3 )

θ = 120° + 180°n, 2π ___ 3 + πn

18. 2si n 2 θ - sin θ = 0sin θ (2sin θ - 1) = 0

sin θ = 0 or 1 __ 2

θ = 0°, 30°, 150°, 180°

19. 2co s 2 θ + 3sin θ = 0 2(1 - sin 2 θ ) + 3sin θ = 0 2si n 2 θ - 3sin θ - 2 = 0(2sin θ + 1)(sin θ - 2) = 0

sin θ = - 1 __ 2 or 2

θ = 7π ___ 6 , 11π ____

6

20. V(t) = 110 = 156sin 2π(60t)

55 ___ 78

= sin 2π(60t)

t = π _____ 2(60)

si n -1 ( 55 ___ 78

)

t ≈ 0.0021 or 0.0063

The period is 2π ______ 2π(60)

= 1 ___ 60

, so all solutions are

≈ 0.0021 + 1 ___ 60

n s and ≈ 0.0063 + 1 ___ 60

n s.


CS10_A2_MESK710389_C11.indd 355 4/11/11 1:17:47 PM

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