Hungerford T.W., Shaw D.J. Contemporary Precalculus: A Graphing Approach
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69.
1
cos
s
3
i
x
n x
cos
3
1
x
(1
s
in
s
2
i
x
n x)
cos
3
x(
c
1
os
2
x
sin x)
cos x(1 sin x)
1
sec
si
x
n x
71.
1
cos
si
x
n x
cos
1
x
(1
s
in
2
si
x
n x)
cos x
c
(1
os
2
x
sin x)
1
cos
si
x
n x
co
1
s x
c
s
o
in
s
x
x
sec x tan x
73. Use Strategy 6: Prove that (cos x cot x)(cos x cot x)
(cot x cos x)(cot x cos x).
(cot x cos x)(cot x cos x) cot
2
x cos
2
x
c
s
o
in
s
2
2
x
x
cos
2
x
cos
2
x(
s
1
in
2
x
sin
2
x)
c
s
o
in
s
2
2
x
x
(1 sin
2
x) cot
2
x cos
2
x
(cos x cot x)(cos x cot x)
Therefore,
co
c
t
o
x
s x
c
c
o
o
t
s
x
x
co
c
t
o
x
s x
c
c
o
o
t
s
x
x
75. cot x
ta
1
n x
, so log
10
(cot x) log
10
ta
1
n x
log
10
(tan x)
77. csc x cot x (csc x cot x)
(
(
c
c
s
s
c
c
x
x
c
c
o
o
t
t
x
x
)
)
c
c
s
s
c
c
2
x
x
c
c
o
o
t
t
2
x
x
csc x
1
cot x
;
so, log
10
(csc x cot x) log
10
csc x
1
cot x
log
10
(csc x cot x)
79. tan x tan y(cot x cot y)
tan y(tan x cot x) tan x(tan y cot y)
tan y tan x tan x tan y
81.
c
c
o
o
s
s
x
y
s
s
i
i
n
n
y
x
c
c
o
o
s
s
x
x
s
s
i
i
n
n
y
y
c
c
o
o
s
s
y
x
s
s
i
i
n
n
x
y
Section 7.2, page 530
1.
2
4
6
3.
2
4
6
5. 2 3
7. 2 3
9.
2
4
6
11.
6
4
2
13. cos x 15. sin x 17.
c
os
1
x
(cos y sin x)(cos y sin x)
(cos y sin x)(cos x sin y)
cos
2
y sin
2
x
(cos y sin x)(cos x sin y)
(1 sin
2
x) (1 cos
2
y)
(cos y sin x)(cos x sin y)
cos
2
x sin
2
y
(cos y sin x)(cos x sin y)
cos
2
x cos
2
x sin
2
x
sin
2
x
cos x(1 sin x)
(1 sin x)(1 sin x)
cos
3
x(1 sin x)
(1 sin x)(1 sin x)
1016 ANSWERS
19. sin 2 21. cos x 23. 2 sin x sin y
25.
4
6
2
27.
3
1
0
26
29.
4
1
3
0
3
31.
4
1
3
0
3
33. From the figure, sin t
4
5
and cos t
3
5
. So,
5 sin(x t) 5(sin x cos t cos x sin t)
sin x
5 cos t cos x
5 sin t sin x
5
3
5
cos x
5
4
5
3 sin x 4 cos x
35.
f (x h
h
) f (x)
cos(x h
h
) cos x
cos x
cos h
h
1
sin x
sin
h
h
37. sin(x y) sin(x (y)) sin x cos(y)
cos x sin(y) sin x cos y cos x (sin y)
sin x cos y cos x sin y
39.
c
c
o
o
s
s
(
x
x
c
os
y
y
)
c
c
o
o
s
s
x
x
c
c
o
o
s
s
y
y
c
s
o
in
s
x
x
s
c
i
o
n
s
y
y
1 tan x tan y
41.
s
s
i
i
n
n
(
x
x
s
in
y
y
)
s
s
i
i
n
n
x
x
c
s
o
in
s
y
y
c
s
o
in
s
x
x
s
s
i
i
n
n
y
y
cot y cot x
43.
s
s
i
i
n
n
(
x
x
c
os
y
y
)
s
s
i
i
n
n
x
x
c
c
o
o
s
s
y
y
c
si
o
n
s
x
x
c
s
o
in
s
y
y
1 cot x tan y
45. sin(x y) 44/125; tan(x y) 44/117; x y lies in
quadrant III.
47. cos(x y)
5
6
6
5
; tan(x y)
3
5
3
6
; x y lies in quad-
rant III.
49. sin u cos v cos w cos u sin v cos w
cos u cos v sin w sin u sin v sin w
51. If x y
p
2
, then y
p
2
x and sin y sin
p
2
x
cos x. Thus, sin
2
x sin
2
y sin
2
x cos
2
x 1.
