pre-calc 1e ssm chapter 6 part1 final
TRANSCRIPT
361
CHAPTER 6
Section 6.1 Solutions--------------------------------------------------------------------------------
1. ( ) ( ) ( )sin 60 cos 90 60 cos 30= − =� � � � 3. ( ) ( )cos sin 90x x= −�
5. ( ) ( ) ( )csc 30 sec 90 30 sec 60= − =� � � �
7. ( ) ( )( ) ( )sin cos 90 cos 90x y x y x y+ = − + = − −� �
9. ( ) ( )( ) ( )cos 20 sin 90 20 sin 70A A A+ = − + = −� � � �
11. ( ) ( )( ) ( )cot 45 tan 90 45 tan 45x x x− = − − = +� � � �
13. ( ) ( )( ) ( )csc 60 sec 90 60 sec 30θ θ θ− = − − = +� � � �
15. sin csc sinx x x=1
sin x1
=
17. 1 cos 1
sec( ) cotcos( ) sin cos( )
xx x
x x x
− = =
−
cos x 1csc
sin sin( )x
x x
= =
19. 1 1
csc( )sin sinsin( ) sin
x x xx x
− = =− −
sin x 1= −
21. 2 1
sec cos( ) tancos
x x xx
− + = cos x 2 2 2tan 1 tan secx x x+ = + =
23. ( )2 2 2 2 2sin cot 1 sin csc sinx x x x x+ = =2
1
sin x1
=
25. ( )( ) 2 2sin cos sin cos sin cosx x x x x x− + = −
27.
1
csc sin
cot
x x
x=
cos
sin
x
x
1sec
cosx
x= =
29. 1 cot( ) 1 cot
11 cot 1 cot
x x
x x
− − += =
+ +
31. ( ) ( )2 2
4
2
1 cos 1 cos1 cos
1 cos
x xx
x
− +−=
+ 21 cos x+
2 21 cos sinx x= − =
33 ( ) ( )2 2
4
2
1 cot 1 cot1 cot
1 cot
x xx
x
+ −−=
− 21 cot x−
2 21 cot cscx x= + =
35.
( )2 2 1 cossin 1 cos1 1 1
1 cos 1 cos
xx x
x x
−−− = − = −
− −
( )1 cos
1 cos
x
x
+
−( )1 1 cos 1 1 cos cosx x x= − + = − − = −
362
37. 2 2
2 2
sin cossin cos
tan cot sin coscos sin2cos 2cossin costan cot
cos sin
x xx x
x x x xx xx xx xx x
x x
−−
−+ = + =
+ +2 2sin cos
sin cos
x x
x x
+2
2 22 2 2 2
2 2
1
2 2
2cos
sin cos2cos sin cos 2cos
sin cos
sin cos 1
x
x xx x x x
x x
x x
=
+
−= + = − +
+
= + =
�������
39.
( ) ( )2 2
2
sin cos sin cos
sin 2sin cos
x x x x
x x x
+ + −
= +( )2 2
What remains after cancellation = 1
cos sin 2sin cosx x x x+ + −�������������
( )2
What remains after cancellation = 1
cos 2,x+ =�������������
as claimed.
41. ( )( ) ( )2 2 2csc 1 csc 1 csc 1 1 cot 1 cotx x x x x+ − = − = + − = , as claimed.
43. 2 2sin cos sin cos 1 1 1
tan cot csc seccos sin sin cos sin cos sin cos
x x x xx x x x
x x x x x x x x
+ + = + = = = =
,
as claimed.
45. ( )22 2 21 1 sin2 sin 1 cos 1 cos
sec coscos cos cos cos cos
xx x xx x
x x x x x
+ −− += = = + = +
47. [ ][ ] [ ][ ] 2 2cos( ) 1 1 cos cos 1 cos 1 cos 1 sinx x x x x x− − + = − + = − = −
49.
1 cos
sec( )cot sin( ) cos sin coscos( ) sin1
1csc( ) cos( ) sin cos sin
sin( )
x
x x x x x xx x
x x x x x
x
− −−= = = − = −
− −
−
51. 2 2
2 2
2 2
1 1 1 1sin cos 1
1 1csc sec
sin cos
x xx x
x x
+ = + = + = , as claimed.
53. ( ) ( )( )( )
2
2 2
1 sin 1 sin1 1 2 22sec
1 sin 1 sin 1 sin 1 sin 1 sin cos
x xx
x x x x x x
+ + −+ = = = =
− + + − −, as claimed.
55. ( )2 2 1 cossin 1 cos
1 cos 1 cos
xx x
x x
−−= =
− −
( )1 cos
1 cos
x
x
+
−1 cos x= + , as claimed.
363
57.
