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S–52 Chapter 2 Trigonometry Instructor’s Solutions Manual

Chapter 2 Trigonometry Instructor’s Solutions Manual

Exercise Set 2.1, p. 85

1. a.

–1

1

x

y

(0, 1)�2( ), 0

(�, –1)

(2�, 1)

( ), 02

3�

� 3� 4�2

3�

y = cos xb.

–1

1

x

y

�2( ), 1

(�, 0)(2�, 0)

( ), –12

3�

y = sin x

–2� 2�–� �

(0, 0)

2. a–b. Since A = −1, the graph is reflected across the x-axis.

a.

–1

1

x

y

(0, 0)

�2( ), –1

(2�, 0)(�, 0) ( ), 1

23�

4�3�2��

y = –sin x b.

–1

1

x

y

(0, –1)

�2( ), 0

(2�, –1)

(�, 1)

( ), 02

3�

–�–2� �

y = –cos x

3. Since A = 6, we have a vertical stretch.

–6

6

x

y

(0, 6)

�2( ), 0

(�, –6)

(2�, 6)

( ), 02

3�

�–2� 2�–�

y = 6 cos x

amplitude = |A | = 6Plot the critical points for y = 6 cos x on 0 ≤ x ≤ 2π :

• The x-intercepts are the same as y = cos x.• The maximum and minimum y-coordinates of the critical points are

6 and −6, respectively.

Connect the critical points with four identical arc sections in the cosineform of a sinusoidal wave. Repeat the pattern for one more cycle.

range: | y | ≤ 6, x-intercepts: −3π

2,−π

2,π

2,

3π

2or x = π

2+ kπ


–4

4

x

y

(0, 0)

�2( ), 4

(2�, 0)

(�, 0)

( ), –42

3�

�–2� 2�

y = 4 sin x

amplitude = |A | = 4Plot the critical points for y = 4 sin x on 0 ≤ x ≤ 2π :

• The x-intercepts are the same as y = sin x.• The maximum and minimum y-coordinates of the critical points are

4 and −4, respectively.

Connect the critical points with four identical arc sections in the sine formof a sinusoidal wave. Repeat the pattern for one more cycle.

range: | y | ≤ 4, x-intercepts: −2π,−π, 0, π, 2π or x = kπ


Exercise Set 2.1 S–53

5. Since A = −π , we have a vertical stretch and a reflection across the x-axis.

x

y

(0, 0)

�2( ), –�

(2�, 0)

(�, 0) ( ), �2

3�

� 2�–2�

y = –� sin x

–�

�amplitude = |A | = π

Plot the critical points for y = −π sin x on 0 ≤ x ≤ 2π :

• The x-intercepts are the same as y = sin x.• The maximum and minimum y-coordinates of the critical points are

π and −π , respectively.

Connect the critical points with four identical arc sections in the sine form of asinusoidal wave. Repeat the pattern for one more cycle.

range: | y | ≤ π , x-intercepts: −2π,−π, 0, π, 2π or x = kπ

6. Since A = −2

3, we have a vertical shrink and a reflection across the x-axis.

x

y

–1

1�2( ), 0

3�2( ), 0

�–2� –� 2�

( )23�,

( )232�, –

( )230, –

y = – cos x23amplitude = |A | = 2

3

Plot the critical points for y = −2

3cos x on 0 ≤ x ≤ 2π :

• The x-intercepts are the same as y = cos x.• The maximum and minimum y-coordinates of the critical points

are2

3and −2

3, respectively.

Connect the critical points with four identical arc sections in the cosine formof a sinusoidal wave. Repeat the pattern for one more cycle.

range: | y | ≤ 2

3, x-intercepts: −3π

2,−π

2,π

2,

3π

2or x = π

2+ kπ


–7

7

x

y

(0, 0)

�2( ), 7

(2�, 0)

(�, 0)

�–2� –� 2�

y = 7 sin x

( ), –72

3�

amplitude = |A | = 7Plot the critical points for y = 7 sin x on 0 ≤ x ≤ 2π :

• The x-intercepts are the same as y = sin x.• The maximum and minimum y-coordinates of the critical points

are 7 and −7, respectively.


range: | y | ≤ 7, x-intercepts: −2π,−π, 0, π, 2π or x = kπ

8. Since A = 2.5, we have a vertical stretch.

x

y

(0, 0)

�2( ), 2.5

3�2

(2�, 0)

(�, 0)

( ), –2.5

�–2� –� 2�

y = 2.5 sin x

2

3amplitude = |A | = 2.5Plot the critical points for y = 2.5 sin x on 0 ≤ x ≤ 2π :


are 2.5 and −2.5, respectively.


range: | y | ≤ 2.5, x-intercepts: −2π,−π, 0, π, 2π or x = kπ



9. Since A = −0.4, we have a vertical shrink and a reflection across the x-axis.

x

y

0.40.6

1

–0.6

–1

(0, –0.4)

�2( ), 0

(2�, –0.4)

(�, 0.4)

( ), 02

3�

�–2� –� 2�

y = –0.4 cos xamplitude = |A | = 0.4Plot the critical points for y = −0.4 cos x on 0 ≤ x ≤ 2π :




range: | y | ≤ 0.4, x-intercepts: −3π

2,−π

2,π

2,

3π

2or x = π

2+ kπ

10. Since A = −7

5, we have a vertical stretch and a reflection across the x-axis.

x

y

�2( ), 0

�–2� –� 2�

( )75�,

( )752�, –( )7

50, –

y = – cos x75

1

2

–2

( ), 02

3�amplitude = |A | = 7

5

Plot the critical points for y = −7

5cos x on 0 ≤ x ≤ 2π :


are7

5and −7

5, respectively.


range: | y | ≤ 7


2,−π

2,π

2,

3π

2or x = π

2+ kπ

11. Since A = 3

2, we have a vertical stretch.

x

y

–1

2

–2

�2( ), 0

�–2� –� 2�

( )32�, –

( )322�,

( )320,

y = cos x32

( ), 02

3�

amplitude = |A | = 3

2

Plot the critical points for y = 3

2cos x on 0 ≤ x ≤ 2π :


are3

2and −3

2, respectively.


range: | y | ≤ 3


2,−π

2,π

2,

3π

2or x = π

2+ kπ

12. Since A = −0.9, we have a vertical shrink and a reflection across the x-axis.

x

y

(0, 0)

�2( ), –0.9

(–2�, 0)(�, 0)

( ), 0.92

3�

�–� 2�

y = –0.9 sin x

–1

1amplitude = |A | = 0.9Plot the critical points for y = −0.9 sin x on 0 ≤ x ≤ 2π :



Connect the critical points for four identical arc sections in the sine form of asinusoidal wave. Repeat the pattern for one more cycle.range: | y | ≤ 0.9, x-intercepts: −2π,−π, 0, π, 2π or x = kπ



13. The amplitude is 2 and the sine graph has been reflected across the x-axis, so y = −2 sin x.

14. The amplitude is 3 in the sine form, so y = 3 sin x.

15. The amplitude is 3 in the cosine form, y = 3 cos x.

16. The amplitude is 1.5 and the cosine graph has been reflected across the x-axis, so y = −1.5 cos x.

17. The amplitude is 0.3 in the sine form, so y = 0.3 sin x.

18. The amplitude is 6 in the cosine form, so y = 6 cos x.

19. No, the period or cycle should be 2π .

20. No, the cycle should be 2π ; x-intercepts are incorrect.

21. No, the graph should be arc shaped.

22. No, there should be four equal arc sections for the cycle.

23. No, the x-intercept is incorrect. There should be four equal arcs, and the period should be 2π .

24. No, it is not a complete cycle; there should be one more arc from3π

2to 2π to complete the four equal arcs for each

cycle.

25. a. The distance along the x-axis from −4π to 6π is | 6π − (−4π) | = 10π . Since each cycle is 2π , there are fivecomplete cycles.

b. The distance along the x-axis from −3π to 2π is | 2π − (−3π) | = 5π . Since each cycle is 2π , there are twocomplete cycles.

26. a. The distance along the x-axis from −8π to −6π is | − 6π − (−8π) | = 2π . Since each cycle is 2π , there isone complete cycle.

b. The distance along the x-axis from π to 10π is | 10π − π | = 9π . Since each cycle is 2π , there are fourcomplete cycles.

27. We can use the negative identity cos(−x) = cos x to verify that the cosine function is even.

28. We can use the negative identity sin(−x) = − sin x to verify that the sine function is odd.

29. a. Since A = 3, there is a vertical stretch, with maximum

–3

3

t

y

� 3�2� 4�

e

(2.3, ≈2.2)d(t) = 3 sin t

and minimum y-coordinates at 3 and −3, respectively.b. By inspection, we see d(t) (or y) is zero when t = 0, π, 2π, 3π, 4π .c. The maximum displacement is 3 in.d. The maximum displacement is equal to the amplitude.e. d(2.3) = 3 sin(2.3) ≈ 2.2 in. (see graph in part (a))

30. a. Since A = −5, there is a vertical stretch, and a reflection across the

–5

5

t

y

� 3�2� 4�

e(6, ≈ –4.8)

d(t) = –5 cos t

x-axis, with maximum and minimum y-coordinates at 5 and −5, respectively.

b. By inspection, we see d(t) (or y) is zero when t = π

2,

3π

2,

5π

2,

7π

2.

c. The minimum displacement is −5 cm.d. The absolute value of the minimum displacement is equal to the amplitude.e. d(6) = −5 cos(6) ≈ −4.8 cm (see graph in part (a))



31. a. The arc on the unit circle isπ

2, and the cosine value is 0. The point on the cosine graph that corresponds to(π

2, 0)

is B.

b. The arc on the unit circle isπ

2, and the sine value is 1. The point on the sine graph that corresponds to

(π

2, 1)

is F .

