Download Section 5.4 Sum and Difference Identities for Sine

Section 5.4 Sum and Difference Identities for Sine and Tangent
82. Since 270° is a quadrantal angle whose
terminal side lies along the y-axis, follow the
reasoning in Case 2 in the text. If θ is a small
positive angle, then 270° − θ lies in quadrant
III and tan θ and cot θ are positive. Thus,
tan ( 270° − θ ) = cot θ.
Section 5.4
Sum and Difference
Identities for Sine and
Tangent
1. sin15° = sin ( 45° − 30°)
= sin 45° cos 30° − cos 45° sin 30°
2 3
2 1
=
⋅
−
⋅
2 2
2 2
6
2
6− 2
=
−
=
4
4
4
Thus, the correct choice is C.
2. sin105° = sin (60° + 45°)
= sin 60° cos 45° + cos 60° sin 45°
3 2 1 2
6
2
=
⋅
+ ⋅
=
+
2 2 2 2
4
4
6+ 2
=
4
Thus, the correct choice is A.
3. tan15° = tan (60° − 45°)
tan 60° − tan 45°
=
1 + tan 60° tan 45°
3 −1
3 −1 1− 3
=
=
⋅
1 + 3 (1) 1 + 3 1 − 3
sin ( −105°) =
6
2 − 6− 2
−
=
4
4
4
Thus, the correct choice is B.
6. tan (−105°) = − tan105° = − tan (60° + 45°)
tan 60° + tan 45°
=−
1 − tan 60° tan 45°
3 +1
3 +1 1+ 3
=−
=−
⋅
1 − 3 (1)
1− 3 1+ 3
3 + 3 +1+ 3
1− 3
4+2 3
=−
= 2+ 3
−2
Thus, the correct choice is D.
=−
7.−8.
9. sin
3 + 3 +1+ 3 4 + 2 3
=
= −2 − 3
1− 3
−2
Thus, the correct choice is F.
5. sin ( −105°) = sin ( 45° − 150°)
= sin 45° cos150° − cos 45° sin150°
Answers will vary.
5π
⎛π π ⎞
= sin ⎜ + ⎟
⎝4 6⎠
12
= sin
π
cos
π
+ cos
π
sin
4
6
4
2 3
2 1
=
⋅
+
⋅
2 2
2 2
6
2
6+ 2
=
+
=
4
4
4
10. sin
3 − 3 − 1 + 3 −4 + 2 3
=
= 2− 3
1− 3
−2
Thus, the correct choice is E.
=
2⎛
3⎞
2 1
−
−
⋅
2 ⎜⎝ 2 ⎟⎠ 2 2
=−
=
4. tan105° = tan (60° + 45°)
tan 60° + tan 45°
=
1 − tan 60° tan 45°
3 +1
3 +1 1+ 3
=
=
⋅
1 − 3 (1) 1 − 3 1 + 3
233
11. tan
π
6
13π
⎛ π 5π ⎞
= sin ⎜ +
⎟
⎝4
12
6 ⎠
π
π
5π
5π
= sin cos
+ cos sin
4
6
4
6
2⎛
3⎞
2 ⎛1⎞
=
−
+
⎜ ⎟
2 ⎜⎝ 2 ⎟⎠ 2 ⎝ 2 ⎠
2− 6
=
4
π
12
⎛π
π⎞
= tan ⎜ − ⎟ =
⎝4 6⎠
3
1−
1−
3 =
=
3
1+
1+
3
tan
π
4
1 + tan
π
6
tan
π
6
3
3 ⋅ 3 = 3− 3
3 3 3+ 3
3
(
4
)
( )
3− 3
3− 3 3− 3
=
⋅
=
3 + 3 3 − 3 32 − 3
=
π
− tan
2
2
9 − 6 3 + 3 12 − 6 3
=
= 2− 3
9−3
6
Copyright © 2013 Pearson Education, Inc.
234
Chapter 5 Trigonometric Identities
5π
⎛π π ⎞
= tan ⎜ + ⎟ =
12. tan
⎝6 4⎠
12
π
tan
6
1 − tan
π
+ tan
π
4
tan
π
6
4
3
3
+1
+1
3
3 +3
= 3
= 3
⋅ =
3
3 3 3− 3
1−
1−
3
3
(
)
( )
3+ 3
3 +3 3+ 3
=
⋅
=
3 − 3 3 + 3 32 − 3
=
13. sin
⎛ 5π ⎞
⎛ π π⎞
= tan ⎜ − − ⎟
17. sin ⎜ −
⎟
⎝ 12 ⎠
⎝ 6 4⎠
⎛ π⎞
⎛ π⎞
tan ⎜ − ⎟ + tan ⎜ − ⎟
⎝ 6⎠
⎝ 4⎠
=
⎛ π⎞
⎛ π⎞
1 − tan ⎜ − ⎟ tan ⎜ − ⎟
⎝ 6⎠
⎝ 4⎠
=
2
= sin
π
cos
4
π
+ cos
π
3
4
2 1
2
3
=
⋅ +
⋅
=
2 2
2 2
=
π
sin
3
2+ 6
4
=
=
⎛π π ⎞
= sin ⎜ − ⎟
14. sin
⎝4 6⎠
12
π
= sin
π
cos
π
− cos
π
sin
4
6
4
2 3
2 1
=
⋅
−
⋅
2 2
2 2
6
2
6− 2
=
−
=
4
4
4
3
2
6
=−
4
=−
π
3
cos
π
6
− cos
π
sin
= − sin
π
=
=
π
⎛ 5π ⎞
⎛ π π⎞
16. sin ⎜ −
= sin ⎜ − − ⎟
⎝ 12 ⎟⎠
⎝ 6 4⎠
π
π
⎛ π⎞
⎛ π⎞
= sin ⎜ − ⎟ cos − cos ⎜ − ⎟ sin
⎝ 6⎠
⎝ 6⎠
4
4
π
1 − tan
π
6
π
4
π
tan
− 3 −3 3+ 3
⋅
3− 3 3+ 3
−3 3 − 3 − 9 − 3 3
32 −
( 3)
2
=
−12 − 6 3
9−3
−12 − 6 3
= −2 − 3
6
⎛ π⎞
⎛ π⎞
tan ⎜ − ⎟ + tan ⎜ − ⎟
⎝ 4⎠
⎝ 3⎠
=
⎛ π⎞
⎛ π⎞
1 − tan ⎜ − ⎟ tan ⎜ − ⎟
⎝ 4⎠
⎝ 3⎠
4
3
4
2
1 2
⋅− ⋅
2
2 2
2 − 6− 2
−
=
4
4
π
6
− tan
⎡ π ⎛ π ⎞⎤
⎛ 7π ⎞
18. tan ⎜ −
= tan ⎢ − + ⎜ − ⎟ ⎥
⎝ 12 ⎟⎠
⎣ 4 ⎝ 3 ⎠⎦
π
⎛ 7π ⎞
⎛ π π⎞
15. sin ⎜ −
= sin ⎜ − − ⎟
⎝ 12 ⎟⎠
⎝ 3 4⎠
π
π
⎛ π⎞
⎛ π⎞
= sin ⎜ − ⎟ cos − cos ⎜ − ⎟ sin
⎝ 3⎠
⎝ 3⎠
4
4
= − sin
π
4
3
−
−1
− 3 −3
3
=
=
⎛ 3⎞
3− 3
1− ⎜
⎟ (1)
3
⎝
⎠
2
9 + 6 3 + 3 12 + 6 3
=
= 2+ 3
9−3
6
7π
⎛π π ⎞
= sin ⎜ + ⎟
⎝4 3⎠
12
− tan
π
cos − cos sin
6
4
6
4
1 2
3 2
=−
⋅−
⋅
2 2
2 2
2
6 − 2− 6
=−
−
=
4
4
4
=
=
(
−1 + − 3
(
)
1 − ( −1) − 3
)
=
−1 − 3 1 + 3
⋅
1− 3 1+ 3
−1 − 3 − 3 − 3
12 −
( 3)
2
−1 − 3
1− 3
=
−4 − 2 3
1− 3
−4 − 2 3
= 2+ 3
−2
19. sin 76° cos 31° − cos 76° sin 31° = sin (76° − 31°)
= sin 45° =
2
2
20. sin 40° cos 50° + cos 40° sin 50° = sin ( 40° + 50°)
= sin 90° = 1
21.