53. sin(x p) sin x cos p cos x sin p sin x(1)
cos x(0) sin x
55. cos(p x) cos p cos x sin p sin x (1)cos x
(0)sin x cos x
57. sin(x p) sin x cos p cos x sin p sin x
(1)
cos x
(0) sin x
sin x cos y cos x sin y
sin x cos y
sin x cos y cos x sin y
sin x sin y
cos x cos y sin x sin y
cos x cos y
cos x(cos h 1) sin x sin h
h
cos x cos h sin x sin h cos x
h
59. tan(x p)
1
ta
n
t
x
a
n x
ta
ta
n
n
p
p
1
t
an
ta
x
n
x
0
0
tan x
61.
1
2
[cos(x y) cos(x y)]
1
2
[cos x cos y sin x sin y (cos x cos y sin x sin y)]
1
2
2 sin x sin y sin x sin y
63. cos(x y)cos(x y)
(cos x cos y sin x sin y)(cos x cos y sin x sin y)
cos
2
x cos
2
y sin
2
x sin
2
y
65.
c
si
o
n
s(
x
x
c
os
y
y
)
c
s
o
in
s
x
x
c
s
o
in
s
y
y
cot x tan y
67. This is not an identity.
69.
s
s
i
i
n
n
(
(
x
x
y
y
)
)
1
1
/
/
c
c
o
o
s
s
x
x
c
c
o
o
s
s
y
y
t
t
a
a
n
n
x
x
t
t
a
a
n
n
y
y
71. Not an identity 73. Not an identity
Special Topics 7.2.A, page 535
1. 3/4 3. 4/5 5. Undefined
7. 1.37 or 1.77 9. p/4 or 3p/4 11. 1.39 or 1.75
13. .142 or 3.000
Section 7.3, page 542
1. sin 2x
1
1
2
6
0
9
, cos 2x
1
1
1
6
9
9
, tan 2x
1
1
2
1
0
9
3. sin 2x
2
2
4
5
, cos 2x
2
7
5
, tan 2x
2
7
4
5. sin 2x
2
2
4
5
, cos 2x
2
7
5
, tan 2x
2
7
4
7. sin 2x
8
15
, cos 2x 7/8, tan 2x
7
15
9. (a) Length 2x, height y, area 2xy
(b) x 3 cos t, y 3 sin t
(c) A 2xy
A 2(3 cos t)(3 sin t)
A 9(2 sin t cos t)
A 9 sin 2t
11.
2
2
2
13.
2
2
2
15. 2 3
17.
2
2
3
19.
2
2
2
21. 1 2
c
s
o
in
s
x
x
c
s
o
in
s
y
y
c
s
o
in
s
x
x
c
s
o
in
s
y
y
sin x cos y cos x sin y
sin x cos y cos x sin y
cos x cos y sin x sin y
sin x cos y
ANSWERS 1017
23. 25.
27.
sin
2
x
1
3
0
0
, cos
2
x
1
7
0
0
, tan
2
x
3
7
29. sin
2
x
1
10
, cos
2
x
3
10
, tan
2
x
1
3
31. sin
2
x
50
10
2
05
, cos
2
x
50
10
2
05
,
tan
2
x
5
2
33.
1
2
sin 10x
1
2
sin 2x 35.
1
2
cos 6x
1
2
cos 2x
37.
1
2
cos 20x
1
2
cos 14x 39. 2 sin 4x cos x
41. 2 cos 7x sin 2x
43. 2 cos
7
2
x
cos
2
3x
2 cos
7
2
x
cos
3
2
x
45. .96 47. .28
49. .316 51. 4 cos
3
x 3 cos x
53. cos x 55. sin 4y 57. 1
59. Use the fourth product to sum identity with
1
2
(x y) in
place of x and
1
2
(x y) in place of y.
2 cos
1
2
(x y)
sin
1
2
(x y)
2
1
2
sin
1
2
(x y)
1
2
(x y)
sin
1
2
(x y)
1
2
(x y)
sin x sin y
61. Use the second product to sum identity with
1
2
(x y) in
place of x and
1
2
(x y) in place of y.
2 sin
1
2
(x y)
sin
1
2
(x y)
2
1
2
cos
1
2
(x y)
1
2
(x y)
cos
1
2
(x y)
1
2
(x y)
(cos y cos x)
cos x cos y
63. sin 16x sin 2(8x) 2 sin 8x cos 8x
65. cos
4
x sin
4
x (cos
2
x sin
2
x)(cos
2
x sin
2
x)
1(cos
2
x sin
2
x) cos 2x
67. This is not an identity.
69.