32. a. The arc on the unit circle is7π

4, and the cosine value is

√2

2. The point on the cosine graph that corresponds to(

7π

4,

√2

2

)is D.

b. The arc on the unit circle is7π

4, and the sine value is −

√2

2. The point on the sine graph that corresponds to(

7π

4,−√

2

2

)is H .

33. True 34. True

35. True 36. False; the y-intercept is (0, 0).

37. True 38. False; the amplitude is |A |.39. False; the graph is symmetric with respect to the origin.

40. True

41. b. The period of y = sin(2x) is π , and the x-intercepts are 0,π

2, π,

3π

2, 2π .

d. The period of y = sin

(1

2x

)is 4π , and the x-intercepts are 0, 2π, 3π, 4π .

e. The period of y = sin

(1

4x

)will be 8π , which can be verified using Xmin = 0 and Xmax = 8π .

42. b. The period of y = cos(2x) is π , and the x-intercepts areπ

4,

3π

4,

5π

4,

7π

4.

d. The period of y = cos

(1

2x

)is 4π , and the x-intercepts are π, 3π, 5π .

e. The period of y = cos

(1

4x

)will be 8π , which can be verified using Xmin = 0 and Xmax = 8π .

43. b. The graph of y = sin(x + π

4

)has different locations for all the critical points, as each represents a critical

point of y = sin x shifted leftπ

4units.

d. The graph of y = cos(x − π

4

)has different locations for all the critical points, as each represents a critical

point of y = cos x shifted rightπ

4units.

e. The graph of y = sin (x + π) represents the graph of y = sin x shifted left π units.

44. b. The graph of y = sin(x)+ 2 has different locations for all the critical points, as each represents a critical pointof y = sin x shifted up two units.

d. The graph of y = cos(x)− 1 has different locations for all the critical points, as each represents a critical pointof y = cos x shifted down one unit.

e. The graph of y = sin(x)− 2 represents the graph of y = sin x shifted down two units.



45. a. Yes, the x-intercepts are the same for both graphs.b. The amplitude is 3.c. The amplitude of y = x sin(x) varies.d. If A is a variable, the amplitude will vary.e. The graphs of y = x and y = −x are the boundary graphs.f. Since −1 ≤ sin(x) ≤ 1, then −x ≤ x sin(x) ≤ x. So y = x sin x is bounded by y = −x and y = x.

46. a. The boundary functions are y = 1

x + 1and y = − 1

x + 1.

47. c. Using the table function and the trace key, it appears they converge on the interval (−2.07, 2.07).e. The interval of convergence appears to become larger when more terms from the infinite series are added.

f. Using y = 1− x2

2! +x4

4! −x6

6! + . . .− x34

34! , the graphs appear to converge on [−4π, 4π ].48. The powers of x are odd, which can be represented by 2n− 1. So if n = 1 represents the first term, then the power

of x in the first term is 2(1)− 1 = 1; if n = 2 represents the second term, then the power of x in the second term is2(2)− 1 = 3, and so on. The factorial in the denominator of the term is the same as the power, and since the termsalternate from positive to negative (with the first term positive, if n = 1(−1)1−1 = (−1)0 = 1), we get:

sin x = x − x3

3! +x5

5! −x7

7! + . . .+ (−1)n−1 x2n−1

(2n− 1)! − . . . .

49. The sine function is an odd function, so its infinite series has odd numbers as powers of x and as factorials. The sinefunction oscillates from 1 to −1, and its infinite series oscillates from a positive value to a negative value.


For 1–22: Period = pure period

|B | = 2π

|B |

1. A = 1, B = 1

2

–1

1

x

y

�–4� –�

(�, 1)

(2�, 0)

(3�, –1)

(4�, 0)

( )y = sin x12

(0, 0)

amplitude: |A | = 1

period:2π(

12

) = 4π

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe sine form, plot the critical points.By inspection the x-intercepts are at x = −4π,−2π, 0, 2π or x = k · 2π .

2. Using negative identities, y = sin(−2x)→ y = − sin(2x), so A = −1, B = 2.

–1

1

x

y

�4( ), –1

( ), 14

3�y = sin (–2x)

(0, 0)

�2

�–�

amplitude: |A | = 1

Since A < 0, reflect the graph across the x-axis.

period:2π

2= π

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe sine form reflected across the x-axis, plot the critical points.

By inspection the x-intercepts are at x = −π,−π

2, 0,

π

2, π or x = k · π

2.



3. Using negative identities, y = cos(−4x)→ y = cos(4x), so A = 1, B = 4.

–1

1

x

y

�4( ), –1

y = cos (–4x)(0, 1)

�2

�4

�8

�4

–�2

–

( ), 1�2amplitude: |A | = 1

period:2π

4= π

2Start at the origin, mark off one period, and divide it into four subintervals. Usingthe cosine form, plot the critical points.

x-intercepts: −3π

8,−π

8,π

8,

3π

8or x = π

8+ k · π

4

4. amplitude: |A | = 1

–1

1

x

y

( )y = cos x34

32�

34�

38�

34�–

38�–

period:2π(3

4

) = 2π · 4

3= 8π

3

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe cosine form, plot the critical points.

x-intercepts: −2π,−2π

3,

2π

3, 2π or x = 2π

3+ k · 4π

3

5. Using negative identities, y = sin

(−1

3x

)→ y = − sin

(1

3x

), so

–1

x

y13

y = sin – x( )

23�

29�

23�– 6�

A = −1, B = 1

3.

amplitude: |A | = 1

period:2π(1

3

) = 2π · 3

1= 6π

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe sine form reflected across the x-axis, plot the critical points.x-intercepts: −6π,−3π, 0, 3π, 6π or x = k · 3π

6. Using negative identities, y = − sin(−πx)→ y = sin(πx), so A = 1, B = π .

–1

1

x

y

12

3 2–2 12

y = – sin (–�x)

12–3

2–

amplitude: |A | = 1

period:2π

π= 2

x-intercepts: −2,−1, 0, 1, 2 or x = k

7. A = 3, B = 1

4

x

y

–3

3

–8� –4� 4� 8�

y = 3 cos x14( )

amplitude: |A | = 3

period:2π(1

4

) = 8π

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe cosine form, plot the critical points.x-intercepts: −6π,−2π, 2π, 6π or x = 2π + k · 4π



8. A = 3

2, B = 2

x

y

–1

–2

2y = cos (2x)3

2

�2

�4

�2

––� �

amplitude: |A | = 3

2

period:2π

2= π

x-intercepts: −3π

4,−π

4,π

4,

3π

4or x = π

4+ k · π

2

9. A = −4, B = 2

x

y

–4

4y = –4 sin (2x)

�4

�2

–� �

amplitude: |A | = 4

period:2π

2= π

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe sine form reflected across the x-axis, plot the critical points.

x-intercepts: −π,−π

2, 0,

π

2, π or x = k · π

2

10. y = 3.5 sin

(−1

4x

)→ y = −3.5 sin

(1

4x

)

x

y

–4

4

–6� –2� 2� 6� 8�

y = 3.5 sin x14( )–

A = −3.5, B = 1

4amplitude: |A | = 3.5

period:2π(1

4

) = 8π

x-intercepts: −8π,−4π, 0, 4π, 8π or x = k · 4π

11. A = 0.5, B = π

x

y

–1

1

0.5

12

2–2 –1 1

y = 0.5 sin (�x)amplitude: |A | = 0.5

period:2π

π= 2

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe sine form, plot the critical points.

x-intercepts: −2,−1, 0, 1, 2 or x = k

12. A = 6, B = π

3

–6

6

x

y ( )y = 6 sin x�3

3–3–6 632

92

32–

amplitude: |A | = 6

period:2π(π

3

) = 2π · 3

π= 6

x-intercepts: −6,−3, 0, 3, 6 or x = k · 3



13. A = −2, B = π

4

–2

1

2

x

y

–8 –4 4 8

( )y = –2 cos x�4

Since A < 0, the graph reflects across the x-axis.

amplitude: |A | = 2

period:2π(π

4

) = 8

Start at the origin, mark off one period, and divide it into four subintervals. Usingthe cosine form, plot the critical points.

x-intercepts: −6,−2, 2, 6 or x = 2+ k · 4

14. y = −5 cos(−π

2x)→ y = −5 cos

(π

2x)

–5

5

x

y

–4 –2 21 4

( )y = –5 cos – x�2

A = −5, B = π

2amplitude: |A | = 5

period:2π(π

2

) = 4

x-intercepts: −3,−1, 1, 3 or x = 1+ k · 2

15. amplitude: 2period: 2π

}Dot in y = 2 cos x.

phase shift: right π

range: | y | ≤ 2

–1

1

2

x

y

(0, –2)

(0, 2) (�, 2)

� 2� 3�

y = 2 cos (x – �)


}Dot in y = 5 sin x.

phase shift: left π

range: | y | ≤ 5

x

y

�–� 2�

y = 5 sin (x + �)

( ), –52

3�

�2( ), 5–

�2( ), 5

�2( ), –5

–1


}Dot in y = 4 sin

(1

2x

).

phase shift: leftπ

2range: | y | ≤ 4

–4

4

x

y

�2

–� � 3� 4��2

–

(�, 4)

�2( ), 4

[ ([(y = 4 sin x + �2

12

18. amplitude: 3period: π

}Dot in y = 3 cos(2x).

phase shift: rightπ

8range: | y | ≤ 3

–3

x

y

�8

�2

�

[ ([(y = 3 cos 2 � – �8

�8( ), 3

�2( ), –3

( ), 32

3�

( ), 38

7�

( ), –38

5�

(0, 3)



19. amplitude: 3

period:π

2

Dot in y = −3 cos(4x).

phase shift: leftπ

4range: | y | ≤ 3

–3

–1

1

x

y

(0, 3)�4( ), 3

�2

�4

�4

–

�2( ), –3�

4( ), –3

y = –3 cos 4 x +�4[ ( )]

20. amplitude: 1

period:2π

3

Dot in y = − sin(3x).

phase shift: leftπ

2range: | y | ≤ 1

1

x

y

�2( ), 1

�2

�6

�3

�3

–

�2

–

�6( ), –1( ), –1

32�

(0, 1)y = –sin 3x +( )23�

21. amplitude: 2period: π

}Dot in y = 2 sin(2x).

vertical shift: up 3range: 1 ≤ y ≤ 5

–2–1

5

x

y �4( ), 5

�2

(0, 3)

(0, 0)

( ), 14

3�

( ), –24

3�

(�, 3)

(�, 0)

��4( ), 2

y = 2 sin (2x) + 3


phase shift: left π

Dot in y = cos(x + π).

vertical shift: down 2range: −3 ≤ y ≤ −1

1

2

x

y

�2( ), 0

�2( ), –2(0, –3)

(�, 1)(–�, 1)

(–�, –3)

2�

y = cos (x + �) – 2

�–�

23. The graph can represent y = sin x reflected across the x-axis (g), or the graph can represent y = cos x shifted

leftπ

2(b).