tan 80° + tan 55°
= tan (80° + 55°)
1 − tan 80° tan 55°
= tan135° = −1
Copyright © 2013 Pearson Education, Inc.
Section 5.4 Sum and Difference Identities for Sine and Tangent
22.
tan 80° − tan (−55°)
= tan ⎣⎡80° − (−55°)⎦⎤
1 + tan 80° tan (−55°)
= tan135° = −1
23.
tan100° + tan 80°
= tan (100° + 80°)
1 − tan100° tan 80°
= tan180° = 0
24.
tan 512π + tan π4
1 − tan 512π tan π4
⎛ 5π π ⎞
= tan ⎜
+
⎝ 12 4 ⎟⎠
= tan
π
3π
3π
⎛ 3π
⎞
cos x + sin
sin x
− x ⎟ = cos
31. cos ⎜
⎝ 4
⎠
4
4
⎛
⎛ 2⎞
2⎞
cos x + ⎜
= ⎜−
⎟
⎟ sin x
⎝ 2 ⎠
⎝ 2 ⎠
2
=
(− cos x + sin x )
2
2 (sin x − cos x )
=
2
32. sin ( 45° + θ ) = sin 45° cos θ + sin θ cos 45°
2π
=− 3
3
2
2
sin θ +
cos θ
2
2
2 (sin θ + cos θ )
=
2
=
3π
π
3π
+ cos sin
5
10
5
10
⎛ π 3π ⎞
⎛π ⎞
= sin ⎜ +
= sin ⎜ ⎟ = 1
⎝ 5 10 ⎟⎠
⎝2⎠
25. sin
cos
tan θ + tan 30°
1 − tan θ tan 30°
1
tan θ +
3
=
⎛ 1 ⎞
1− ⎜
tan θ
⎝ 3 ⎟⎠
3 tan θ + 1
=
3 − tan θ
33. tan (θ + 30°) =
26. sin100° cos10° − cos100° sin10°
= sin (100° − 10°) = sin 90° = 1
27. cos (30° + θ ) = cos 30° cos θ − sin 30° sin θ
3
1
cos θ − sin θ
2
2
1
=
3 cos θ − sin θ
2
3 cos θ − sin θ
=
2
=
(
)
⎛π
⎞
34. tan ⎜ + x ⎟ =
⎝4
⎠
=
(
)
29. cos (60° + θ ) = cos 60° cos θ − sin 60° sin θ
1
3
cos θ −
sin θ
2
2
1
= cos θ − 3 sin θ
2
cos θ − 3 sin θ
=
2
=
(
)
30. cos (45° − θ ) = cos 45° cos θ + sin 45° sin θ
2
2
cos θ +
sin θ
2
2
2
=
(cos θ + sin θ )
2
2 (cos θ + sin θ )
=
2
=
tan
π
4
1 − tan
+ tan x
π
tan x
4
1 + tan x
1 + tan x
=
=
1 − 1 ⋅ tan x 1 − tan x
28. cos (θ − 30°) = cos θ cos 30° + sin θ sin 30°
3
1
cos θ + sin θ
2
2
1
3 cos θ + sin θ
=
2
3 cos θ + sin θ
=
2
235
π
π
⎛π
⎞
35. sin ⎜ + x ⎟ = sin cos x + cos sin x
⎝4
⎠
4
4
2
2
=
cos x +
sin x
2
2
2 (cos x + sin x )
=
2
3π
3π
⎛ 3π
⎞
− x ⎟ = sin
cos x − cos
sin x
36. sin ⎜
⎝ 4
⎠
4
4
⎛ 2⎞
⎛
2⎞
=⎜
⎟ (cos x ) − ⎜ − 2 ⎟ (sin x )
2
⎝
⎠
⎝
⎠
⎛ 2⎞
⎛ 2⎞
=⎜
⎟ (cos x ) + ⎜ 2 ⎟ (sin x )
⎝ 2 ⎠
⎝
⎠
2 (cos x + sin x )
=
2
37. sin ( 270° − θ ) = sin 270° cos θ − cos 270° sin θ
= (−1)(cos θ ) − (0)(sin θ )
= − cos θ
Copyright © 2013 Pearson Education, Inc.