1
sin
co
2
s
x
2x
1
2s
2
in
c
x
os
c
2
o
x
s
x
1
2
2
c
s
o
in
s
x
x
c
c
o
o
s
s
x
x
c
s
o
in
s
x
x
cot x
2
2
3
2
2
2
2
2
71. sin 3x sin(2x x) sin 2x cos x cos 2x sin x
2 sin x cos x cos x (1 2 sin
2
x)sin x 2 sin x cos
2
x
sin x 2 sin
3
x 2 sin x(1 sin
2
x) sin x 2 sin
3
x
sin x(2 2 sin
2
x 1 2 sin
2
x) sin x(3 4 sin
2
x)
73. This is not an identity.
75. csc
2
2
x
1
2
cos x
77.
sin
2
5x
co
s
s
4
i
x
n3x
sin x Proof:
sin
2
5x
co
s
s
4
i
x
n3x
2c
2
os
co
4
s
x
4
s
x
in x
sin x.
79.
cos
2
x
c
os
c
2
o
x
s3x
cos x Proof:
cos
2
x
c
os
c
2
o
x
s3x
2 cos
2
2
c
x
o
c
s
o
2
s
x
(x)
cos (x) cos x
81.
c
s
o
in
s
3
x
x
co
si
s
n
3
x
x
tan x Proof:
c
s
o
in
s
3
x
x
co
si
s
n
3
x
x
co
s
s
in
(
x
x)
c
s
o
in
s
x
x
tan x
83.
c
s
o
in
s
x
x
s
c
i
o
n
s
3
3
x
x
c
s
o
in
s
(
(
x
x
)
)
co
si
s
n
x
x
tan x
85.
c
s
o
in
s
4
4
x
x
s
c
i
o
n
s
6
6
x
x
2
2
s
s
i
i
n
n
5
5
x
x
c
s
o
in
s
x
x
c
s
o
in
s
x
x
cot x
87.
c
s
o
in
s
x
x
s
c
i
o
n
s
y
y
cot
x
2
y
89. (a) Use Strategy 6: Prove that (1 cos x)(1 cos x)
sin x sin x.
(1 cos x)(1 cos x) 1 cos
2
x sin
2
x sin x sin x
Therefore
1
sin
co
x
s x
1
sin
co
x
s x
(b) On page 539, we proved that tan
2
x
1
sin
co
x
s x
.
Hence, by part (a) tan
2
x
1
sin
co
x
s x
1
sin
co
x
s x
.
cos
x
2
y
sin
x
2
y
2 sin
x
2
y
cos
x
2
y
2 sin
x
2
y
sin
x
2
y
2 cos
x
2
3x
sin
x
2
3x
2 cos
x
2
3x
cos
x
2
3x
2 cos (2x) sin x
2 cos (2x) cos (x)
2 cos
8
2
x
sin
2
2
x
2 cos 4x
1
1
2
cos x
1
1
2
co
s x
2
1
sin
2
2
x
1018 ANSWERS
91.
Section 7.4, page 553
1. p/2 3. p/4 5. 0
7. p/6 9. p/4 11. p/3
13. 2p/3 15. .3576 17. 1.2728
19. .7168 21. .8584 23. 2.2168
25. cos u 1/2; tan u 3
27. p/2 29. 5p/6 31. p/3
33. p/3 35. p/6 37. 4/5
39. 4/5 41. 2.4 43. 22/5
45. 2/13 47. 1/21
49. cos(sin
1
v)
1 v
2
(1 v 1)
51. tan(sin
1
v) (1 v 1)
53. cos(tan
1
v)
55. sin(2 cos
1
v) 2v
1 v
2
(1 v 1)
57. tan(cos
1
v) tan
p
2
sin
1
v
cot(sin
1
v)
59.
61.
63. (a) t
2p
1
f
sin
1
V
m
V
ax
n
f
(n 0, 1, 2, . . .)
(b) t 5.8219 10
4
sec
65. (a) u sin
1
(40/x)
(b) u .16 radians or 9.2°
y
x
−2π 2π
−1
−2
−2
−1
−π π
y
x
f
(x) = cos
−1
(x + 1)
−2 −1
π
1
1 v
2
v
1 v
2
2
2
2
2
2
67. (a) t tan
1
h
4
(b) t .0624 radians; t .2450 radians; t .4636 radians
(c) about 1.6912 mi
69. (a) t tan
1
3
x
0
tan
1
6
x
(b) about 13.4 ft
71. (a) u tan
1
2
x
5
tan
1
1
x
0
(b) 15.8 ft
73. No horizontal line intersects the graph of f (x) csc x more
than once when p/2 x p/2 (see Figure 6–85). Hence,
the restricted cosecant function has an inverse function, as
explained in Section 3.7.
75. Let cos u w with 0 u p. Then u cos
1
w and
cos
1
(cos u) cos
1
w u. Let u cos
1
v. Then
cos u v and cos(cos
1
v) cos u v.