24. The graph y = cos x can be represented using negative identities as y = cos(−x) (e), or the graph can represent

y = sin x shifted leftπ

2(c).

25. The graph can represent y = cos x reflected across the x-axis (h), or the graph can represent y = sin x shifted

rightπ

2(d).

26. The graph y = sin x can be represented using negative identities as y = − sin(−x) (f), or as y = cos x shifted

rightπ

2(a), which is the same equation as (i) since y = cos

(x − π

2

)= cos

(−x + π

2

).



27–32. There are many possible answers.

27. By inspection, amplitude = 2, P = π = 2π

|B | → B = 2.

–2–1

12

x

y

� 2�

Since the graph has the cosine form, y = 2 cos(2x).

28. By inspection, amplitude = 3, P = π = 2π

|B | → B = 2.

–3

–1

1

3

x

y

� 2�

Since the graph has the sine form, y = 3 sin(2x).

29. The amplitude =∣∣∣∣ 5− 1

2

∣∣∣∣ = 2, so A = 2. And since P = 2π, B = 1.

–3

–1

1

3

5

x

y

� 2�

Since the sine form has been shifted up three, y = 2 sin x + 3.

30. The amplitude =∣∣∣∣ (−2)− (−6)

2

∣∣∣∣ = 2, so A = 2. And since P = 2π, B = 1.

–7

–5

–3

–1

1

x

y

� 2�Since the cosine form has been shifted down four, y = 2 cos x − 4.

31. By inspection, amplitude = 4, P = 4π = 2π

|B | → B = 1

2.

–5

–3

–1

1

3

5

x

y

� 2� 3� 4�

Since the cosine form has been reflected across the x-axis, y = −4 cos

(1

2x

).



32. By inspection, amplitude = 2, P = 8π = 2π

|B | → B = 1

4.

–3–2–1

123

x

y

2� 4� 6� 8�

Since the sine form has been reflected across the x-axis, y = −2 sin

(1

4x

).

33–38. Make sure that B > 0, and that the values of C stay within the restrictions: −π < C < 0 and 0 < C < π (whichmeans that all graphs must be the result of a phase shift).

33. y = sin[2(x + π

4

)], or y = sin

[2

(x − 3π

4

)]34. y = cos

[2(x − π

4

)], or y = cos

[2

(x + 3π

4

)]

35. y = cos

[1

2

(x − π

2

)], or y = sin

[1

2

(x + π

2

)]

36. y = sin

[1

2

(x − π

2

)]37. y = sin

(x − π

2

)

38. y = cos[2(x + π

4

)], y = sin

[2(x + π

2

)], y = sin

[2(x − π

2

)], or y = cos

[2

(x − 3π

4

)]

39. y = −6 sin

(1

2x

), y = 6 cos

[1

2(x + π)

]40. y = 3 cos x, y = 3 sin

(x + π

2

)

41. Rewrite as y = 1

4sin

[4

(x − 1

2

)]. amplitude = |A | = 1

4, P = 2π

4= π

2, phase shift: right

1

2, range: | y | ≤ 1

4

42. Rewrite as y = 5.7 cos [3(x + 4)]. amplitude = 5.7, P = 2π

3, phase shift: left 4, range: | y | ≤ 5.7

43. amplitude: 2.3, P = 12π , vertical shift: up π , range: −2.3+ π ≤ y ≤ 2.3+ π

44. amplitude: = √2, P = 4π ,vertical shift: down 5, range: −√2− 5 ≤ y ≤ √2− 5

45. a. y = sin x + 2 shifts the graph of y = sin x up 2 units; y = sin(x + 2) shifts the graph of y = sin x left 2 units.b. y = sin(2x) has an amplitude of 1 and a period of π ; y = 2 sin x has an amplitude of 2 and a period of 2π .

c. y = 1

2sin 2x has an amplitude of

1

2and a period of π ; y = sin x has an amplitude of 1 and a period of 2π .

d. y = sin(x + π

2

)shifts y = sin x left

π

2units; y = sin x + sin

π

2is the same as y = sin x + 1, which shifts

y = sin x up 1 unit.

46. a. y = cos

(1

2x

)has a period of 4π ; y = 1

2cos x has an amplitude of

1

2.

b. y = cos x + cosπ

2= cos x + 0 = cos x; y = cos

(x + π

2

)shifts y = cos x left

π

2.

c. y = cos(−2x) = cos(2x) These are the same.d. y = cos x − 5 shifts y = cos x down 5; y = cos(x − 5) shifts y = cos x right 5.

47. a. amplitude = 1

2b. P = π

c. vertical shift: up1

2d. y = 1

2cos(2x)+ 1

2



48. a. amplitude = 1

2b. P = π

c. vertical shift: up1

2d. y = −1

2cos(2x)+ 1

2

49. No 50. Yes

51. a. Yes, insert them around the entire arc: y = cos(

2(x − π

2

)).

b. They are already around the entire arc since(

2(x − π

2

))= (2x − π).

52. a. Yes, by inspection the period is 1. b. amplitude =∣∣∣∣ 0.9999− 0

2

∣∣∣∣ ≈ 0.5

c. −5,−4,−3,−2,−1, 0, 1, 2, 3, 4, 5 d. No, it is not continuous on the domain.

53. a. Yes, the period is 2. b. amplitude =∣∣∣∣ 0.9999− 0

2

∣∣∣∣ ≈ 0.5

c. −4,−2, 0, 2, 4 d. No, it is not continuous on the domain.

54. a. y = 1

4x −

[1

4x

]b. y = −2x − [−2x]


1. a. maximum = 88.7, minimum = 80.1 b. amplitude =∣∣∣∣ 88.7− 80.1

2

∣∣∣∣ = 4.3

2. a. maximum = 83.7, minimum = 29 b. amplitude =∣∣∣∣ 83.7− 29

2

∣∣∣∣ = 27.35


2

∣∣∣∣ = 8


2

∣∣∣∣ = 21.4

5. Yes

5

10

x

y

0 2 4 6 8 10 12

6. Yes

5

10

x

y

0 5 10

7. No

5

15

25

x

y

0 2 4 6 8



8. No

–10

–2

2

10

20

x

y

5 10

9. Yes

–2

–1

1

2

3

x

y

5 10 15 20 25

10. Yes

–60–50–40–30–20–10

10

x

y

1 2 3 4 5 6 7

11–14. Note: For the sine or cosine form, the consecutive maximum and minimum critical points occur one-half periodapart (or max to max points are a period apart).

11. amplitude =∣∣∣∣ 21− 10.5

2

∣∣∣∣ = 5.25 We let A = 5.25.

As a result of the repetition of maximum (or minimum) critical points occurring 15 units apart along the x-axis, theperiod is 15.

P = 15 = 2π

|B | → B = 2π

15

With period 15, each arc occurs in a length of15

4= 3.75 units along the x-axis. Using the sine form, and knowing

a maximum critical point occurs at x = 2, the phase shift is 2− 3.75 = −1.75 (left). So C = −1.75.If A = 5.25 and the minimum y-coordinate is 10.5, the shift up is

D = 10.5+ 5.25 = 15.75.

Therefore, y = 5.25 sin

[2π

15(x + 1.75)

]+ 15.75.

12. amplitude =∣∣∣∣ 9− (−1)

2

∣∣∣∣ = 5 We let A = 5.

As a result of the repetition of maximum (or minimum) critical points occurring 12 units apart along the x-axis, theperiod is 12.

P = 12 = 2π

|B | → |B | =π

6


4= 3 units along the x-axis.

Using the sine form, and knowing a maximum critical point occurs at x = 0, the phase shift is three units left. SoC = −3.If A = 5 and the minimum y-coordinate is −1, the shift up is

D = −1+ 5 = 4.

Therefore, y = 5 sin[π

6(x + 3)

]+ 4.



13. amplitude =∣∣∣∣ 90− 71

2

∣∣∣∣ = 9.5 We let A = 9.5.

As a result of the minimum y-coordinate occurring at x = 2, and the maximum y-coordinate at x = 8, we use theperiod 12.

P = 12 = 2π

|B | → B = π

6


4= 3 units along the x-axis. Using the sine form and knowing a

maximum critical value occurs at x = 8, the phase shift is 8− 3 = 5 units right. So C = 5.If A = 9.5 and the minimum y-coordinate is 71, the shift up is

D = 71+ 9.5 = 80.5.