236
Chapter 5 Trigonometric Identities
tan180° + tan θ
1 − tan180° tan θ
0 + tan θ
=
= tan θ
1 − 0 ⋅ tan θ
38. tan (180° + θ ) =
tan 2π − tan x
1 + tan 2π tan x
0 − tan x
=
= − tan x
1 + 0 ⋅ tan x
39. tan ( 2π − x ) =
40. sin (π + x ) = sin π cos x + cos π sin x
= 0 ⋅ cos x + ( −1) sin x
= − sin x
tan π − tan x
1 + tan π tan x
0 − tan x
=
= − tan x
1 + 0 ⋅ tan x
41. tan (π − x ) =
42. To follow the method of Example 2 to find
tan ( 270° − θ ) , we need to use the tangent of a
difference formula:
tan 270° − tan θ
tan (270° − θ ) =
1 + tan 270° tan θ
However, tan 270º is undefined.
43. Answers will vary.
44. If A, B, and C are angles of a triangle, then
A + B + C = 180°. Therefore, we have
sin ( A + B + C ) = sin180° = 0.
3
5
, sin t = , and s and t are in
5
13
quadrant I.
First find the values of sin s, tan s, cos t, and
tan t. Because s and t are both in quadrant I,
the values of sin s and cos t, tan s, and tan t
will be positive.
45. cos s =
2
9
16 4
⎛3⎞
=
=
sin s = 1 − ⎜ ⎟ = 1 −
⎝5⎠
25
25 5
2
25
144 12
⎛5⎞
=
=
cos t = 1 − ⎜ ⎟ = 1 −
⎝ 13 ⎠
169
169 13
tan s =
sin s
=
cos s
sin t
=
tan t =
cos t
4
5
3
5
5
13
12
13
=
4
3
5
=
12
(a) sin ( s + t ) = sin s cos t + cos s sin t
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞
= ⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
48 15 63
=
+
=
65 65 65
4
+ 5
tan s + tan t
= 3 4 12 5
1 − tan s tan t 1 −
3 12
48 + 15 63
=
=
36 − 20 16
(b) tan ( s + t ) =
( )( )
(c) From parts (a) and (b), sin (s + t) > 0 and
tan (s + t) > 0. The only quadrant in
which the values of both the sine and the
tangent are positive is quadrant I, so s + t
is in quadrant I.
3
12
and sin t = −
, s is in quadrant I
5
13
and t is in quadrant III.
First find the values of cos s, cos t, tan s, and
tan t. Because s is in quadrant I and t is in
quadrant III, the values of cos s, tan s, and tan
t will be positive, while cos t will be negative.
46. sin s =
2
9
16 4
⎛3⎞
cos s = 1 − ⎜ ⎟ = 1 −
=
=
⎝5⎠
25
25 5
2
144
⎛ 12 ⎞
cos t = − 1 − ⎜ − ⎟ = − 1 −
⎝ 13 ⎠
169
25
5
=−
=−
169
13
3
5
4
5
tan s =
sin s
=
cos s
tan t =
12
sin t − 13 12
= 5 =
cos t − 13
5
=
3
4
(a) sin ( s + t ) = sin s cos t + cos s sin t
⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞ ⎛ 12 ⎞
= ⎜ ⎟ ⎜− ⎟ + ⎜ ⎟ ⎜− ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠
15 ⎛ 48 ⎞
63
=−
+ ⎜− ⎟ = −
65 ⎝ 65 ⎠
65
(b) tan ( s + t ) =
3
4
+ 125
( )( )
=
1−
63
63
=
=−
−16
16
Copyright © 2013 Pearson Education, Inc.
3
4
12
5
15 + 48
20 − 36
Section 5.4 Sum and Difference Identities for Sine and Tangent
(c) From parts (a) and (b), sin (s + t) < 0 and
tan (s + t) < 0. The only quadrant in
which the values of sine and tangent are
both negative is quadrant IV, so s + t is in
quadrant IV.
8
3
and cos t = − , s and t are in
17
5
quadrant III
First find the values of sin s, sin t, tan s, and
tan t. Because s and t are both in quadrant III,
the values of sin s and sin t will be negative,
while tan s and tan t will be positive.
47. cos s = −
⎛ 8⎞
sin s = − 1 − cos 2 s = − 1 − ⎜ − ⎟
⎝ 17 ⎠
= − 1−
2
64
225
15
=−
=−
289
289
17
⎛ 3⎞
sin t = − 1 − cos 2 t = − 1 − ⎜ − ⎟
⎝ 5⎠
2
9
16
4
=−
=−
25
25
5
15
sin s − 17 15
=
=
tan s =
cos s − 178
8
= − 1−
tan t =
4
sin t − 5 4
= 3 =
cos t − 5 3
(a) sin ( s + t ) = sin s cos t + cos s sin t
⎛ 15 ⎞ ⎛ 3 ⎞ ⎛ 8 ⎞ ⎛ 4 ⎞
= ⎜− ⎟ ⎜− ⎟ + ⎜− ⎟ ⎜− ⎟
⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 17 ⎠ ⎝ 5 ⎠
45 32 77
=
+
=
85 85 85
(b) tan ( s + t ) =
15
8
+
4
3
( )( )
=
1−
77
77
=
=−
−36
36
15
8
4
3
45 + 32
24 − 60
15
4
, sin t = , s is in quadrant II, and
17
5
t is in quadrant I.
First find the values of sin s, cos t, tan s, and
tan t. Because s is in quadrant II and t is in
quadrant I, the values of sin s, cos t, and tan t
will be positive, while tan s will be negative.
64
8
=
289 17
2
3
⎛4⎞
cos t = 1 − ⎜ ⎟ =
⎝5⎠
5
8
tan s =
sin s
8
= 1715 = −
cos s − 17
15
tan t =
sin t
=
cos t
4
5
3
5
=
4
3
(a) sin ( s + t ) = sin s cos t + cos s sin t
⎛ 8 ⎞ ⎛ 3 ⎞ ⎛ 15 ⎞ ⎛ 4 ⎞
= ⎜ ⎟ ⎜ ⎟ + ⎜− ⎟ ⎜ ⎟
⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 17 ⎠ ⎝ 5 ⎠
24 60
36
=
−
=−
85 85
85
(b) tan(s + t) =
(− 158 ) + 34 = −24 + 60 = 36
8
1 − (− 15
)( 43 ) 45 + 32 77
(c) To find the quadrant of s + t, notice from
36
<0
the preceding that sin( s + t ) = −
85
36
and tan ( s + t ) =
> 0 . The only
77
quadrant in which the value of sine is
negative and tangent is positive is
quadrant III, so s + t is in quadrant III.
2
1
and sin t = − , s is in quadrant II
3
3
and t is in quadrant IV.
First find the values of cos s, cos t, tan s, and
tan t. Because s is in quadrant II and t is in
quadrant IV, the values of cos s, tan s, and tan t
will be negative, while cos t will be positive.