77. Let u sin
1
(x). Then
x sin u (p/2 u p/2)
x sin u
x sin(u)
sin
1
x u (p/2 u p/2)
sin
1
x sin
1
(x)
sin
1
(x) sin
1
x
79. Let u cos
1
(x). Then x cos u. Note that
0 u p; hence, p u 0, which implies that
0 p u p. Hence,
x cos u
x cos(p u) (from the stated identity)
p u cos
1
x
u p cos
1
x
cos
1
(x) p cos
1
x
81. Let u cot x. Then
tan(p/2 x) cot x u (0 u p,
hence, p/2 p/2 u p/2)
p/2 x tan
1
(u) tan
1
(cot x)
83. Let u sin
1
x. Then sin u x, (p/2 u p/2)
tan u
c
s
o
in
s
u
u
x
1 x
2
sin u
1
si
n
2
u
123−1−2−3
2
π
2
π
−
y
x
y = csc
−1
x
⏐x⏐≥ 1
ANSWERS 1019
(positive square root, since u lies in quadrant I or IV)
u tan
1
sin
1
x tan
1
85. The statement is false. For example, let x 1/2.
tan
1
1
2
.4636, but
c
s
o
in
s
1
1
(
(
1
1
/
/
2
2
)
)
p
p
/
/
6
3
1
2
Section 7.5, page 564
In the answers for this section k 0, 1, 2, 3, . . .
1. x .4836 2kp or 2.6580 2kp
3. x 2.1700 2kp or 2.1700 2kp
5. x .3402 kp 7. x .4101 kp
9. x 2.2459 2kp 11. x .1193 or 3.0223
13. x 1.3258 or 4.4674
15. x
p
3
2kp or
2
3
p
2kp
17. x
p
3
kp
19. x 5p/6 2kp or 5p/6 2kp
21. x
p
6
2kp or
7
6
p
2kp
23. x
p
6
2kp or
5
6
p
2kp
25. u 82.8° or 262.8° 27. u 114.8° or 245.2°
29. u 210°, 270°, 330° 31. u 60°, 120°, 240°, 300°
33. u 120°, 240° 35. a 65.4°
37. a 30° 39. u 27.6°
41. u 14.2°
43. x
p
6
kp or
2
3
p
kp
45. x
p
2
4kp 47. x
p
9
k
3
p
49. x .7381
2k
3
p
51. x 2.2143 2kp
53. t 1.8546; A 11.18 55. t 1.6961; A .35
57. t 1.9346, L 16.47 59. t 1.6236, L 11.61
61. (a) A 2x(2 cos 2x) 4x cos 2x
(b) x .3050 and x .5490
63. x 5.9433 and 3.4814
65. x
3
4
p
,
7
4
p
, 1.9513, 5.0929
67. x p/2, 3p/2, p/4, 5p/4
69. x p/2, 3p/2, p/6, 5p/6
71. x .8481, 2.2935, 1.7682, 4.9098
x
1 x
2
x
1 x
2
73. x .8213, 2.3203.
75. x 1.2059, 4.3475, .3649, and 3.5065
77. x 1.0591, 4.2007, 2.8679, and 6.0095
79. No solution
81. x p/2, 3p/2, 7p/6, 11p/6
83. x p/6, 5p/6
85. x 7p/6, 11p/6, p/2
87. x 0, p/3, 5p/3
89. x .5275 kp or x 1.6868 kp
91. x .4959, 1.2538, 1.5708, 1.8877, 2.6457, or 4.7124.
Since the function is periodic with period 2p, all solutions
are given by each of these six roots plus 2kp.
93. x .1671, 1.8256, 2.8867, or 4.5453. Since the function is
periodic with period 2p, all solutions are given by each of
these four roots plus 2kp.
95. x 1.2161 2kp or x 5.0671 2kp
97. x 2.4620 2kp or x 3.8212 2kp
99. x .5166 2kp or x 5.6766 2kp
101. (a) days 60 and 282 (March 2 and October 10)
(b) day 171 (June 21)
103. u .5475 or 1.0233 105. 13.25° or 76.75°
107. The basic equations sin x c and cos x c have no solu-
tions if c 1 or c 1.
109. The solutions x 0 and x p are missed due to dividing
by sin x (which is 0 when x 0 or p).
Chapter 7 Review, page 568
1. 1/3 cot t 3. sin
4
x
5. sin
4
t cos
4
t (sin
2
t cos
2
t)(sin
2
t cos
2
t)
1(sin
2
t (1 sin
2
t)) sin
2
t 1
sin
2
t 2 sin
2
t 1
7. Use Strategy 6. Prove that sin t sin t (1 cos t)(1 cos t)
is an identity.
sin t sin t sin
2
t 1 cos
2
t (1 cos t)(1 cos t)
Therefore,
1
sin
co
t
s t
1
sin
co
t
s t
.