Therefore, y = 9.5 sin[π

6(x − 5)

]+ 80.5.

14. amplitude =∣∣∣∣ 14.5− 2

2

∣∣∣∣ = 6.25 We let A = 6.25.

As a result of the minimum y-coordinate occurring at x = 6 and the maximum y-coordinate occurring at x = 12,we use the period 12.

P = 12 = 2π

|B | → B = π

6


4= 3 units along the x-axis.

Using the sine form and knowing the maximum occurs at x = 12, would mean the maximum also occurs at x = 0.So the phase shift is C = 0− 3 = −3, or 3 units left. So C = −3.If A = 6.25 and the minimum y-coordinate is 2, the shift up is

D = 6.25+ 2 = 8.25.

Therefore, y = 6.25 sin[π

6(x + 3)

]+ 8.25.

15–18. Sine regression values may vary due to the calculator model or manufacturer. The following values are from aTI-83 Plus, using only one period of values.

15. a ≈ 5.1767, b ≈ 0.3748,c ≈ 1.4498, d ≈ 16.1445

y = 5.1767 sin(0.3748x + 1.4498)+ 16.1445

16. a ≈ 4.7462, b ≈ 0.5343,c ≈ 1.6139, d ≈ 3.5409

y = 4.7462 sin(0.5343x + 1.6139)+ 3.5409



17. a. Let x = 0 represent 12 midnight: a ≈ 1.4346, b ≈ 0.5664, c ≈ −1.8332, d ≈ 5.3372So y = 1.4346 sin(0.5664x − 1.8332)+ 5.3372, where y represents the water depth in feet.

b. At 5am, x = 5 and the water depth is:

y = 1.4346 sin(0.5664(5)− 1.8332)+ 5.3372

= 6.5 ft (to the nearest half-foot)

At 4pm, x = 16 and depth ≈ 6.5 ft.c. At 9pm, x = 21 and depth ≈ 4.5 ft, which means the depth is not enough to bring the boat in since it needs a

depth of 5 ft.

18. a. Let x = 0 represent 7/1992.

110

8

0

b. a ≈ 1.7115, b ≈ 0.3719, c ≈ 2.1857, d ≈ 6.0930So y = 1.7115 sin(0.3719x − 2.1857)+ 6.0930, where y represents the percent unemployment.

c. Since B = 0.3719, P = 2π

0.3719≈ 16.9 yrs.

d. For 7/2012, x = 20, and using the equation in part (b) we get y ≈ 5.8. So based on this model, in 2012 theunemployment rate would be 5.8 percent.

19. a. Since P1 has a frequency of 10, with cosine form, F = 10 = 1

P→ P = 1

10= 2π

|B | → B = 20π .

P1 : y = cos(20πx)

Since P2 has a frequency of 12, with reflected cosine form, F = 12 = 1

P→ P = 1

12= 2π

|B | → B = 24π .

P2 : y = − cos(24πx)

b. No

c. By inspection, period = 1

2, which means the frequency is 2.

20. Using a TI-83 Plus,sin(4000π) = −2× 10−10

(a number close to zero).

21. Graph (b) is correct, period too small in graph (a), and minimum height below ground level in graph (c).


Function Domain Range Period Asymptotes x-intercepts y-intercepts Graph

1. y = sin x x ∈ R | y | ≤ 1 2π — x = kπ (0, 0)

–1

1

x

y

–� � 2�



Function Domain Range Period Asymptotes x-intercepts y-intercepts Graph

2. y = cos x x ∈ R | y | ≤ 1 2π — x = π

2+ kπ (0, 1)

–1

1

x

y

–� � 2�

3. y = tan x x �= π

2+ kπ y ∈ R π x = π

2+ kπ x = kπ (0, 0)

–4

4

x

y

�2

�2

3��2–

4. y = cot x x �= kπ y ∈ R π x = kπ x = π

2+ kπ —

–4

4

x

y

�2

� 2�

5. y = sec x x �= π

2+ kπ | y | ≥ 1 2π x = π

2+ kπ — (0, 1)

–1

1

x

y

�2

�–� 2�

6. y = csc x x �= kπ | y | ≥ 1 2π x = kπ — —

–2

1

x

y

�2

�–� 2�

7. y = sin x, y = tan x 8. y = cos x, y = sec x

9. y = tan x, y = cot x, y = csc x, y = sin x 10. y = sec x, y = csc x

11. all 12. y = tan x, y = cot x, y = sec x, y = csc x

13. y = tan x, y = cot x, y = sec x, y = csc x 14. y = cos x, y sin x



For 15–28, P = pure period

|B | .

15. Step 1. B = 1

2, so P = π(

1

2

) = 2π .

Step 2. asymptotes: x =π

2+ kπ(1

2

) = π + k · 2π For k = −1, 0 we get x = −π, π .

Step 3. x-intercepts: x = kπ(1

2

) = k · 2π For k = 0 we get x = 0.

Steps 4–6. x y

−π

2−1

π

21

–4

4

x

y

2��–� 3�

y = tan

(1

2x

)

16. Step 1. B = 1

4, so P = π(

1

4

) = 4π

Step 2. asymptotes: x = kπ(1

4

) = k · 4π For k = 0, 1, we get x = 0, 4π .

Step 3. x-intercept: x =π

2+ kπ(1

4

) = 2π + k · 4π For k = 0, we get x = 2π .

Steps 4–6. x y

π 1

3π −1

–2

4

x

y

2�–2�–4� 4�

y = cot

(1

4x

)



17. Rewrite as y = tan(x −

(−π

2

)).

Step 1. B = 1, so P = π

Step 2. Asymptotes for y = tan x are x = −π

2and x = π

2.

Since C = −π

2, the graph of y = tan x shifts left

π

2. The asymptotes for y = tan

(x −

(−π

2

))are

x = −π

2+(−π

2

)= −π, x = π

2+(−π

2

)= 0, or x = kπ .

Step 3. x-intercept: x = −π

2,π

2(one-half the distance between the asymptotes) or x = π

2+ kπ .

Steps 4–6. x y

−3π

4−1

−π

41

–4

–1

1

x

y

�2

�2

––� �

y = tan(x + π

2

)

18. For y = cot(x − π

4

), B = 1, C = π

4.

Step 1. P = π .

Step 2. Asymptotes for y = cot x are at x = 0 and x = π . Since C = π

4, the asymptotes for y = cot

(x − π

4

)are x = 0+ π

4= π

4, x = π + π

4= 5π

4, or x = π

4+ kπ .

Step 3. x-intercept: x = 3π

4(one-half the distance between the asymptotes) or x = 3π

4+ kπ

Steps 4–6. x y

π

21

π −1

–4

4

x

y

�4 4

3�4

5�4

9��4

–

y = cot(x − π

4

)



19. Since A = −2, we have a reflection across the x-axis.

Step 1. B = 2, so P = π

2

Step 2. Asymptotes: x = k · π2

For k = 0, 1 we get x = 0, and x = π

2.

Step 3. x-intercepts:

π

2+ kπ

2= π

4+ k · π

2

For k = 0 we get x = π

4(one-half the distance between the asymptotes).

Steps 4–6. x y

π

8−2

3π

82

–4

–1

1

4

x

y

�2

�4

�4

3�

y = −2 cot(2x)

20. Rewrite: y = tan[2(x − π)]

Step 1. B = 2, so P = π

2

Step 2. The asymptotes for y = tan(2x) are x =π

2+ kπ

2. For k = −1, 0 we get x = −π

4and x = π

4.

Since C = π , the asymptote for y = tan[2(x − π)] are x = −π

4+ π = 3π

4, and x = π

4+ π = 5π

4or

x = π

4+ k · π

2.

Step 3. x-intercept: x = 0,π

2, π (one-half the distance between the asymptotes, and a period apart) or x = k · π

2

Steps 4–6. x y

−π

8−1

π

81

1

2

3

x

y

�2

�4

��2

–

y = tan(2x − 2π)



21. For y = − cot x + 1, since A < 1, we have a reflection; D = 1 indicates a vertical shift of y = − cot x up 1.

Step 1. P = π

Step 2. Asymptotes: x = kπ

Step 3. Since the graph shifts up 1 and the x-intercepts are where y = 0, then 0 = − cot x + 1→ cot x = 1, so

x = π

4,

5π

4, . . . or x = π

4+ kπ .

Step 4–6. x − cot x − cot x + 1π

4−1 0

3π

41 2

–4

–1

4

x

y

�2

�2

3� 2�

y = − cot x + 1

22. For y = tan x − 1, D = −1 indicates the graph of y = tan x shifts vertically down 1.

Step 1. P = π

Step 2. asymptotes: x = π

2+ k · π

Step 3. Since the graph shifts down 1 and the x-intercepts are where y = 0, then 0 = tan x − 1→ tan x = 1, or

x = π

4,

5π

4, . . . or x = π

4+ k · π .

Steps 4–6. x tan x tan x − 1

−π

4−1 −2

π

41 0

–4

–1

1

4

x

y

�2

�2

3��2–

y = tan x − 1

23. Step 1. Dot in the graph of the guide function y = −4 cos x.

–1

1

4

x

y

–� � 2��2

y = −4 sec xStep 2. At each x-intercept, plot the asymptotes.

Step 3. Draw the four arc sections in the form of twoseparate U-shaped branches between the asymptotes.

P = 2π , range: | y | ≥ 4



24. Step 1. Dot in y = −1

2sin x as the guide function.

–2

–1

2

x

y

� 2�–�2

3��2–

y = −1

2csc xStep 2. Plot the asymptote at each x-intercept.

Step 3. Draw the four arcs in the form of the twoU-shaped branches between the asymptotes.