49. sin s =
2
4
5
5
⎛2⎞
cos s = − 1 − ⎜ ⎟ = − 1 − = −
=−
⎝3⎠
9
9
3
2
(c) From parts (a) and (b), sin (s + t) > 0 and
tan (s + t) < 0. The only quadrant in
which the value of the sine is positive and
the value of the tangent is negative is
quadrant II, so s + t is in quadrant II.
48. cos s = −
2
225
⎛ 15 ⎞
=
sin s = 1 − ⎜ − ⎟ = 1 −
⎝ 17 ⎠
289
237
1
⎛ 1⎞
cos t = 1 − ⎜ − ⎟ = 1 − =
⎝ 3⎠
9
8
9
8 2 2
=
3
3
2
sin s
2
2 5
tan s =
= 3 =−
=−
cos s − 5
5
5
=
3
−1
sin t
1
2
tan t =
= 3 =−
=−
2
2
cos t
4
2 2
Copyright © 2013 Pearson Education, Inc.
3
(continued on next page)
238
Chapter 5 Trigonometric Identities
(continued)
(a) sin ( s + t ) = sin s cos t + cos s sin t
5 ⎞⎛ 1⎞
⎛2⎞⎛2 2 ⎞ ⎛
= ⎜ ⎟⎜
+ ⎜−
−
⎟
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠⎟ ⎝⎜ 3 ⎠⎟
=
4 2
5 4 2+ 5
+
=
9
9
9
(b) Different forms of tan (s + t) will be
obtained depending on whether tan s and
tan t are written with rationalized
denominators.
( ) = −8 5 − 5 2
tan ( s + t ) =
1 − (−
) (− ) 20 − 2 10
− 2 55 + −
2
4
2 5
5
=
=
2
4
−8 5 − 5 2 20 + 2 10
⋅
20 − 2 10 20 + 2 10
−160 5 − 16 50 − 100 2 − 10 20
400 − 40
−160 5 − 80 2 − 100 2 − 20 5
=
360
−180 5 − 180 2 − 5 − 2
=
=
360
2
or
( )
tan ( s + t ) =
1 − ( − ) (−
)
−
2
5
+ −
2
5
1
2 2
2
1
⎛ 1⎞
sin s = 1 − ⎜ − ⎟ = 1 −
=
⎝ 5⎠
25
24 2 6
=
=
5
5
24
25
2
9
⎛3⎞
cos t = − 1 − ⎜ ⎟ = − 1 −
⎝5⎠
25
16
4
=−
=−
25
5
2 6
tan s =
sin s
= 5 = −2 6
cos s − 15
tan t =
sin t
3
= 54 = −
cos t − 5
4
3
(a) sin ( s + t ) = sin s cos t + cos s sin t
⎛ 2 6 ⎞ ⎛ 4⎞ ⎛ 1⎞ ⎛3⎞
=⎜
⎟ ⎜− ⎟ + ⎜− ⎟ ⎜ ⎟
⎝ 5 ⎠⎝ 5⎠ ⎝ 5⎠ ⎝5⎠
=
1
−8 6 3 −8 6 − 3
−
=
25
25
25
2 2
−4 2 − 5 4 2 + 5
=
=
2 10 − 2
2 − 2 10
4 2 + 5 2 + 2 10
=
⋅
2 − 2 10 2 + 2 10
8 2 + 8 20 + 2 5 + 2 50
=
4 − 40
8 2 + 16 5 + 2 5 + 10 2
=
−36
18 2 + 18 5 − 2 − 5
=
=
2
−36
(c) To find the quadrant of s + t, notice from the
preceding that sin ( s + t ) =
4 2+ 5
>0
9
−8 5 − 5 2
≈ −1.8 < 0 .
20 − 2 10
The only quadrant in which the values of sine
are positive and tangent is negative is
quadrant II. Therefore, s + t is in quadrant II.
and tan ( s + t ) =
1
3
, sin t = , and s and t are in
5
5
quadrant II.
First find the values of sin s, tan s, cos t, and
tan t. Because s and t are both in quadrant II,
the values of sin s is positive. Cos t, tan s, and
tan t are all negative
50. cos s = −
(b) tan ( s + t ) =
=
tan s + tan t
1 − tan s tan t
−2 6 + − 34
(
1 − −2
( ) = −8 46 − 3
4− 6 6
6 ) (− 34 )
4
−8 6 − 3
4−6 6
−8 6 − 3 4 + 6 6
=
⋅
4−6 6 4+6 6
−32 6 − 288 − 12 − 18 6
=
16 − 216
6 +6
−50 6 − 300
=
=
4
−200
Note that other forms of tan (s + t) will be
obtained depending on whether tan s and
tan t are written with rationalized
deonominators.
=
(c) From parts (a) and (b), sin (s + t) < 0 and
tan (s + t) > 0. The sine is negative in
quadrants III and IV, while the tangent is
positive in quadrants I and III. Therefore,
s + t is in quadrant III.
Copyright © 2013 Pearson Education, Inc.
Section 5.4 Sum and Difference Identities for Sine and Tangent
51. sin165° = sin (180° − 15°)
= sin180° cos15° − cos180° sin15°
= (0) cos15° − ( −1) sin15° = 0 + sin15°
= sin15°
Now use a difference identity to find sin 15°.
sin15° = sin ( 45° − 30°)
= sin 45° cos 30° − cos 45° sin 30°
2 3
2 1
=
⋅
−
⋅
2 2
2 2
6
2
6− 2
=
−
=
4
4
4
52. sin 255° = sin (270° − 15°)
= sin 270° cos15° − cos 270° sin15°
= ( −1) cos15° − (0) sin15°
= − cos15° − 0 = − cos15°
Now use a difference identity to find cos15°.
cos15° = cos ( 45° − 30°)
= (cos 45° cos 30° + sin 45° sin 30°)
2 3
2 1
6
2
⋅
+
⋅ =
+
2 2
2 2
4
4
6+ 2
=
4
⎛ 6+ 2⎞
Thus, sin 255° = − cos15° = − ⎜
⎟
4
⎝
⎠
− 6− 2
.