9. This is not an identity.
11. (sin x cos x)
2
sin 2x sin
2
x 2 sin x cos x
cos
2
x sin 2x sin
2
x cos
2
x 2 sin x cos x
2 sin x cos x 1 0 1
13.
tan
2
x
t
an
s
x
in x
sin
2
x
(s
/
i
c
n
o
x
s
/
x
c
os
s
x
i
)
n x
sin x
2s
s
i
i
n
n
x
x cos x
sin x(
2
1
s
in
c
x
os x)
1
2
cos x
sin
2
(x/2)
sin x
co
si
s
n
x
x cos x
2
co
si
s
n
x
x
1020 ANSWERS
15. cos(x y)cos(x y)
(cos x cos y sin x sin y)(cos x cos y sin x sin y)
cos x
2
cos
2
y sin
2
x sin
2
y
cos
2
x (1 sin
2
y) (1 cos
2
x)sin
2
y
cos
2
x cos
2
x sin
2
y sin
2
y cos
2
x sin
2
y
cos
2
x sin
2
y
17. Use Strategy 6. Prove that (sec x 1)(sec x 1)
tan x tan x is an identity.
(sec x 1)(sec x 1) sec
2
x 1 tan
2
x tan x tan x
Therefore,
sec
ta
x
n
x
1
sec
ta
x
n
x
1
is an identity.
19.
1
tan
ta
2
x
n
2
x
tan
1
2
x
t
t
a
a
n
n
2
2
x
x
cot
2
x 1 csc
2
x
21. tan
2
x sec
2
x sec
2
x 1 sec
2
x 1
csc
2
x 1 csc
2
x cot
2
x csc
2
x
23. 120/169 25.
5
6
6
5
27.
45
8
1
or
35
8
1
29. Yes. If sin x 0, sin 2x 2 sin x cos x
2
0
cos x 0.
31. Using the half-angle identity,
cos
1
p
2
cos
1
2
p
6
1 co
2
s(p/6
)
1
2
3
/2
2
4
3
2
2
3
Using the subtraction identity for cosine,
cos
1
p
2
cos
p
3
p
4
cos
p
3
cos
p
4
sin
p
3
sin
p
4
1
2
2
2
2
3
2
2
2
4
6
Hence,
2
2
3
2
4
6
and 2
3
2
2
6
.
33.
6
4
2
35. (a) 37. .96
39. u 2.24076 kp (k 0, 1, 2, . . .)
41. p/4 43. p/4 45. p/6
47. 5p/6 49. .75 51. p/3
53.
−
−
−π
2
π
2
3π
y
x
55. 15/4
In the following answers, k 0, 1, 2, 3, . . . .
57. x
p
6
2kp or
5
6
p
2kp
59. x
p
4
2kp 61. x p/4 kp
63. x
p
9
2k
3
p
or
4
9
p
2k
3
p
65. x .4115 2kp or 2.7301 2kp
67. x 1.5042 kp
69. x
p
6
2kp or
5
6
p
2kp
71. x
p
6
2kp or
7
6
p
2kp
73. x
p
3
kp
75. x .8959 2kp or 2.2457 2kp 77. x kp
79. x .8419, 2.2997, 4.1784, or 5.2463. Since the function is
periodic with period 2p, all solutions are given by x
.8419 2kp, x 2.2997 2kp, and so on.
81. u 314.5° and 225.5° 83. u 9.06° or 80.94°
Chapter 7 Test, page 569
1. 1 sin x cos x tan x 1 sin x cos x
c
s
o
in
s
x
x
1 sin
2
x cos
2
x.
2.
6
4
2
3.
4. cos
2
x cot
2
x cos
2
x
c
s
o
in
s
2
2
x
x
cos
2
x (
s
s
i
i
n
n
2
2
x
x 1)
c
s
o
in
s
2
2
x
x
(sin
2
x 1) cot
2
x (cos
2
x)
cos
2
x cot
2
x.
5. cos y
6. (a)
1
1
5
6
1
8
9
1
(b)
1
7
6
2
8
0
1
(c)
1
7
5
2
1
0
9
7. (sin
2
x 1) (cot
2
x 1) sin
2
x cot
2
x sin
2
x
cot
2
x 1 sin
2
x
c
s
o
in
s
2
2
x
x
sin
2
x cot
2
x 1
cos
2
x sin
2
x cot
2
x 1 1 cot
2
x 1 cot
2
x.
8.
3
8
15
9. cos x
10. Several possible proofs, including:
1
co
t x
ta
n
1
x
co
1
t x
c
s
o
in
s
x
x
tan x.
1
c
s
o
in
s
x
x
co
c
t
o
x
t
x
1
cot x 1
1
co
1
t x
cot x 1
cos
2
x sin
2
x cos
2
x
sin
2
x
2
1
2
1
ANSWERS 1021
Alternatively, use Strategy 6 of Section 7.1 to prove the
equivalent identity
(cot x 1) tan x 1 tan x
as follows:
(cot x 1) tan x cot x tan x tan x
c
s
o
in
s
x
x
c
s
o
in
s
x
x
tan x 1 tan x.
11. sin x sin (p x) sin x (sin p cos x cos p sin x)
sin x (0 cos x (1) sin x) sin
2
x 1 cos
2
x.
12. (a)
(b)
13.