P = 2π , range: | y | ≥ 1

2

25. Step 1. Dot in y = sin(−2x) = − sin(2x) as the guide

–3

–2

–1

1

2

3

x

y

�2

3�2

2�

function. y = csc(−2x)

Step 2. Plot the asymptotes at each x-intercept.

Step 3. Draw the four arc sections in the form of twoU-shaped branches between the asymptotes.

P = π , range: | y | ≥ 1

26. Step 1. Dot in y = 3 cos

(1

2x

)as the guide function.

x

y

1

–1

–3

–5

3

5

2�� 4� 6�

y = 3 sec

(1

2x

)Step 2. Plot the asymptotes at each x-intercept.

Step 3. Draw four arc sections in the form of twoU-shaped branches between the asymptotes.

P = 4π , range: | y | ≥ 3

27. Step 1. Dot in y = 4 cos(x + π

2


–4

–1

1

4

x

y

�–� 3�

y = 4 sec(x + π

2



P = 2π , range: | y | ≥ 4

28. Step 1. Dot in y = sin[2(x − π

4

)]as the guide function.

1

2

3

x

y

23�

47��

4– �4

�2

y = csc 2(x − π

4



P = π , range: | y | ≥ 1

29. c. Use y = cos

(1

2x

)+ 2 as a guide function. Since B = 1

2, P = 4π , and since D = 2, there is a vertical shift

up 2.



30. d. Use y = − sin x + 2 as a guide function. Since A = −1 and B = 1 (P = 2π), y = sin x is reflected across thex-axis. With D = 2, there is a vertical shift up 2.

31. a. Since B = 2, P = π

2in cotangent form.

32. b. Since A = −1, the tangent graph with P = π

2is reflected across the x-axis.

33. The graph does not represent a function since it does not pass the vertical line test. The graph should not cross itsasymptotes.

34. The graph represents y = − tan x; the asymptotes and x-intercepts are incorrect for y = cot x.

35. One branch is missing. The cycle is incomplete.

36. The graph (used as a guide) of y = sin x should not be included (it should be only dotted in). The graph does notrepresent a function.

37. a, c;Since y = tan x has period π , then y = − tan x = − tan(x + π) (a).

Also y = cot x shifted rightπ

2, is the same as y = − tan x (c).

38. a, b, or d;y = − sec x = csc

(x − π

2

), using the guide y = − cos x, which is the same as y = sin x shifted

rightπ

2(a).

y = − sec x = − csc(x + π

2

), using y = − sin x as a guide shifted left

π

2(b).

y = − sec x = sec(x − π), using y = cos x as a guide shifted right π (d).

39–42. Variations on Ymin and Ymax are possible.

39. a. Period = π

.01π= 10, range: R

b. WINDOW Xmin = −10, Xmax = 10,Ymin = −20,Ymax = 20

40. a. Period = π(1

5

) = 5π , range: R

b. WINDOW Xmin = −5π, Xmax = 5π,Ymin = −20,Ymax = 20

41. a. Period = 2π

2= π , range: | y | ≥ 5

b. WINDOW Xmin = −π, Xmax = π,Ymin = −20,Ymax = 20

42. a. Period = 2π , range: | y | ≥ 1

b. WINDOW Xmin = −2π, Xmax = 2π,Ymin = −10,Ymax = 10

43. a. asymptotes: x = −π,−π

2, 0,

π

2, π or x = k · π

2b. The asymptotes are a combination of those for y = tan x and y = cot x.c. Nod. Where y = tan x would have an x-intercept, y = cot x has an asymptote, and vice versa. (Where one function

is zero, the other is undefined.)e. The graphs are the same.



f. tan x + cot x = csc x sec x

sin x

cos x+ cos x

sin x= 1

sin x· 1

cos x

--------------------

sin x

cos x· sin x

sin x+ cos x

sin x· cos x

cos x=

sin2 x + cos2 x

sin x · cos x=

1

sin x · cos x=

1

sin x· 1

cos x=

44. a. The graphs appear to coincide, so the statement appears to be an identity.b. The graphs do not coincide. The statement is not an identity.c. The graphs do not coincide. The statement is not an identity.

45. From Exercise 37, we obtainthe following identities:

cot(x + π

2

)= − tan x

− tan x = − tan(x + π)

− tan x = cot(x − π

2

)

46. From Exercise 38, we obtainthe following identities:

− sec x = csc(x − π

2

)− sec x = − csc

(x + π

2

)− sec x = sec(x − π)


1. y = arcsin

√3

2←→ sin y =

√3

2, − π

2≤ y ≤ π

2Since y = π

3, the only arc with this sine value in the restricted range is y = π

3.

2. y = arctan√

3←→ tan y = √3, − π

2< y <

π

2Since y = π

3, the only arc with this tangent value in the restricted range is

y = π

3.

3. y = cos−1 1←→ cos y = 1, 0 ≤ y ≤ π

The only arc with this cosine value in the restricted range is y = 0.



4. y = sin−1(−1)←→ sin y = −1, −π

2≤ y ≤ π

2The only arc with this sine value in the restricted range is y = −π

2.

5. y = arccsc (−√2)←→ csc y = −√2, − π

2≤ y < 0, 0 < y ≤ π

2

←→ sin y = − 1√2

Since y = π

4, the only arc in the restricted range with this sine value y = −π

4.

6. y = sec−1 2√3←→ sec y = 2√

3, 0 ≤ y <

π

2,π

2< y ≤ π

←→ cos y =√

3

2

Since y = π

6, the only arc in the restricted range with this cosine value is y = π

6.

7. tan−1 1←→ tan y = 1, −π

2< y <

π

2

y = π

4, so y = π

4

8. cos−1

(−√

2

2

)←→ cos y = −

√2

2, 0 ≤ y ≤ π

y = π

4, so y = 3π

4

9. arccos1

2←→ cos y = 1

2, 0 ≤ y ≤ π

y = π

3, so y = π

3

10. arcsin

(−1

2

)←→ sin y = −1

2, −π

2≤ y ≤ π

2

y = π

6, so y = −π

6



11. arcsin 0←→ sin y = 0, −π

2≤ y ≤ π

2so y = 0 12. arccos 0←→ cos y = 0, 0 ≤ y ≤ π so y = π

2

13. sec−1(−1)←→ sec y = −1, 0 ≤ y <π

2,

π

2< y ≤ π

←→ cos y = −1 so, y = π

14. sin−1 1←→ sin y = 1, −π

2≤ y ≤ π

2so, y = π

2

15. cot−1 1←→ cot y = 1, 0 < y < π

←→ tan y = 1, so y = π

4

16. cot−1

(√3

3

)←→ cot y =

√3

3, 0 < y < π

←→ tan y = √3 so y = π

3

17. arctan (−√3)←→ tan y = −√3, −π

2< y <

π

2

y = π

3, so y = −π

3

18. arccsc 2←→ csc y = 2, − π

2≤ y < 0, 0 < y ≤ π

2

←→ sin y = 1

2

y = π

6, so y = π

6

19. cos−1

(−√

3

2

)←→ cos y = −

√3

2, 0 ≤ y ≤ π

y = π

6, so y = 5π

6

20. tan−1

(−√

3

3

)←→ tan y = −

√3

3, −π

2< y <

π

2

y = π

6, so y = −π

6



21–34. Your calculator should be in radian mode.

21. arccos

(−2

3

)cos−1(−2÷ 3) ENTER ≈ 2.3005

22. arcsin 0.6

sin−1(0.6) ENTER ≈ 0.6435

23. tan−1 4

3

tan−1(4÷ 3) ENTER ≈ 0.9273

24. cos−1(−3

4

)cos−1(−3÷ 4) ENTER ≈ 2.4189

25. sec−1(−1.23)

cos−1(1÷−1.23) ENTER ≈ 2.5201

26. csc−1(−22)

sin−1(1÷−22) ENTER ≈ −0.0455

27. arccsc13

5

cos−1(5÷ 13) ENTER ≈ 0.3948

28. arctan√

7

tan−1(√

(7))

ENTER ≈ 1.2094

29. arcsin 0.6445

sin−1(0.6445) ENTER ≈ 0.7004

30. arccos(−0.2356)

cos−1(−0.2356) ENTER ≈ 1.8086

31. tan−1(−5.967)

tan−1(−5.967) ENTER ≈ −1.4048

32. sec−1 55.6

cos−1(1÷ 55.6) ENTER ≈ 1.5528

33. cos−1

√2

33

cos−1(√

(2÷ 33))