=
4
=
53. tan165° = tan (180° − 15°)
tan180° − tan15°
=
1 + tan180° tan15°
0 − tan15°
=
= − tan15°
1 + 0 ⋅ tan15°
Now use a difference identity to find tan15°.
tan 45° − tan 30°
tan15° = tan (45° − 30°) =
1 + tan 45° tan 30°
3
1−
3 = 3− 3
=
3 3+ 3
1 + 1⋅
3
3− 3 3− 3 9−3 3 − 3 3 +3
=
⋅
=
9−3
3+ 3 3− 3
12 − 6 3
=
= 2− 3
6
Thus,
(
239
54. tan 285° = tan (360° − 75°)
tan 360° − tan 75°
=
1 + tan 360° tan 75°
− tan 75°
0 − tan 75°
=
=
1 + 0 ⋅ tan 75°
1+ 0
= − tan 75°
Now use a sum identity to find tan 75°.
3
+1
tan 75° = tan (30° + 45°) = 3
3
⋅1
1−
3
3 +3
3 +3 3 +3
=
=
⋅
3− 3 3− 3 3+ 3
3 + 3 3 + 3 3 + 9 12 + 6 3
=
=
9−3
6
= 2+ 3
Thus,
(
)
tan 285° = − tan 75° = − 2 + 3 = −2 − 3.
55. tan
π ⎞
11π
⎛
= tan ⎜ π − ⎟
⎝
12
12 ⎠
π
tan π − tan 12
π
=
= − tan
π
12
1 + tan π tan 12
Now use a difference identity to find tan
π
⎛π π ⎞
tan
= tan ⎜ − ⎟ =
⎝4 6⎠
12
tan
π
− tan
4
1 + tan
π
π
π
12
6
π
tan
4
6
3
3 3− 3
1−
1−
3 =
3 =
3
=
3
3 3+ 3
1 + 1⋅
1+
3
3
3
3− 3 3− 3 3− 3
=
=
⋅
3+ 3 3+ 3 3− 3
9 − 6 3 + 3 12 − 6 3
=
=
9−3
6
6 2− 3
=
= 2− 3
6
Thus,
11π
π
tan
= − tan
= − 2 − 3 = −2 + 3.
12
12
(
)
tan165° = − tan15° = − 2 − 3 = −2 + 3.
Copyright © 2013 Pearson Education, Inc.
)
(
)
.
240
Chapter 5 Trigonometric Identities
π ⎞
⎛ 13π ⎞
⎛ 13π ⎞
⎛
= − sin ⎜
= − sin ⎜ π + ⎟
56. sin ⎜ −
⎟
⎟
⎝ 12 ⎠
⎝ 12 ⎠
⎝
12 ⎠
π
π ⎞
⎛
= − ⎜ sin π cos + cos π sin ⎟
⎝
12
12 ⎠
π
π⎤
⎡
= − ⎢(0) cos + (−1) sin ⎥
12
12
⎣
⎦
π ⎞
π ⎞
⎛
⎛
= − ⎜ 0 − sin ⎟ = − ⎜ − sin ⎟
⎝
⎝
12 ⎠
12 ⎠
= sin
π
12
⎛π π ⎞
= sin ⎜ − ⎟
⎝4 6⎠
= sin
π
⎛π
⎞
59. tan ⎜ + θ ⎟ appears to be equivalent to
⎝2
⎠
− cot θ .
π
12
Now use a difference identity to find sin
sin
3π
3π
⎛ 3π
⎞
sin ⎜
cos θ + cos
sin θ
+ θ ⎟ = sin
⎝ 2
⎠
2
2
= ( −1) cos θ + (0) sin θ = − cos θ
cos
π
− cos
π
4
6
4
2 3
2 1
=
⋅
−
⋅ =
2 2
2 2
sin
π
12
π
.
⎛π
⎞
tan ⎜ + θ ⎟ =
⎝2
⎠
6
6− 2
4
=
⎛π
⎞
sin ⎜ + θ ⎟
⎝2
⎠
π
⎛
⎞
cos ⎜ + θ ⎟
⎝2
⎠
sin
π
2
π
cos θ + cos
2
sin θ
π
sin θ
2
2
1 ⋅ cos θ + 0 ⋅ sin θ
cos θ
=
=
0 ⋅ cos θ − 1 ⋅ sin θ − sin θ
= − cot θ
The graphs in exercises 57−60 are shown in the
following window. This window can be obtained
through the Zoom Trig feature.
cos
cos θ − sin
π
⎛π
⎞
60. tan ⎜ − θ ⎟ appears to be equivalent to
⎝2
⎠
cot θ .
⎛π
⎞
57. sin ⎜ + θ ⎟ appears to be equivalent to
⎝2
⎠
cos θ .
⎛π
⎞
tan ⎜ − θ ⎟ =
⎝2
⎠
π
π
⎛π
⎞
sin ⎜ + θ ⎟ = sin cos θ + sin θ cos
⎝2
⎠
2
2
= 1 ⋅ cos θ + sin θ ⋅ 0
= cos θ + 0 = cos θ
=
⎛π
⎞
sin ⎜ − θ ⎟
⎝2
⎠
⎛π
⎞
cos ⎜ − θ ⎟
⎝2
⎠
sin
π
2
π
cos θ − cos
π
2
sin θ
π
cos θ + sin sin θ
2
2
1 ⋅ cos θ − 0 ⋅ sin θ cos θ
=
=
0 ⋅ cos θ + 1 ⋅ sin θ sin θ
= cot θ
cos
⎛ 3π
⎞
+ θ ⎟ appears to be equivalent to
58. sin ⎜
⎝ 2
⎠
− cos θ .
61. Verify sin 2 x = 2 sin x cos x is an identity.
sin 2 x = sin ( x + x ) = sin x cos x + cos x sin x
= 2 sin x cos x
Copyright © 2013 Pearson Education, Inc.
Section 5.4 Sum and Difference Identities for Sine and Tangent
62. Verify sin ( x + y ) + sin ( x − y ) = 2 sin x cos y is an identity.
sin ( x + y ) + sin ( x − y ) = (sin x cos y + cos x sin y ) + (sin x cos y − cos x sin y ) = 2 sin x cos y
⎛ 7π
⎞
⎛ 2π
⎞
+ x ⎟ − cos ⎜
+ x ⎟ = 0 is an identity.
63. Verify sin ⎜
⎝ 6
⎠
⎝ 3
⎠
7π
7π
2π
2π
⎛ 7π
⎞
⎛ 2π
⎞ ⎛
⎞ ⎛
⎞
+ x ⎟ − cos ⎜
+ x ⎟ = ⎜ sin
sin ⎜
cos x + cos
sin x ⎟ − ⎜ cos
cos x − sin
sin x ⎟
⎝ 6
⎠
⎝ 3
⎠ ⎝
⎠ ⎝
⎠
6
6
3
3
⎛ 1
⎞ ⎛ 1
⎞
3
3
= ⎜ − cos x −
sin x ⎟ − ⎜ − cos x −
sin x ⎟ = 0
2
2
⎝ 2
⎠ ⎝ 2
⎠
64. Verify tan ( x − y ) − tan ( y − x ) =
tan ( x − y ) − tan ( y − x ) =
65. Verify
2 ( tan x − tan y )
is an identity.