5
6
p
14. x .7752 2kp 15.
p
4
16. x .4789 and 5.8042 17. t .9273
18. x
p
2
4kp and x
3
2
p
4kp (k 0, 1, 2, 3, . . .)
19. By the cofunction identity, sin x cos
p
2
x
. Since
p
2
x
p
2
, it follows that 0
p
2
x p. Therefore,
cos
1
(sin x) cos
1
cos
p
2
x
p
2
x by the fact
in the box on page 549
20. about .2266 radians (An angle of approximately 1.3442
radians is theoretically possible, but seems unlikely when
throwing a baseball.)
Chapter 8
Section 8.1, page 582
sin cos tan csc sec cot
1.
3
13
13
2
13
13
3
2
3
13
2
13
2
3
3.
7
58
5
3
8
3
7
7
58
3
58
7
3
5.
11
22
3
1
1
11
3
2
2
22
3
11
3
2
2
7. sin u
1
2
1
9. sin u
3
7
cos u
3
11
cos u
2
7
tan u
3
2
tan u
2
3
11. sin u
m
h
13. c 36
cos u
m
d
15. c 36
tan u
h
d
17.
c 8.4
1 .84
2
1 .84
2
19. h 252
/2 21. h 300 23. h 503
25. c 43
/3 27. a 103
/3
29. ⭿A 50°, a 7.7, c 6.4
31. ⭿C 76°, b 66.1, c 64.2
33. ⭿C 25°, a 10.7, b 11.8
35. ⭿C 18°, a 3.3, c 1.1
37. u 48.6° 39. u 48.2°
41. ⭿A 28.1°, ⭿C 61.9°
43. ⭿A 44.4°, ⭿C 45.6°
45. ⭿A 48.2°, ⭿C 41.8°
47. ⭿A 60.8°, ⭿C 29.2°
49. (a) A
1
2
ah (b) sin u
h
b
(c) h b sin u follows immediately from (b).
(d) A
1
2
ah
1
2
ab sin u
51. A 29.6 53. A 220
55. (a) sin u a/c cos a
(b) Since u a 90° 180°, u a 90°.
(c) Since u a 90°, a 90° u ; thus, sin u cos a
cos(90° u)
Section 8.2, page 593
1. ⬔A 74°, c 29.0, a 27.9
3. ⬔A 36.9°, ⬔B 53.1°
5. w 100.7 ft 7. About 68° 9. 23.0 m
11. About 15.3 ft 13. 276.3 ft
15. About 3402.8 ft 17. 85.9 ft
19. 27.5 ft. It is not safe for him to jump.
21. No 23. 8598.3 ft (1.6 mi)
25. 30° 27. 351.1 m 29. 1.53 mi
31. 25.4 m 33. 3.52 m 35. 52.5 mph
37. 18.67 ft, u 53.13°
39. 205.7 ft 41. 173.2 mi
43. (a) 56.7 ft (b) 9.7 ft
45. (a) Area 200 sin t cos t
(b) t 45°; approximately 14.14 by 7.07 ft
47. 449.1 ft
Section 8.3, page 603
1. a 6.5; B 95.9°; C 44.1°
3. c 15.7; A 19.7°; B 42.3°
5. a 24.4; B 18.4°; C 21.6°
7. c 21.5; A 33.5°; B 67.9°
9. A 120°; B 21.8°; C 38.2°
1022 ANSWERS
11. A 29.7°; B 68.2°; C 82.1°
13. A 38.8°; B 34.5°; C 106.7°
15. A 35.6°; B 53.2°; C 91.2°
17. A 77.5°; B 48.4°; C 54.1°
19. 334.9 km 21. 54.7° 23. 63.7 ft
25. 84.9° 27. 26,205 mi
29. For the left-hand cable, A 34.2°; for the right-hand cable,
B 25.6°
31. 8.4 km 33. 154.5 ft
35. 425.7 ft; 469.4 ft 37. 33.44°
39. 978.7 mi 41. 7.2 in. and 11.5 in.
43. a 19.5; b 21.23; c 24.27.
A 50.2°; B 56.8°; C 73.0°.
45. 16.99 m
Section 8.4, page 614
1. C 114°, b 3.2, c 7.9
3. B 30°, b 6.4, c 8.2
5. A 86°, a 8.9, c 7.1
7. C 41.5°, b 9.7, c 10.9
9. No solution.
11. Case 1: A 154.7°, C 5.3°, c 3.3
Case 2: A 25.3°, C 134.7°, c 24.9
13. No solution.
15. Case 1: B 95.8°, A 28.2°, a 5.7
Case 2: B 84.2°, A 39.8°, a 7.7
17. b .84, c 5.6, C 132.3°
19. a 9.8, B 23.3°; C 81.7°
21. A 18.6°, B 39.6°, C 121.8°
23. c 13.9, A 60.1°, B 72.9°
25. A 77.7°, C 39.8°, a 18.9
27. No solution.
29. a 18.5, B 36.7°, C 78.3°
31. C 126°, b 193.8, c 273.3
33. 135.5 m 35. a 5.4° 37. 5.0 ft
39. 5.3° 41. 30.1 km 43. 9.99 m
45. About 9642 ft
47. (a) Solve triangle ABC to find ⬔BAC, the angle at A.