( )( ) 2 2
2
sec tan sec tan sec tansec tan
sec tan sec tan
1 tan
x x x x x xx x
x x x x
x
+ − −+ = =
− −
+=
( ) 2tan x− 1,
sec tan sec tanx x x x=
− −
as claimed.
59.
2cos sin1 sin
csc tan sin cossin cos1 cossec cot
cos sin
x xx
x x x xx xxx x
x x
−−
−= =
+ +2sin cos
sin cos
x x
x x
+
2
2
cos sin
sin cos
x x
x x
−=
+, as claimed.
61.
( )( )
( ) ( )
2 22
2 22
1 sin 1 sin sin sin 2cos 1 sin
cos 3 sin 41 sin 3
sin 1 sin 2
x x x xx x
x xx
x x
− + + − − −+ + = =+ − −− +
− + −=
(sin 2)x− −
sin 1 1 sin,
sin 2 2 sin(sin 2)
x x
x xx
+ += =
+ ++
as claimed.
63.
( )
( )
1
2 2
2 2 2
1 sin cos 1 sin cos 1 1sec tan cot
cos cos sin cos cos sin cos cos sin
1 1 1 csccsc ,
cos sin cos cos
x x x xx x x
x x x x x x x x x
xx
x x x x
= +
+ = + = =
= = =
�������
as claimed.
65. Observe that the left-side of the equation simplifies as follows:
( )( ) ( )
( )( )
2 2 2 2
2 2 2 2
cos tan sec tan sec cos tan sec
cos tan 1 tan cos
x x x x x x x x
x x x x
− + = −
= − + = −
Since 2cos x− is, in fact, never equal to 1 (being a non-positive quantity), the given
equation is a conditional.
67. Observe that
3
cot cot
1cot
csc cot 1 1 1 1 cos 1sin cot cot cos cot cot1 1sec tan sin sin tan sin tan
tan tancos cos x x
xx x xx x x x x xx x x x x x x
x xx x = =
= = = = =
So, the given equation is an identity.
69. The equation sin cos 2x x+ = is a conditional since, for instance, it does not hold if
x = 0 (since the left-side reduces to 1).
71. Observe that
( )2 2 2 2tan sec tan 1 tan 1 1x x x x− = − + = − ≠ .
So, the given equation is a conditional (which, incidentally, is never true).
364
73. The equation 2sin 1 cosx x= − is a conditional. To see this, observe that the right-
side reduces as follows: 2 21 cos sin sinx x x− = = . As such, the given equation is not
true for 32
x π= , for instance.
(Note: The equation IS an identity if one restricts attention to only those x-values for
which sin 0x > .)
75. Observe that 2 2sin cos 1 1x x+ = = . So, the given equation is an identity.
77. Conditional. Note that it is false when 0x = .
79. ( ) ( )2 2 2sec 1 tan tanx x xπ π π π= + = +
81. 2 2 2 2 2 21 tan 1 tan sec sech hθ θ θ θ+ = ⇒ = + = = .
83. Simplified the two fractions in Step 2 incorrectly. The correct computations are: 2
2
cos cos cos
sin cos sin cos sin1
cos cos
sin sin sin
cos sin cos sin cos1
sin sin
x x x
x x x x x
x x
x x x
x x x x x
x x
= =− −
−
= =− −−
85. Need to observe that 2tan 1x = is a conditional, which does not hold for all values of
x. Verifying the truth of the equation for a particular value of x (here, 4
x π= ) is not
enough to prove that it is an identity.
87. False. The equation 2sin 1x = is not an identity since even though it is true for
2x nπ π= + (n is an integer), it is false for all other values of x for which sin x is defined.
89. The equation 2cos 1 sinθ θ= − is true whenever cos 0θ > (since 21 sin θ− =
2cos cosθ θ= .) This occurs for those angles θ whose terminal side is in QI or QIV.
91. 2csc cscθ θ= − occurs for those angles θ whose terminal side is in QIII or QIV.
93. The equation sin( ) sin sinA B A B+ = + is not true in general. For instance, let
30A = � and 60B = �
. Then, ( ) ( )sin 30 60 sin 90 1+ = =� � � , whereas
( ) ( ) 1 3 1 3sin 30 sin 60 1
2 2 2
++ = + = ≠� �
.
95. No. For instance, take 4
A π= . Note that tan(2 )A is not defined, whereas
( )42 tan 2π = .
97.
