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May 4, 2004 17:13 K95-ch02 Sheet number 1 Page number 52 black
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. Since A = 4, we have a vertical stretch.
–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
May 4, 2004 17:13 K95-ch02 Sheet number 2 Page number 53 black
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. Since A = 7, we have a vertical stretch.
–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.
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 | ≤ 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 :
• The x-intercepts are the same as y = sin x.• The maximum and minimum y-coordinates of the critical points
are 2.5 and −2.5, 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 | ≤ 2.5, x-intercepts: −2π,−π, 0, π, 2π or x = kπ
May 4, 2004 17:13 K95-ch02 Sheet number 3 Page number 54 black
S–54 Chapter 2 Trigonometry Instructor’s Solutions Manual
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 :
• The x-intercepts are the same as y = cos x.• The maximum and minimum y-coordinates of the critical points
are 0.4 and −0.4, 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 | ≤ 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 :
• The x-intercepts are the same as y = cos x.• The maximum and minimum y-coordinates of the critical points
are7
5and −7
5, 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 | ≤ 7
5, x-intercepts: −3π
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 :
• The x-intercepts are the same as y = cos x.• The maximum and minimum y-coordinates of the critical points
are3
2and −3
2, 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 | ≤ 3
2, x-intercepts: −3π
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 :
• The x-intercepts are the same as y = sin x.• The maximum and minimum y-coordinates of the critical points
are 0.9 and −0.9, respectively.
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
May 4, 2004 17:13 K95-ch02 Sheet number 4 Page number 55 black
Exercise Set 2.1 S–55
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))
May 4, 2004 17:13 K95-ch02 Sheet number 5 Page number 56 black
S–56 Chapter 2 Trigonometry Instructor’s Solutions Manual
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.
May 4, 2004 17:13 K95-ch02 Sheet number 6 Page number 57 black
Exercise Set 2.2 S–57
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.
Exercise Set 2.2, p. 104
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.
May 4, 2004 17:13 K95-ch02 Sheet number 7 Page number 58 black
S–58 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
May 4, 2004 17:13 K95-ch02 Sheet number 8 Page number 59 black
Exercise Set 2.2 S–59
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
May 4, 2004 17:13 K95-ch02 Sheet number 9 Page number 60 black
S–60 Chapter 2 Trigonometry Instructor’s Solutions Manual
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 – �)
16. amplitude: 5period: 2π
}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
17. amplitude: 4period: 4π
}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)
May 4, 2004 17:13 K95-ch02 Sheet number 10 Page number 61 black
Exercise Set 2.2 S–61
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
22. amplitude: 1period: 2π
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
).
May 4, 2004 17:13 K95-ch02 Sheet number 11 Page number 62 black
S–62 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
).
May 4, 2004 17:13 K95-ch02 Sheet number 12 Page number 63 black
Exercise Set 2.2 S–63
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
May 4, 2004 17:13 K95-ch02 Sheet number 13 Page number 64 black
S–64 Chapter 2 Trigonometry Instructor’s Solutions Manual
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]
Exercise Set 2.3, p. 115
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
3. a. maximum = 74.5, minimum = 58.5 b. amplitude =∣∣∣∣ 74.5− 58.5
2
∣∣∣∣ = 8
4. a. maximum = 68.4, minimum = 25.6 b. amplitude =∣∣∣∣ 68.4− 25.6
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
May 4, 2004 17:13 K95-ch02 Sheet number 14 Page number 65 black
Exercise Set 2.3 S–65
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
With period 12, each arc occurs in a length of12
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.
May 4, 2004 17:13 K95-ch02 Sheet number 15 Page number 66 black
S–66 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
With period 12, each arc occurs in a length of12
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
With period 12, each arc occurs in a length of12
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
May 4, 2004 17:13 K95-ch02 Sheet number 16 Page number 67 black
Exercise Set 2.4 S–67
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).
Exercise Set 2.4, p. 129
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�
May 4, 2004 17:13 K95-ch02 Sheet number 17 Page number 68 black
S–68 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
May 4, 2004 17:13 K95-ch02 Sheet number 18 Page number 69 black
Exercise Set 2.4 S–69
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
)
May 4, 2004 17:13 K95-ch02 Sheet number 19 Page number 70 black
S–70 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
)
May 4, 2004 17:13 K95-ch02 Sheet number 20 Page number 71 black
Exercise Set 2.4 S–71
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)
May 4, 2004 17:13 K95-ch02 Sheet number 21 Page number 72 black
S–72 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
May 4, 2004 17:13 K95-ch02 Sheet number 22 Page number 73 black
Exercise Set 2.4 S–73
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
)as the guide function.
–4
–1
1
4
x
y
�–� 3�
y = 4 sec(x + π
2
)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 = 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
)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
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.
May 4, 2004 17:13 K95-ch02 Sheet number 23 Page number 74 black
S–74 Chapter 2 Trigonometry Instructor’s Solutions Manual
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.
May 4, 2004 17:13 K95-ch02 Sheet number 24 Page number 75 black
Exercise Set 2.5 S–75
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)
Exercise Set 2.5, p. 143
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.