ENTER ≈ 1.3221

34. cot−1 0.71π

2− tan−1(0.71) ENTER ≈ 0.9534

35. sin

(cos−1 1

2

)= sin

(π

3

)=√

3

236. cos

(sin−1 1

2

)= cos

(π

6

)=√

3

2

37. cos(cos−1 1

) = cos(0) = 1 38. sin(sin−1 0) = sin(0) = 0

39. tan(

arctan√

3)= tan

(π

3

)= √3 40. sin(tan−1(−1)) = sin

(−π

4

)= −√

2

2

41. cos(arcsin 1) = cos(π

2

)= 0 42. cos(arctan 0) = cos(0) = 1

43. csc

(cos−1

(−√

3

2

))= csc

(5π

6

)= 1

sin

(5π

6

) = 2

44. tan

(sin−1

(−√

2

2

))= tan

(−π

4

)= −1 45. cos

(arcsin

(−√

2

2

))= cos

(−π

4

)=√

2

2

46. sec

(cos−1

(−√

2

2

))= sec

(3π

4

)= 1

cos

(3π

4

) = −√2

47. cot

(cos−1

(−1

2

))= cot

(2π

3

)= 1

tan

(2π

3

) = 1

−√3= −√

3

3



48. csc(cos−1 0) = csc(π

2

)= 1(

sinπ

2

) = 1 49. sin−1(

cos7π

6

)= sin−1

(−√

3

2

)= −π

3

50. sin−1(

tan3π

4

)= sin−1(−1) = −π

251. arctan

(cos

π

2

)= arctan(0) = 0

52. arccsc (cos π) = arccsc(−1) = arcsin

(1

−1

)= −π

2

53. sin−1(

cos(−π

3

))= sin−1

(1

2

)= π

654. tan−1

(sin

7π

2

)= tan−1(−1) = −π

4

55. arcsec

(tan

3π

4

)= arcsec(−1) = arccos(−1) = π

56. arccot

(sin

3π

2

)= arccot(−1) = π

2− arctan(−1) = π

2+ π

4= 3π

4

57. sin

(cos−1 12

13

)Let y = cos−1 12

13←→ cos y = 12

13, 0 ≤ y ≤ π

So we want sin y, where cos y = 12

13. Use the Pythagorean identity.

(12

13

)2

+ (sin y)2 = 1

(sin y)2 = 1− 144

169= 25

169

sin y = 5

13, since sin y > 0 for 0 ≤ y ≤ π

58. cos

(sin−1 3

5

)Let y = sin−1 3

5←→ sin y = 3

5, −π

2≤ y ≤ π

2

So we want cos y, where sin y = 3


(cos y)2 +(

3

5

)2

= 1

(cos y)2 = 1− 9

25= 16

25

cos y = 4

5, since cos y > 0 for − π

2≤ y ≤ π

2



59. cos

(sin−1 8

17

)Let y = sin−1 8

17←→ sin y = 8

17, −π

2≤ y ≤ π

2

So we want cos y, where sin y = 8


(cos y)2 +(

8

17

)2

= 1

(cos y)2 = 1− 64

289= 225

289

cos y = 15


2≤ y ≤ π

2

60. sin

(arccos

7

25

)Let y = arccos

7

25←→ cos y = 7

25, 0 ≤ y ≤ π



(7

25

)2

+ (sin y)2 = 1

(sin y)2 = 1− 49

625= 576

625

sin y = 24

25, since sin y > 0 for 0 ≤ y ≤ π

61. tan(sin−1 0.2657)

tan(sin−1(0.2657)) ENTER ≈ 0.2756

62. sin(tan−1 55.67)

sin(tan−1(55.67)) ENTER ≈ 0.9998

63. sin(cos−1 0.4675)

sin(cos−1(0.4675)) ENTER ≈ 0.8840

64. cos(sin−1(−0.7652))

cos(sin−1(−0.7652)) ENTER ≈ 0.6438

65. arctan(sin 3.95)

tan−1(sin(3.95)) ENTER ≈ −0.6261

66. arccos(tan 47.9)

cos−1(tan(47.9)) ENTER ≈ 0.1922

67. csc

(arcsin

(−5

9

))

1÷ sin(sin−1(−5÷ 9)) ENTER = −9

5= −1.8000

68. sec

(arccos

(− 2

15

))

1÷ cos(cos−1(−2÷ 15)) ENTER = −15

2= −7.5000

69. sec−1

(tan

√7

2

)

cos−1(1÷ tan

(√(7)÷ 2

))ENTER ≈ 1.3149

70. arccot

(cos

(−√

3

2

))

tan−1(1÷ cos(−√(3)÷ 2)) ENTER ≈ 0.9959

71. a. y = tan x, ii. y = arctan x b. y = sin x, i. y = arcsin x c. y = cos x, iii. y = arccos x

72. a. y = arcsec x b. y = arccot x c. y = arccsc x



73. 5 sin(6x) = −5

sin(6x) = −1 Multiply each side by1

5.

6x = sin−1(−1) Use the definition of inverse sine function.

6x = −π

2sin−1(−1) = −π

2

x = − π

12Multiply each side by

1

6.

74. 4 cos(3x) = 2√

3

cos(3x) =√

3


1

4.

3x = cos−1

(√3

2

)Use the definition of inverse cosine function.

3x = π

6cos−1

(√3

2

)= π

6

x = π


1

3.

75. 4 tan(9x) = 7

tan(9x) = 7


1

4.

9x = tan−1(

7

4

)Use the definition of inverse tangent function.

x = 1

9tan−1

(7

4

)Multiply each side by

1

9.

≈ 0.1169 (Find a calculator approximation.)

76. −2 tan(x

2

)= 5

tan(x

2

)= −5

2Multiply each side by −1

2.

x

2= tan−1

(−5

2

)Use the definition of inverse tangent function.

x = 2 tan−1(−5

2

)Multiply each side by 2.

≈ −2.3806 (Find the calculator approximation.)

77. sin−1 x is the inverse function of sin x, whereas (sin x)−1 is the reciprocal function of sin x. Another name forsin−1 x is arcsin x, and another name for (sin x)−1 is csc x.



78. a. The domain of the arcsin x is −1 ≤ x ≤ 1, and3

2is not in the domain.

b. Rewrite arcsin3

2= y as sin y = 3

2, which should help Larry see that the sin y cannot be larger than 1 since we

know −1 ≤ sin y ≤ 1.

c. cos−1 5→ cos y = 5, which is not possible since −1 ≤ cos y ≤ 1.

sec−1(

1

2

)→ cos−1(2)→ cos y = 2, which is also not possible.

sin−1(−2)→ sin y = −2, which is not possible.

79. a. Only one answer is possible (sin−1 x is a function).

b. Even though sin3π

2= −1, his answer is incorrect since

3π

2is not in the required range of −π

2≤ y ≤ π

2.

80. a–c. Graphs obtained will be the same as those in Exercise 72.d. By inspection, it is true.e. By inspection, it is true.

Chapter 2 Review Exercises, p. 151

1. amplitude = 2

–2

–1

1

2

x

y

�–� 2�–2�

y = 2 sin x

2. amplitude = 3

–3

1

2

3

x

y

�–� 2�–2�

y = –3 cos x

3. amplitude = 0.7

–1–0.8

0.60.8

1

x

y

�–� 2�–2�

y = –0.7 cos x

4. amplitude = 1

2

–1

0.5

1

x

y

�–� 2�–2�

y = sin x12

5. amplitude = 3

2

–2

–1

2

x

y

�–� 2�–2�

y = – sin x32

6. amplitude = 2π

x

y

�–� 2�–2�

–2�

2�

�

y = –2� cos x

7. Since A = 4 and B = 1

2,

amplitude: 4

period:2π(1

2

) = 4π

range: | y | ≤ 4

x-intercepts: x = π + k · 2π

–4

1

4

x

y

(0, 4)

(2�, –4)

�–2� 2� 4� 6�

( )y = 4 cos x12

(4�, 4)


Chapter 2 Review Exercises S–83

8. Since A = 1.5 and B = 2,

amplitude: 1.5

period:2π

2= π

range: | y | ≤ 1.5, x-intercepts: x = k · π2

–2

–1

2

x

y

�2

�4

( ), –1.54

3�

� 2�2

3�

�4( ), 1.5

y = 1.5 sin (2x)

9. Rewrite: y = −3 sin

(−1

4x

)

as y = 3 sin

(1

4x

)

→ A = 3, B = 1

4amplitude: 3

period:2π(1

4

) = 8π

range: | y | ≤ 3, x-intercept: x = k · 4π

–4–3–2–1

1234

x

y

(2�, 3)

(6�, –3)

2� 6� 10� 14�

( )y = –3 sin x14–

10. Rewrite: y = −2 cos

(−1

2x

)

as y = −2 cos

(1

2x

)

→ A = −2, B = 1

2amplitude: 2

period:2π(1

2

) = 4π

range: | y | ≤ 2, x-intercept: x = π + k · 2π

–3

–2

–1

1

2

3

x

y

(2�, 2)

(3�, 0)

(�, 0)

(4�, –2)

2� 4� 6� 8�

( )y = –2 cos x12–

(0, –2)

11. Rewrite: y = − cos(−πx)

as y = − cos(πx)

→ A = −1, B = π

amplitude: 1

period:2π

π= 2

range: | y | ≤ 1, x-intercept: x = 1

2+ k

–1

1

x

y

–2 –1 1 2

(1, 1)

(0, –1) (2, –1)

32( ), 0

12( ), 0

y = –cos (–�x)



12. Rewrite: y = 5 sin

(−2π

5x

)

as y = −5 sin

(2π

5x

)

→ A = −5, B = 2π

5amplitude: 5

period:2π(2π

5

) = 5,

range: | y | ≤ 5, x-intercept: x = k · 5

2

–5

–3

–1

1

3

5

x

y

54

52

154( ), 5

54( ), –5

5 10

y = 5 sin x( )52�–

13. Since A = 1, B = 1, and C = π

2,

period: 2π , phase shift: rightπ

2

–1

1

x

y

–2� 2�–� �

y = cos x – �2( )

14. Since A = −1, B = 1, and C = −π

2,

period: 2π , phase shift: leftπ

2

–1

1

x

y

–2� 2�–� �

y = –sin x + �2( )

15. Since A = −4, B = 1

2, and C = −π ,

period: 4π , phase shift: left π

–4

2

5

x

y

–4� –2� 2� 4�

y = –4 sin [ (x + �)]12

16. Since A = 2.5, B = 2, and C = π

4,

period: π , phase shift: rightπ

4

–3

–1

1

2

3

x

y

�2

�–� �2

–

y = 2.5 cos [ 2(x – ]�4 )

17. Since A = 2, B = 1

3, C = 0, and D = 4,

period: 6π , vertical shift: up 4

x

y

–2

2

4

–3� 3� 6� 9�

( )y = 2 cos x + 413

18. Since A = 1

2, B = 2, C = 0, and D = −1

2,

period: π , vertical shift: down1

2

–1

–0.5

0.5

x

y

y = sin (2x) –12

�4

�2

12

� 2�



19. y = 5 sin x 20. y = −4 cos x

21. y = 0.6 cos x 22. y = −1

2sin x

23–24. There are many possible answers. Answers different from the ones given can be checked with a graphing utility.