1 + tan x tan y
tan x − tan y
tan y − tan x
tan x − tan y − tan y + tan x 2 ( tan x − tan y )
−
=
=
1 + tan x tan y 1 + tan y tan x
1 + tan x tan y
1 + tan x tan y
cos (α − β )
= tan α + cot β is an identity.
cos α sin β
cos (α − β ) cos α cos β + sin α sin β cos α cos β sin α sin β cos β sin α
=
=
+
=
+
= cot β + tan α
cos α sin β
cos α sin β
cos α sin β cos α sin β sin β cos α
sin ( s + t )
= tan s + tan t is an identity.
cos s cos t
sin ( s + t ) sin s cos t + cos s sin t sin s cos t cos s sin t sin s sin t
=
=
+
=
+
= tan s + tan t
cos s cos t
cos s cos t
cos s cos t cos s cos t cos s cos t
66. Verify
67. Verify that
sin ( x − y ) tan x − tan y
=
is an identity
sin ( x + y ) tan x + tan y
sin x cos y cos x sin y
sin x cos y cos x sin y
−
⋅
−
⋅
sin ( x − y ) sin x cos y − cos x sin y cos x cos y cos x cos y cos x cos y cos x cos y
=
=
=
sin x cos y cos x sin y
sin ( x + y ) sin x cos y + cos x sin y sin x cos y cos x sin y
+
⋅
+
⋅
cos x cos y cos x cos y cos x cos y cos x cos y
sin x sin y
sin x
sin y
⋅1 − 1⋅
−
cos x
cos y cos x cos y tan x − tan y
=
=
=
sin x
sin y
sin x sin y tan x + tan y
⋅1 + 1⋅
+
cos x
cos y cos x cos y
sin ( x + y )
cot x + cot y
=
is an identity.
cos ( x − y ) 1 + cot x cot y
Working with the right side, we have
cos x cos y
cos x cos y
+
+
cot x + cot y
sin x sin y
sin x sin y sin x sin y cos x sin y + sin x cos y sin ( x + y )
=
=
⋅
=
=
cos x cos y
cos x cos y sin x sin y sin x sin y + cos x cos y cos ( x − y )
1 + cot x cot y
1+
1+
sin x sin y
sin x sin y
68. Verify
Copyright © 2013 Pearson Education, Inc.
241
242
Chapter 5 Trigonometric Identities
sin ( s − t ) cos ( s − t )
sin s
+
=
is an identity.
sin t
cos t
sin t cos t
sin ( s − t ) cos ( s − t ) sin s cos t − sin t cos s cos s cos t + sin t sin s
+
=
+
sin t
cos t
sin t
cos t
sin s cos 2 t − sin t cos t cos s sin t cos t cos s + sin 2 t sin s sin s cos 2 t + sin s sin 2 t
=
+
=
sin t cos t
sin t cos t
sin t cos t
2
2
sin s cos t + sin t
sin s
=
=
sin t cos t
sin t cos t
69. Verify
(
)
tan (α + β ) − tan β
= tan α is an
1 + tan (α + β ) tan β
identity.
tan (α + β ) − tan β
= tan ⎣⎡(α + β ) − β ⎦⎤
1 + tan (α + β ) tan β
= tan α
70. Verify
71. ∠β and ∠ABC are supplementary, so
m∠ABC = 180° − β .
72. α + (180° − β ) + θ = 180° ⇒ α − β + θ = 0 ⇒
θ = β −α
73. tan θ = tan ( β − α ) =
tan β − tan α
1 + tan β tan α
74. Substituting m1 for tan α and m2 for tan β
into the expression in exercise 69, we have
tan β − tan α
m − m1
= 2
tan θ =
1 + tan α tan β 1 + m1m2
75. x + y = 9, 2 x + y = −1
Convert each equation to slope-intercept form
to find the slopes:
x + y = 9 ⇒ y = − x + 9 ⇒ m1 = −1
2 x + y = −1 ⇒ y = −2 x − 1 ⇒ m2 = −2
−2 − ( −1)
m2 − m1
1
=
=− ⇒
1 + m1m2 1 + ( −1)( −2)
3
1
⎛
⎞
θ = tan −1 ⎜ − ⎟ ≈ −18.4°
⎝ 3⎠
Note that the angle must be positive, so
θ ≈ 18.4°
tan θ =
76. 5 x − 2 y + 4 = 0, 3 x + 5 y = 6
Convert each equation to slope-intercept form
to find the slopes:
5 x − 2 y + 4 = 0 ⇒ −2 y = −5 x − 4 ⇒
5
5
y = x + 2 ⇒ m1 =
2
2
3x + 5 y = 6 ⇒ 5 y = −3x + 6 ⇒
3
6
3
y = − x + ⇒ m2 = −
5
5
5
tan θ =
− 35 − 52
m2 − m1
=
1 + m1m2 1 + 5 − 3
2
5
− 53 −
( )( )
5
2
10 −6 − 25
⋅ =
1 + 52 − 53 10 10 − 15
−31
=
= 6.2 ⇒
−5
−1
θ = tan 6.2 ≈ 80.8°
=
( )( )
0.6W sin (θ + 90°)
sin12°
0.6 (170) sin (30 + 90) °
=
sin12°
102 sin120°
=
≈ 425 lb
sin12°
(This is a good reason why people
frequently have back problems.)
77. (a) F =
(b) F =
=
=
=
=
0.6W sin (θ + 90°)
sin12°
0.6W (sin θ cos 90° + sin 90° cos θ )
sin12°
0.6W (sin θ ⋅ 0 + 1 ⋅ cos θ )
sin12°
0.6W (0 + cos θ )
sin12°
0.6
W cos θ ≈ 2.9W cos θ
sin12°
(c) F will be maximum when cos θ = 1 or
θ = 0° . ( θ = 0° corresponds to the back
being horizontal which gives a maximum
force on the back muscles. This agrees
with intuition since stress on the back
increases as one bends farther until the
back is parallel with the ground.)