May 4, 2004 17:13 K95-ch02 Sheet number 25 Page number 76 black
S–76 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
May 4, 2004 17:13 K95-ch02 Sheet number 26 Page number 77 black
Exercise Set 2.5 S–77
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
May 4, 2004 17:13 K95-ch02 Sheet number 27 Page number 78 black
S–78 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
May 4, 2004 17:13 K95-ch02 Sheet number 28 Page number 79 black
Exercise Set 2.5 S–79
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
5. Use the Pythagorean identity.
(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
May 4, 2004 17:13 K95-ch02 Sheet number 29 Page number 80 black
S–80 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
17. Use the Pythagorean identity.
(cos y)2 +(
8
17
)2
= 1
(cos y)2 = 1− 64
289= 225
289
cos y = 15
17, since cos y > 0 for − π
2≤ y ≤ π
2
60. sin
(arccos
7
25
)Let y = arccos
7
25←→ cos y = 7
25, 0 ≤ y ≤ π
So we want sin y, where cos y = 7
25. Use the Pythagorean identity.
(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
May 4, 2004 17:13 K95-ch02 Sheet number 30 Page number 81 black
Exercise Set 2.5 S–81
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
2Multiply each side by
1
4.
3x = cos−1
(√3
2
)Use the definition of inverse cosine function.
3x = π
6cos−1
(√3
2
)= π
6
x = π
18Multiply each side by
1
3.
75. 4 tan(9x) = 7
tan(9x) = 7
4Multiply each side by
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.
May 4, 2004 17:13 K95-ch02 Sheet number 31 Page number 82 black
S–82 Chapter 2 Trigonometry Instructor’s Solutions Manual
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)
May 4, 2004 17:13 K95-ch02 Sheet number 32 Page number 83 black
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)
May 4, 2004 17:13 K95-ch02 Sheet number 33 Page number 84 black
S–84 Chapter 2 Trigonometry Instructor’s Solutions Manual
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�
May 4, 2004 17:13 K95-ch02 Sheet number 34 Page number 85 black
Chapter 2 Review Exercises S–85
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
)]
May 4, 2004 17:13 K95-ch02 Sheet number 35 Page number 86 black
S–86 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
May 4, 2004 17:13 K95-ch02 Sheet number 36 Page number 87 black
Chapter 2 Review Exercises S–87
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
May 4, 2004 17:13 K95-ch02 Sheet number 37 Page number 88 black
S–88 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
May 4, 2004 17:13 K95-ch02 Sheet number 38 Page number 89 black
Chapter 2 Review Exercises S–89
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
)as the guide function.
–2
–1
2
y
x2�–2�–4� 4�
y = – csc (12
12
x)
Step 2. Plot the asymptotes at each x-intercept.
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 2. Plot the asymptotes at each x-intercept.
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.
May 4, 2004 17:13 K95-ch02 Sheet number 39 Page number 90 black
S–90 Chapter 2 Trigonometry Instructor’s Solutions Manual
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 2. Plot the asymptotes at each x-intercept.
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.
May 4, 2004 17:13 K95-ch02 Sheet number 40 Page number 91 black
Chapter 2 Review Exercises S–91
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
17, since cos y > 0 for − π
2≤ y ≤ π
2.
So, tan
(sin−1 8
17
)= tan y = sin y
cos y=
(8
17
)(
15
17
) = 8
15.
May 4, 2004 17:13 K95-ch02 Sheet number 41 Page number 92 black
S–92 Chapter 2 Trigonometry Instructor’s Solutions Manual
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
4Multiply each side by
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.
≈ 2.5714 Find the calculator approximation.
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
)Multiply each side by
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.
May 4, 2004 17:13 K95-ch02 Sheet number 42 Page number 93 black
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
May 4, 2004 17:13 K95-ch02 Sheet number 43 Page number 94 black
S–94 Chapter 2 Trigonometry Instructor’s Solutions Manual
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).
May 4, 2004 17:13 K95-ch02 Sheet number 44 Page number 95 black
Chapter 2 Test S–95
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
)
May 4, 2004 17:13 K95-ch02 Sheet number 45 Page number 96 black
S–96 Chapter 2 Trigonometry Instructor’s Solutions Manual
i. sin
(arccos
3
5
)= sin y, where arccos
3
5= y for 0 ≤ y ≤ π .
So we want sin y, where cos y = 3
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
2Multiply each side by
1
6.
8x = sin−1(
1
2
)Use the definition of arcsine.
8x = π
6sin
π
6= 1
2
x = π
48Multiply each side by
1
8.
≈ 0.0654 Find the calculator approximation.
b. 4 tan 2x = y
tan 2x = y
4Multiply each side by
1
4.
2x = tan−1(y
4
)Use the definition of arctangent.
x = 1
2tan−1
(y
4
)Multiply each side by
1
2.
May 4, 2004 17:13 K95-ch02 Sheet number 46 Page number 97 black
Chapter 2 Test S–97
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.