180° ⬔BAC ⬔EAB: Solve triangle ABD to find
⬔ABD, the angle at B. 180° ⬔ABD ⬔EBA. Now
solve triangle EAB using the two angles and included
side to find EA.
(b) About 94 ft
49. 13.36 m
51. (a) See Example 1 of Section 8.3.
(b) 92.2
(c) 27.8
If you did not get the correct answer in part (b), look at
Example 5 and the Note following it.
Special Topics 8.4.A, page 618
1. 9.6 square units 3. 38.9 square units
5. 82.3 square units 7. 31.4 square units
9. 6.5 square units 11. 7691 square units
13. 235.9 square ft 15. 5.8 gal
17. 11.18 square units
19. (a) 50 (b) 63
21. Since 12 20 36, this violates the Triangle Inequality,
which states that the sum of two sides of a triangle is greater
than the third side. This is not a triangle, and the area is un-
defined.
Chapter 8 Review, page 621
1. sin u
3
34
; cos u
5
34
; tan u
3
5
3. (d) 5. (e)
7. A 35.5°; C 54.5°, b 17.2
9. b 14.65; c 8.40; A 55°
11. 225.9 ft 13. 1.52°
15. A 58.8°, B 34.8°, C 86.4°
17. b 20.4; A 26.4°; C 38.6°
19. 301 mi
21. A 16°, a 1.5, b 4.5
23. A 44.2°, B 73.8°, b 103.3
25. Case 1: B 81.8°, C 38.2°, c 2.5
Case 2: B 98.2°, C 21.8°, c 1.5
27. 147.4 square units 29. 13.4 km
31. C 57°, a 38.5, c 43.4
33. b 12.0; A 75.1°; C 28.9°
35. Joe is 217.9 m from the flagpole. Alice is 240 m from the
flagpole.
37. (a) 3617.65 ft
(b) 4018.71 ft
(c) 3642.19 ft
39. 71.89°
41. 10 square units
43. 37.95 square units
ANSWERS 1023
Chapter 8 Test, page 622
1. (a) (b) (c) 2b
2. 2.51 3. 9
4. 256.94 ft 5. 123
6. 180.08 ft
7. A 53°, b 4.38, c 2.64
8. 1787.06 ft 9. 36.58
10. A 64.16, C 25.84
11. 243.78 ft
12. A 37, a 3.62, c 4.57
13. b 10.78, A 16.65, C 136.75
14. A 66.41, C 39.59, b 23.08
15. A 15.29, B 20.59, C 144.11
16. B 51.61, C 84.39, a 13.47
17. A
1
45.98, C
1
96.82, a 14.63 or A
2
59.62,
C
2
83.18, a 17.55
18. A 16.59, B 43.67, C 119.74
19. 1639.74 mi 20. 4.7 m 21. 260.64 ft
22. (a) 9.56 Square units (b) 21.48 Square units
Chapter 9
Section 9.1, page 630
1–7 odd.
9. 13 11. 3
13. 12
15. Many correct answers, including z 1, w i
17.
y
1
1
x
y
3 + 2i
(1 + i)(1 − i)
(2i)(3 − )
8
3
−
5
3
−
i
i
5
2
i
1
x
1
1 4b
2
2b
1 4b
2
19.
21.
23.
25.
2
i2
27. 0 i i
29.
5
2
i
5
2
3
31.
3
4
3
3
4
i
33. 1.3073 1.5136i 35. 1.6646 3.6372i
37. 32
cos
p
4
i sin
p
4
39. 4
cos
p
3
i sin
p
3
41. 6
cos
11
6
p
i sin
11
6
p
43. 6
cos
5
4
p
i sin
5
4
p
45. 5(cos(.9273) i sin(.9273))
47. 13(cos(5.1072) i sin(5.1072))
49. 5
(cos(1.1071) i sin(1.1071))
51.
2
74
(cos(2.191) i sin(2.191))
53. cos
3
2
p
i sin
3
2
p
0 i(1) i
55. 12
cos
p
3
i sin
p
3
6 63
i
57. 36
cos
p
2
i sin
p
2
36i
y
1
1
x
y
1
2
x
y
2
6
x
1024 ANSWERS
59. cos
p
3
i sin
p
3
1
2
2
3
i
61. 2
cos
p
6
i sin
p
6
3
i
63.