23. amplitude = 4

period = π = 2π

|B | → B = 2

If we consider the sine form shifted rightπ

2:

y = 4 sin[2(x − π

2

)]

or shifted leftπ

2:

y = 4 sin[2(x + π

2

)].

If we consider the cosine form shifted leftπ

4:

y = 4 cos[2(x + π

4

)]

or reflected and shifted rightπ

4:

y = −4 cos[2(x − π

4

)].

24. amplitude = 1

period = π = 2π

|B | → B = 2

If we consider the sine form shifted leftπ

8:

y = sin[2(x + π

8

)]

or shifted right7π

8:

y = sin

[2

(x − 7π

8

)].

If we consider the cosine form shifted rightπ

8:

y = cos[2(x − π

8

)]

or shifted left7π

8:

y = cos

[2

(x + 7π

8

)]

or reflected and shifted left3π

8:

y = − cos

[2

(x + 3π

8

)]



25. See answers for Exercise 23 or check your answer with a graphing utility.

26. See answers for Exercise 24 or check your answer with a graphing utility.

27. a. A = 4, B = π → amplitude = |A | = 4, and P = 2π

π= 2

–4–3

–1

1

34

t

y

1 2 3 4

d(t) = 2

y = 4 cos (�t)

b. Each time the point is in the rest position, or when d(t) = 0, it is at ant-intercept. By inspection this occurs when

t = 1

2,

3

2,

5

2,

7

2.

c. See the graph in part (a) for the times when the displacement is 2 inchesfrom the rest position. (Here displacement is positive, so the positions areabove the rest position.)

28. a. A = 4.5 P = 2π(1

2

) = 4π

–5–4–3–2–1

12345

x

y

4�� 8�

d(x) = –3

y = 4.5 sin ( x)12

b. See the graph in part (a). Here the displacement is negative, so the positionsare below the rest position.

29. a. Amplitude =∣∣∣∣ 14.75− 9.5

2

∣∣∣∣ = 25

8. We let A = 2

5

8.

As a result of the maximum y-coordinate occurring at x = 6 and the minimum y-coordinate occurring atx = 12, we use the period 12.

P = 12 = 2π

|B | → B = π

6

With period 12, each arc occurs in a length of 3 along the x-axis. Using the sine form and knowing themaximum occurs at x = 6, the phase shift is 6− 3 = 3 units right.So C = 3.

If A = 25

8and the minimum y-coordinate is 9

1

2,

D = 91

2+ 2

5

8= 12

1

8.

Therefore,

y = 25

8sin[π

6(x − 3)

]+ 12

1

8.

b. c. For y = a sin(bx + c)+ d:

a ≈ 2.6327, b ≈ 0.5073,

c ≈ −1.4729, c ≈ 12.1466



30. a. Amplitude =∣∣∣∣ 17.5− 7

2

∣∣∣∣ = 51

4. We let A = 5

1

4. Since this table reflects yearly occurences, we let

P = 12 = 2π

|B | → B = π

6.

With period 12, each arc occurs in a length of 3 along the x-axis. Using the sine form and knowing themaximum occurs at x = 7, the phase shift is 7− 3 = 4 units right.So C = 4.

If A = 51

4and the minimum y-coordinate is 7,

D = 7+ 51

4= 12

1

4.

Therefore,

y = 51

4sin[π

6(x − 4)

]+ 12

1

4.

b. c. For y = a sin(bx + c)+ d:

a ≈ 5.1516, b ≈ 0.5094,

c ≈ −1.8199, d ≈ 12.1474

31. Step 1. B = 1

3, so P = π(

1

3

) = 3π

Step 2. asymptotes: x = kπ(1

3

) = k · 3π

For k = 0, 1 we get x = 0, 3π .

Step 3. x-intercepts: x =π

2+ kπ(1

3

) = 3π

2+ k · 3π

For k = 0 we get x = 3π

2(one-half the distance between the asymptotes) or x = 3π

2+ k · 3π

(one period apart).

Steps 4–6. x y

3π

43

9π

4−3

–5

–3

–1

12345

x

y

6�3�

y = 3 cot ( x)13

23�

29�

range: R



32. Step 1. B = 2, so P = π

2

Step 2. asymptotes: x =π

2+ kπ

2= π

4+ k · π

2

For k = −1, 0 we get x = −π

4,

π

4.

Step 3. x-intercepts: x = k · π2= k · π

2

For k = 0 we get x = 0 (one-half the distance between the asymptotes) or x = 0 + k · π

2= k · π

2(one period apart).

Steps 4–6. Since A < 0, the graph will reflect across the x-axis.

x y

−π

81

π

8−1

–3

–1

3

x

y

�4

�4

�2 4

3�–

y = –tan (2x)

range: R

33. Step 1. B = 1

4, so P = π(

1

4

) = 4π

Step 2. asymptote for y = tan1

4x are x =

π

2+ kπ(1

4

) = 2π + k · 4π

For k = −1, 0 we get x = −2π, 2π . Since C = π , we shift each asymptote right π for

y = tan

[1

4(x − π)

]and get

x = −2π + π = −π, x = 2π + π = 3π, or x = −π + k · 4π.

Step 3. x-intercept: x = π + k · 4π (one-half the distance between the asymptotes and a period apart).

Steps 4–6. x y

0 −1

2π 1 1

2

3

x

y

3��–� 5� 7�

y = tan [ (x – �)]14

range: R



34. Step 1. B = 1

2, so P = π(

1

2

) = 2π

Step 2. asymptote for y = cot

(1

2x

)are x = kπ(

1

2

) = k · 2π . For k = 0, 1 we get x = 0, 2π . Since C = −π ,

we shift each asymptote left π for y = cot

[1

2(x + π)

]and get

x = 0− π = −π, x = 2π − π = π, or x = π + k · 2π.

Step 3. x-intercept: x = 0+ k · 2π (one-half the distance between the asymptotes and a period apart).

Steps 4–6. x y

−π

21

π

2−1

–3

–1

3

x

y

2��–� 3�

y = cot [ (x + �)]12

range: R

35. Step 1. Dot in y = −1

2sin

(1

2x


–2

–1

2

y

x2�–2�–4� 4�

y = – csc (12

12

x)


Step 3. Draw the four arcs in the form of two U-shaped branches between theasymptotes.

The period is 4π , and the range is | y | ≥ 1

2.

36. Step 1. Dot in y = cos[2(x + π

2

)]as a guide function.

1

3

5

x

y

�4

3�4

5��2

–

y = 3 sec (2x + �)


Step 3. Draw the four arcs in the form of two U-shaped branches between theasymptotes.

The period is π , and the range is | y | ≥ 3.

37. Step 1. Dot in the guide function y = cos(2x).

–2–1

1

3

5

7

x

y

� 2�

y = sec (2x) + 4

Step 2. Plot the asymptote at each x-intercept.

Step 3. Dot in the four arcs in the form of two U-shaped branches between theasymptotes.

Step 4. Since D = 4, shift each branch up 4.

The period is π , and the range is y ≥ 5 or y ≤ 3.



38. Step 1. Dot in the guide function y = 2 sin x.

–3

2

4

x

y

2�–2�

y = 2 csc x – 2


Step 3. Dot in the four arc sections in the form of two U-shaped branchesbetween the asymptotes.

Step 4. Since D = −2, shift each branch down 2.

The period is 2π , and the range is y ≥ 0 or y ≤ −4.

39–42. Many answers are possible.

39. Using the tangent form that has been reflected with period 2π , we get

A = −1, B = 1

2

(since P = 2π = π

|B |)

.

So y = − tan

(1

2x

). Or, using the cotangent form with period 2π shifted left π , we get y = cot

[1

2(x + π)

].

40. Dotting in the guide function, we get y = 3 sin(2x) (since A = 3 and period P = π = 2π

B→ B = 2).

So y = 3 csc (2x).

41. Dotting in the guide function, we get y = 2 cos

(1

4x

)(

since A = 2, P = 8π = 2π

|B | → B = 1

4

).

So y = 2 sec

(1

4x

).

42. Using the cotangent form with period π shifted leftπ

4, we get

y = cot(x + π

4

).

43. y = arcsin

(√2

2

)←→ sin y =

√2

2, where −π

2≤ y ≤ π

2

So y = π

4, or arcsin

√2

2= π

4.

44. y = arccos (−1)←→ cos y = −1, where 0 ≤ y ≤ π

So y = π , or arccos (−1) = π .

45. y = tan−1(−√3)←→ tan y = −√3, where −π

2< y <

π

2

So y = −π

3, or tan−1(−√3) = −π

3.

46. y = csc−1(− 2√

3

)←→ csc y = −2√

3, −π

2≤ y < 0, 0 < y ≤ π

2

←→ sin y = −√

3

2.

So y = −π

3, or csc−1

(− 2√

3

)= −π

3.



47. y = sec−1(−√2

)←→ sec y = −√2, 0 ≤ y <

π

2, or

π

2< y ≤ π

←→ cos y = − 1√2.

So y = 3π

4, or sec−1

(−√2

)= 3π

4.

48. y = cot−1(0)←→ cot y = 0, 0 < y < π

So y = π

2

(y = π

2− tan−1(0) = π

2− 0 = π

2

), or cot−1(0) = π

2.