0.6W sin (θ + 90°)
, W = 200,
sin12°
θ = 45°
0.6 (200) sin ( 45 + 90) °
F=
sin12°
120 sin135°
=
≈ 408 lb
sin12°
78. (a) F =
Copyright © 2013 Pearson Education, Inc.
Chapter 5 Quiz
(b) The calculator should be in degree mode.
Graph
0.6 ( 200) sin ( x + 90°) 120 sin ( x + 90°)
y=
=
sin12°
sin12°
and y = 400 on the same screen to find the
point of intersection.
243
sin (120π t + φ ) = 0 ⇒
sin ⎡⎣120π (0.0142) + φ ⎤⎦ = 0 ⇒
120π (0.0142) + φ = 0
φ = −120π (0.0142)
≈ −5.353
Thus, V = 50 sin (120 t – 5.353).
(c) 50 sin (120π t − 5.353)
⎡(sin120π t )(cos 5.353)
⎤
= 50 ⎢
− (cos120π t )(sin 5.353)⎥⎦
⎣
A force of 400 lb is exerted when
θ ≈ 46.1°.
⎛πt π ⎞
79. E = 20 sin ⎜ − ⎟
⎝ 4 2⎠
π
πt
π⎞
⎛ πt
= 20 ⎜ sin cos − cos sin ⎟
⎝
4
2
4
2⎠
πt ⎞
⎛ πt
= 20 ⎜ sin (0) − cos (1) ⎟
⎝
⎠
4
4
πt ⎞
πt
⎛
= 20 ⎜ 0 − cos ⎟ = −20 cos
⎝
4 ⎠
4
80. (a) The calculator should be in radian mode.
⎤
≈ 50 ⎡(sin120π t )(0.5977)
⎢
− (cos120π t )(−0.8017 )⎥⎦
⎣
≈ 29.89sin120π t + 40.09 cos120π t
≈ 30sin120π t + 40 cos120π t
81. y ′ = r cos (θ + R ) = r [ cos θ cos R − sin θ sin R ]
= ( r cos θ ) cos R − ( r sin θ ) sin R
= y cos R − z sin R
⎛π
⎞
82. z ′ = r sin (θ + R ) = r cos ⎜ − (θ + R ) ⎟
⎝2
⎠
⎛⎛ π
⎞
⎞
= r cos ⎜ ⎜ − θ ⎟ − R ⎟
⎠
⎝⎝ 2
⎠
⎡
⎤
⎛π
⎞
⎛π
⎞
= r ⎢ cos ⎜ − θ ⎟ cos R + sin ⎜ − θ ⎟ sin R ⎥
⎝2
⎠
⎝2
⎠
⎣
⎦
= r [sin θ cos R + cos θ sin R ]
= ( r sin θ ) cos R + (r cos θ ) sin R
= z cos R + y sin R
Chapter 5 Quiz
(Sections 5.1−5.4)
7
, θ is in quadrant IV
25
In quadrant IV, the cosine and secant function
values are positive. The tangent, cotangent,
and cosecant function values are negative.
1. sin θ = −
(b) Using the maximum and minimum
features on your calculator, the amplitude
appears to be 50. So, let a = 50.
⎛ 7 ⎞
cos θ = 1 − sin 2 θ = 1 − ⎜ − ⎟
⎝ 25 ⎠
= 1−
Estimate the phase shift by approximating
the first t-intercept where the graph of V is
increasing. This is located at t ≈ 0.0142.
tan θ =
49
=
625
2
576 24
=
625 25
7
sin θ − 25
7
= 24 = −
cos θ
24
25
Copyright © 2013 Pearson Education, Inc.
(continued on next page)
244
Chapter 5 Trigonometric Identities
(a) cos ( A + B ) = cos A cos B − sin A sin B
3 ⎛ 12 ⎞ 4 ⎛ 5 ⎞
= ⎜− ⎟ − ⎜− ⎟
5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠
36 20
16
=−
+
=−
65 65
65
(continued)
cot θ =
1
1
24
= 7 =−
tan θ − 24
7
sec θ =
1
1 25
= 24 =
cos θ
24
25
1
1
25
= 7 =−
csc θ =
sin θ − 25
7
2. cot 2 x + csc 2 x =
⎛ 7π
3. sin ⎜ −
⎝ 12
cos 2 x
1
1 + cos 2 x
+
=
2
2
sin x sin x
sin 2 x
⎞
⎛ 7π ⎞
⎛π π ⎞
⎟⎠ = − sin ⎜⎝
⎟⎠ = − sin ⎜⎝ + ⎟⎠
12
3 4
π
π
π⎞
⎛ π
= − ⎜ sin cos + cos sin ⎟
⎝
3
4
3
4⎠
⎡ 3 ⎛ 2 ⎞ 1 ⎛ 2 ⎞⎤
= −⎢
⎜
⎟+ ⎜
⎟⎥
⎣⎢ 2 ⎝ 2 ⎠ 2 ⎝ 2 ⎠ ⎦⎥
⎛ 6
⎛ 6+ 2⎞
2⎞
= −⎜
+
= −⎜
⎟
⎟
4 ⎠
4
⎝ 4
⎝
⎠
− 6− 2
=
4
4. cos (180° − θ ) = − cos θ
3
5
π
, sin B = − , 0 < A < , and
5
13
2
3π
π <B<
2
3 y
cos A = = ⇒ y = 3, r = 5 . Substituting
5 r
into the Pythagorean theorem, we have
x 2 + 33 = 52 ⇒ x = 4 , since sin A > 0. Thus,
4
sin A = .
5
5. cos A =
We will use a Pythagorean identity to find the
value of cos B.
2
25
⎛ 5⎞
cos B = − 1 − ⎜ − ⎟ = − 1 −
⎝ 13 ⎠
169
144
12
=−
=−
169
13
Note that cos B is negative because B is in
quadrant III.
(b) sin ( A + B ) = sin A cos B + cos A sin B
4 ⎛ 12 ⎞ 3 ⎛ 5 ⎞
= ⎜− ⎟ + ⎜− ⎟
5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠
48 15
63
=−
−
=−
65 65
65
(c) Both cos ( A + B ) and sin ( A + B ) are
negative. Thus ( A + B ) is in quadrant III.
3π
⎛ 3π
⎞ tan 4 + tan x
6. tan ⎜
+ x⎟ =
⎝ 4
⎠ 1 − tan 3π tan x
4
−1 + tan x
−1 + tan x
=
=
1 − (−1) tan x 1 + tan x
1 + sin θ
sin θ
=
is an identity.