3
2
cos
p
4
i sin
p
4
3
4
2
3
4
2
i
65. 22
cos
7
1
p
2
i sin
7
1
p
2
67. cos
p
2
i sin
p
2
69. 12
cos
2
3
p
i sin
2
3
p
71. 22
cos
1
1
9
2
p
i sin
1
1
9
2
p
73. Since z r(cos u i sin u) and i cos
p
2
i sin
p
2
we
see that iz r
cos
u
p
2
i sin
u
p
2
. The argu-
ment of iz is 90° more than that of z; hence, iz is equivalent
to rotating z through 90° in the complex plane.
75. (a)
b
a
(b)
d
c
(c) y b
d
c
(x a)
(d) y d
b
a
(x c)
(e)
(f) Substituting (a c, b d) in the equation of line L
yields
y b
d
c
(x a)
b d b
d
c
(a c a)
d d.
Similarly, substituting (a c, b d ) in the equation
of line M yields
y d
b
a
(x c)
b d d
b
a
(a c c)
b b.
Thus, (a c, b d ) lies on both lines L and M.
y
0
K
a + bi
c + di
x
L
M
N
77. (a) r
2
(cos u
2
i sin u
2
)(cos u
2
i sin u
2
)
r
2
(cos
2
u
2
sin
2
u
2
) r
2
1 r
2
(b) r
1
(cos u
1
i sin u
1
)(cos u
2
i sin u
2
)
r
1
(cos u
1
cos u
2
i cos u
1
sin u
2
i sin u
1
cos u
2
i
2
sin u
1
sin u
2
)
r
1
(cos u
1
cos u
2
sin u
1
sin u
2
i(sin u
1
cos u
2
cos u
1
sin u
2
))
r
1
(cos(u
1
u
2
) i sin(u
1
u
2
))
Section 9.2, page 638
1. 0 1i i 3. 128 128i3
5.
243
2
3
24
2
3
i 7. 1 0i 1
9. 64 0i 64
11.
1
2
2
3
i
13. 0 i1 i
15. k 0: 1; k 1: i; k 2: 1; k 3: i
17. k 0: 6
cos
p
6
i sin
p
6
; k 1: 6
cos
7
6
p
i sin
7
6
p
19. k 0: 4
cos
1
p
5
i sin
1
p
5
; k 1: 4
cos
1
1
1
5
p
i sin
1
1
1
5
p
;
k 2: 4
cos
7
5
p
i sin
7
5
p
21. k 0: 3
cos
4
p
8
i sin
4
p
8
; k 1: 3
cos
2
4
5
8
p
i sin
2
4
5
8
p
;
k 2: 3
cos
4
4
9
8
p
i sin
4
4
9
8
p
;
k 3: 3
cos
7
4
3
8
p
i sin
7
4
3
8
p
23. k 0: cos
p
5
i sin
p
5
; k 1: cos
3
5
p
i sin
3
5
p
;
k 2: cos p i sin p 1; k 3: cos
7
5
p
i sin
7
5
p
;
k 4: cos
9
5
p
i sin
9
5
p
25. k 0: cos
1
p
0
i sin
1
p
0
; k 1: cos
p
2
i sin
p
2
;
k 2: cos
9
1
p
0
i sin
9
1
p
0
i; k 3: cos
1
1
3
0
p
i sin
1
1
3
0
p
;
k 4: cos
1
1
7
0
p
i sin
1
1
7
0
p
27. k 0:
4
2
cos
p
8
i sin
p
8
; k 1:
4
2
cos
9
8
p
i sin
9
8
p
29. k 0:2
cos
2
p
4
i sin
2
p
4
; k 1:2
cos
1
2
3
4
p
i sin
1
2
3
4
p
;
k 2:2
cos
2
2
5
4
p
i sin
2
2
5
4
p
; k 3:2
cos
3
2
7
4
p
i sin
3
2
7
4
p
ANSWERS 1025
31. cos
p
6
i sin
p
6
2
3
1
2
i, cos
7
6
p
i sin
7
6
p
2
3
1
2
i, cos
p
2
i sin
p
2
i
cos
3
2
p
i sin
3
2
p
i, cos
5
6
p
i sin
5
6
p
2
3
1
2
i,
cos
11
6
p
i sin
11
6
p
2
3
1
2
i
33. cos
p
6
i sin
p
6
2
3
1
2
i, cos
5
6
p
i sin
5
6
p
2
3
1
2
i, cos
3
2
p
i sin
3
2
p
i
35. 3i,
3
2
3
3
2
i
,
3
2
3
3
2
i
37. 3
cos
p
10
4kp
i sin
p
10
4kp
, k 0, 1, 2, 3, 4
39.
4
2
2
3
1
2
i
,
4
2
1
2
2
3
i
,
4
2
2
3
1
2
i
,
4
2
1
2
2
3
i
41. 1, .6235 .7818i, .2225 .9749i, .9010 .4339i
43. 1, i, .7071 .7071i, .7071 7071i
45. 1, .7660 .6428i, .1736 .9848i, .5 .8660i,
.9397 .3420i
−1
1
−1.5 1.5
−1
1
−1.5 1.5
−1
1
−1.5 1.5