49. cos(arctan 1) = cos(π

4

)=√

2

250. arcsin

(cos

π

3

)= arcsin

(1

2

)= π

6

51. csc

(cos−1

(−√

2

2

))= csc

(3π

4

)= 1

sin

(3π

4

) = √2

52. cot (sin−1(−1)) = cot(−π

2

)= 0 53. arcsin (sec π) = arcsin (−1) = −π

2

54. cos−1(

sin11π

6

)= cos−1

(−1

2

)= 2π

3

55. sin

(cos−1 5

13

)= sin y, where y = cos−1 5

13. So cos y = 5

13, 0 ≤ y ≤ π , and we want sin y. Use the

Pythagorean identity:(cos y)2 + (sin y)2 = 1(

5

13

)2

+ (sin y)2 = 1

(sin y)2 = 1− 25

169= 144

169

sin y = 12

13, since sin y > 0 for 0 ≤ y ≤ π.

So, sin

(cos−1 5

13

)= 12

13.

56. tan

(sin−1 8

17

)= tan y, where y = sin−1 8

17. So sin y = 8

17,−π

2≤ y ≤ π

2, and we want tan y. Use the

Pythagorean identity:

(cos y)2 + (sin y)2 = 1

(cos y)2 +(

8

17

)2

= 1

(cos y)2 = 1− 64

289= 225

289

cos y = 15


2≤ y ≤ π

2.

So, tan

(sin−1 8

17

)= tan y = sin y

cos y=

(8

17

)(

15

17

) = 8

15.



57–62. Use a calculator in radian mode.

57. cos−1(sin 55) ≈ 3.1195 58. cos(tan−1(−3.76)) ≈ 0.2570

59. cos(csc−1 7)

cos(sin−1(1÷ 7)) ≈ 0.989760. sec−1

(22

7

)

cos−1(7÷ 22) ≈ 1.2470

61. tan−1(sin√

(2.6)) ≈ 0.7850 62. arcsin (cot 2.4π)

sin−1(1÷ tan(2.4π)) ≈ 0.3309

63. y = arcsin x 64. y = arctan x

65. 0.5 = 4 sin(2x)

4 sin(2x) = 0.5

sin(2x) = 0.5


1

4.

2x = sin−1(0.125) Use the definition of arcsine.

x = 1

2sin−1(0.125) Multiply each side by

1

2.

≈ 0.0627 Find the calculator approximation.

66. 3.2 cos

(1

2x

)= 0.9

cos

(1

2x

)= 0.9

3.2Divide each side by 3.2.

1

2x = cos−1

(0.9

3.2

)Use the definition of arccosine.

x = 2 cos−1(

0.9

3.2

)Multiply each side by 2.


67. −6 tan(4x) = 2y

tan(4x) = −2y

6= −y

3Divide each side by − 6.

4x = tan−1(−y

3

)Use the definition of arctangent.

x = 1

4tan−1

(−y

3


1

4.

68. π sin(−4x) = y

sin(−4x) = y

πDivide each side by π.

−4x = sin−1( y

π

)Use the definition of arcsine.

x = −1

4sin−1

( y

π

)Divide each side by − 4.


Chapter 2 Test S–93

69. False; the graph of y = sec x is not sinusoidal, for example.

70. False; the period isπ(1

2

) = 2π .

71. False; for y = 3 sin x, the amplitude is 3 and the period is 2π , whereas y = sin 3x has amplitude 1 and period2π

3.

72. True 73. True (since cos y �= −100)

74. False; amplitude =∣∣∣∣ y max−y min

2

∣∣∣∣.75. False; the graph of y = cos(−x) is the same as y = cos x as a result of the negative identities.

76. True 77. True

(since csc y = 5

3is equivalent to sin y = 3

5

)

78. False; cos−1 x is the inverse function of cos x, whereas1

cos xis the reciprocal of cos x.

79. True 80. False; arccos

(−1

2

)= 2π

3.

81. False; sin−1(−1) = −π

2.

(3π

2is not in the range − π

2≤ y ≤ π

2

)

82. False; there are several equations to represent one graph. (See Exercises 25 and 26.)

83. True 84. False; the guide is y = 4 sin(2x).

85. Jo is correct. If you use the guide Jack suggests, you will not be able to locate the asymptote sincey = cos(2x)+ 4 has no x-intercepts.

Chapter 2 Test, p. 155

1. Rewrite y = 5 sin

(−1

2x

)as y = −5 sin

(1

2x

).

–5

–3

–1

6

x

y

4��

y = 5 sin (– x)12

period:2π(1

2

) = 4π , amplitude: |A | = 5, x-intercept: x = k · 2π

2. amplitude = 1

–1

1

x

y

�2

�2

�–� –

y = cos [2(x – )]�2

period:2π

2= π , phase shift: right

π

2, x-intercepts: x = π

4+ k · π

2



3. period:π(1

2

) = 2π , asymptotes: x =π

2+ k · π(

1

2

) = π + k · 2π , range: R

1

3

x

y

3�2��–�

y = tan ( x)12

4. period: 2π , asymptotes: x = π

2+ k · π , range: | y | ≥ 3

3

x

y

2�–2� �–�

y = 3 sec x

�2

(Use y = 3 cos x as a guide.)

5. amplitude = 2

–2 –1 1 2

123

x

y

y = 2 sin (�x) + 2

period: = 2π

π= 2, x-intercepts: x = 3

2+ k · 2, range: 0 ≤ y ≤ 4

6. range: | y | ≥ 5, asymptotes: x = π

2+ kπ

1

3

5

7

x

y

2��–2� –�

)y = 5 csc (x – �2

�2

(Use y = 5 sin(x − π

2

)as a guide.)

7. There are many possible answers.

7. a. Using the reflected form of cosine, with period 2π and amplitude 2, we get

y = −2 cos x.

b. Using the tangent form with periodπ

2, we get

y = tan(2x).



c. Dotting in the guide, we get y = 4 cos

(1

2x

)(since the amplitude is 4 and the period is 4π ).

So, y = 4 sec

(1

2x

).

d. The graph is the sine form, with amplitude 1 and period 2π , with a shift leftπ

4.

So, y = sin(x + π

4

).

8. Using the same amplitude and period, we consider phase shifts of the cosine or sine form. If we do so, a fewpossible answers are:

y = 2 sin(x − π

2

), y = 2 cos(x − π)

y = 2 sin

(x − 3π

2

), y = −2 cos

(x + π

2

)(Check other answers with a graphing utility.)

9. a. amplitude: 5, period:2π(π

2

) = 4

–5

–3

–1

1

3

5

x

yy = 5 cos ( x)�

2

2 4 6 8

b. 5 cm c. See the graph in part (a).

10. Make sure your answers are in the range of the respective inverse function.

10. a. arccos1

2= π

3

(since cos

π

3= 1

2

)b. arctan 0 = 0 (since tan 0 = 0)

c. sin−1

√2

2= π

4

(since sin

π

4=√

2

2

)d. arccsc 2→ arcsin

1

2= π

6

(since sin

π

6= 1

2

)

e. sec−1(−√2

)→ cos−1

(− 1√

2

)= 3π

4

(since cos

3π

4= − 1√

2

)

f. arctan(−1) = −π

4

(since tan

(−π

4

)= −1

)g. arccos(−1) = π (since cos π = −1)

h. cot−1(

1√3

)= π

3

(since cot

π

3= 1√

3

)or

π

2− tan−1

(1√3

)= π

3

(since tan

π

6= 1√

3

)



i. sin

(arccos

3

5

)= sin y, where arccos

3

5= y for 0 ≤ y ≤ π .


5. Use the Pythagorean identity:

(cos y)2 + (sin y)2 = 1(3

5

)2

+ (sin y)2 = 1

(sin y)2 = 1− 9

25= 16

25

sin y = 4

5, since y > 0 for 0 ≤ y ≤ π.

Therefore, sin

(arccos

3

5

)= sin y = 4

5.

j. sec−1(cos π) = sec−1(−1)

= cos−1(−1) = π (since cos π = −1)

11. a. tan−1 13.4

tan−1(13.4) ≈ 1.4963

b. arccos (−0.89)

cos−1(−0.89) ≈ 2.6681

c. sec−1(−13

7

)cos−1(−7÷ 13) ≈ 2.1394

d. sin−1 0.56

sin−1(0.56) ≈ 0.5944

e. tan(arctan(−3.67))

tan(tan−1(−3.67)) = −3.6700

f. cos(arccot 4.65)

cos(π

2− tan−1(4.65)

)≈ 0.9776

12. a. 6 sin 8x = 3

sin 8x = 1


1

6.

8x = sin−1(

1

2

)Use the definition of arcsine.

8x = π

6sin

π

6= 1

2

x = π


1

8.


b. 4 tan 2x = y

tan 2x = y


1

4.

2x = tan−1(y

4

)Use the definition of arctangent.

x = 1

2tan−1

(y

4


1

2.



13. a. y = tan−1 x b. y = cos−1 x

14. a. False; the graph of y = tan x is not a sine wave, for example.

b. True c. True

d. False; the graph of y = A tan(−Bx) is the same as y = −A tan(Bx).

e. False; the graph of y = sin(2x) has amplitude 1 and period π , whereas y = 2 sin x has amplitude 2 andperiod 2π .

f. True g. True

h. False; the graph of y = sin(x + π

2

)shifts y = sin x left

π

2, while y = sin x + sin

π

2= sin x + 1 shifts the

graph of y = sin x up 1.

i. True (sin x �= 25.6)

15. a. The minimum y-coordinate (number of daylight hours) appears to be 6.This occurs between x = 2 and x = 3, or in February.

b. From April to December there will be at least 8 hours of daylight.c. The maximum y-coordinate appears to be 18. This occurs between x = 8 and x = 9, or during August.

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