2
csc θ − 1
cot θ
Working with the right side, we have
sin θ
sin θ
csc θ + 1
=
⋅
csc θ − 1 csc θ − 1 csc θ + 1
sin θ csc θ + sin θ 1 + sin θ
=
=
csc2 θ − 1
cot 2 θ
7. Verify
⎛π
⎞
⎛π
⎞
8. Verify sin ⎜ + θ ⎟ − sin ⎜ − θ ⎟ = sin θ is an
⎝3
⎠
⎝3
⎠
identity.
⎛π
⎞
⎛π
⎞
sin ⎜ + θ ⎟ − sin ⎜ − θ ⎟
⎝3
⎠
⎝3
⎠
π
⎛ π
⎞
= ⎜ sin cos θ + cos sin θ ⎟
⎝
⎠
3
3
π
⎛ π
⎞
− ⎜ sin cos θ − cos sin θ ⎟
⎝
⎠
3
3
π
⎛1⎞
= 2 cos sin θ = 2 ⎜ ⎟ sin θ = sin θ
⎝2⎠
3
sin 2 θ − cos 2 θ
= 1 is an identity.
sin 4 θ − cos 4 θ
sin 2 θ − cos 2 θ
sin 4 θ − cos 4 θ
sin 2 θ − cos 2 θ
1
=
= =1
2
2
2
2
1
sin θ − cos θ sin θ + cos θ
9. Verify
(
Copyright © 2013 Pearson Education, Inc.
)(
)
Section 5.5 Double-Angle Identities
10. Verify
245
cos ( x + y ) + cos ( x − y )
= cot x is an identity.
sin ( x − y ) + sin ( x − y )
cos ( x + y ) + cos ( x − y ) (cos x cos y − sin x sin y ) + (cos x cos y + sin x sin y ) 2 cos x cos y cos x
=
=
=
= cot x
sin ( x + y ) + sin ( x − y ) (sin x cos y + cos x sin y ) + (sin x cos y − cos x sin y ) 2 sin x cos y sin x
Section 5.5
Double-Angle Identities
1. C. 2 cos 2 15° − 1 = cos (2 ⋅ 15°) = cos 30° =
2. E.
3
2
2 tan15°
3
= tan ( 2 ⋅ 15°) = tan (30°) =
3
1 − tan 2 15°
3. B. 2 sin 22.5° cos 22.5° = sin (2 ⋅ 22.5°)
= sin 45° =
4. A. cos 2
5. F. 4 sin
6. D.
π
− sin 2
6
π
3
cos
2 tan π3
1 − tan 2
π
3
π
3
π
6
2
2
π 1
⎛ π⎞
= cos ⎜ 2 ⋅ ⎟ = cos =
⎝ 6⎠
3 2
2π
⎛ π⎞
= 2 sin ⎜ 2 ⋅ ⎟ = 2 sin
⎝ 3⎠
3
3
= 2⋅
= 3
2
2π
⎛ π⎞
= tan ⎜ 2 ⋅ ⎟ = tan
=− 3
⎝ 3⎠
3
2
, cos θ < 0
5
cos 2θ = 1 − 2 sin 2 θ ⇒
7. sin θ =
2
4
8 17
⎛2⎞
cos 2θ = 1 − 2 ⎜ ⎟ 1 − 2 ⋅
= 1−
=
⎝5⎠
25
25 25
289 336
⎛ 17 ⎞
=
sin 2 2θ = 1 − ⎜ ⎟ = 1 −
⎝ 25 ⎠
625 625
Since cos θ < 0, sin 2θ < 0 because
sin 2θ = 2 sin θ cos θ < 0 and sin θ > 0.
336
336
4 21
=−
=−
625
25
25
12
, sin θ > 0
13
2 ( 2)
2 tan x
4
4
=
=
=−
2
2
1− 4
3
1 − tan x 1 − 2
Since both tan x and cos x are positive, x must be
in quadrant I. Since 0° < x < 90°, then
0° < 2 x < 180°. Thus, 2x must be in either
quadrant I or quadrant II. However, tan 2 x < 0 ,
so 2x is in quadrant II, and sec 2 x is negative.
tan 2 x =
2
16 25
⎛ 4⎞
sec2 2 x = 1 + tan 2 2 x = 1 + ⎜ − ⎟ = 1 +
=
⎝ 3⎠
9
9
5
1
3
=−
sec 2 x = − ⇒ cos 2 x =
3
sec 2 x
5
2
2
2
cos 2 x + sin 2 x = 1 ⇒ sin 2 x = 1 − cos 2 2 x ⇒
2
9 16
⎛ 3⎞
sin 2 2 x = 1 − ⎜ − ⎟ ⇒ sin 2 2 x = 1 −
=
⎝ 5⎠
25 25
In quadrants I and II, sin 2 x > 0. Thus, we
have sin 2 x =
16 4
= .
25 5
5
, sin x < 0
3
25 34
⎛5⎞
sec2 x = 1 + tan 2 x = 1 + ⎜ ⎟ = 1 +
=
⎝3⎠
9
9
Since sin x < 0, and tan x > 0, x is in quadrant
III, so cos x and sec x are negative.
34
sec x = −
3
3
3
34
3 34
cos x = −
=−
⋅
=−
34
34
34
34
⎛ 3 34 ⎞
sin x = − 1 − cos 2 θ = − 1 − ⎜ −
⎟
⎝ 34 ⎠
sin θ = 1 − cos 2 θ ⇒
2
9. tan x = 2, cos x > 0
2
2
8. cos θ = −
2
⎛ 12 ⎞
cos 2θ = 2 cos θ − 1 = 2 ⎜ − ⎟ − 1
⎝ 13 ⎠
144
288
119
= 2⋅
−1 =
−1 =
169
169
169
2
10. tan x =
cos 2 2θ + sin 2 2θ = 1 ⇒
sin 2 2θ = 1 − cos 2 2θ ⇒
sin 2θ = −
120
⎛ 5 ⎞ ⎛ 12 ⎞
sin 2θ = 2 sin θ cos θ = 2 ⎜ ⎟ ⎜ − ⎟ = −
⎝ 13 ⎠ ⎝ 13 ⎠
169
144
25
5
⎛ 12 ⎞
sin θ = 1 − ⎜ − ⎟ = 1 −
=
=
⎝ 13 ⎠
169
169 13
= − 1−
2
306
850
5 34
=−
=−
1156
1156
34
(continued on next page)
Copyright © 2013 Pearson Education, Inc.