( ) ( )2 2
2 2
sin cos sin cos
sin 2 sin cos
a x b x b x a x
a x ab x x
+ + −
= + 2 2 2 2cos sin 2 sin cosb x b x ab x x+ + −
( ) ( ) ( )( )
2 2
2 2 2 2 2 2 2 2 2 2 2 2
1
cos
sin cos sin cos
a x
a b x a b x a b x x a b
=
+
= + + + = + + = +�������
365
99. Observe that for any integer n,
( )( ) ( )2
2 2
1 1 1csc 2 sec .
sin 2 sin cosn
n
π
π πθ π θ
θ π θ θ+ + = = = =
+ + +
101. Observe that
[ ]
1 1csc 2 sec sin( ) sin( )
2 2sin 2 cos
2 2
1 1sin( )
sin cos2 2
1 1sin
sin cos2 2
1 1 1sin
cos sin
π ππ θ θ θ θ
π πθ π θ
θπ π
θ θ
θπ π
θ θ
θθ θ
− − ⋅ − ⋅ − = ⋅ ⋅ −
− + + − −
= ⋅ ⋅ −
− + − −
= − ⋅ −
+ −
= ⋅ ⋅ = seccos
θθ
=
103. The correct identity is cos( ) cos cos sin sinA B A B A B+ = − .
Consider the graphs of the following functions. Note that the graphs of 1y (BOLD) and
3y are the same. (The graph of 2
y is dotted.)
1 cos4
y xπ
= +
( ) ( )2 cos cos sin sin4 4
y x xπ π
= +
( ) ( )3 cos cos sin sin4 4
y x xπ π
= −
105. The correct identity is sin( ) sin cos cos sinA B A B A B+ = + .
Consider the graphs of the following functions. Note that the graphs of 1y (BOLD) and
2y are the same. (The graph of 3
y is dotted.)
1 sin4
y xπ
= +
( ) ( )2 sin cos cos sin4 4
y x xπ π
= +
( ) ( )3 sin cos cos sin4 4
y x xπ π
= −
366
107.
( )22 2
2
sin 1 sin
cos
cos ,
a a a
a
a
θ θ
θ
θ
− = −
=
=
since 2 2π πθ− ≤ ≤ .
109.
( )2 2 2
2
sec sec 1
tan
tan ,
a a a
a
a
θ θ
θ
θ
− = −
=
=
since 2
0 πθ≤ < .
Section 6.2 Solutions --------------------------------------------------------------------------------
1. Using the addition formula sin( ) sin cos cos sinA B A B A B− = − yields
sin sin sin cos cos sin12 3 4 3 4 3 4
3 2 1 2 2 3 1 6 2
2 2 2 2 2 2 2 4
π π π π π π π = − = −
− = − = − =
3. Using the even identity for cosine, along with the addition formula
cos( ) cos cos sin sinA B A B A B+ = − yields
5 5cos cos cos cos cos sin sin
12 12 6 4 6 4 6 4
3 2 1 2 2 3 1 6 2
2 2 2 2 2 2 2 4
π π π π π π π π − = = + = −
− = − = − =
5. First, we need to compute both sin12
π
and cos12
π
. Then, we use the definition
of tangent to compute tan12
π
. To this end, we have the following:
Step 1: Using the odd identity for sine and Exercise 1 yields
6 2sin sin
12 12 4
π π − − = − = −
Step 2: Using the even identity for cosine and the addition formula
cos( ) cos cos sin sinA B A B A B− = + yields
cos cos cos cos cos sin sin12 12 3 4 3 4 3 4
1 2 3 2 2 1 3 2 6
2 2 2 2 2 2 2 4
π π π π π π π π − = = − = +
+ = + = + =
Step 3: Using the definition of tangent and the computations in Steps 1 and 2 yields
6 2sin sin
4 6 212 12tan
12 6 2 6 2cos cos
12 12 4
6 2 6 2 6 2 2 6 2 8 4 32 3
6 2 46 2 6 2
π π
π
π π
− − − − − + − = = = = + + −
− + − − + − − += ⋅ = = = − +
−+ −
367
7. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + yields
( ) ( ) ( ) ( ) ( ) ( )sin 105 sin 60 45 sin 60 cos 45 cos 60 sin 45
3 2 1 2 2 6
2 2 2 2 4
= + = + =
+ = + =
� � � � � � �
9. First, we need to compute both ( )sin 105− � and ( )cos 105− � . Then, we use the
definition of tangent to compute ( )tan 105− � . To this end, we have the following:
Step 1: Using the odd identity for sine and Exercise 7 yields
( ) ( ) 2 6sin 105 sin 105
4
+− = − = −
� �
Step 2: Using the even identity for cosine and the addition formula
cos( ) cos cos sin sinA B A B A B+ = − yields
( ) ( ) ( ) ( ) ( ) ( ) ( )cos 105 cos 105 cos 60 45 cos 60 cos 45 sin 60 sin 45
1 2 3 2 2 6
2 2 2 2 4
− = = + = − =
− = − =
� � � � � � � �
Step 3: Using the definition of tangent and the computations in Steps 1 and 2 yields
( )( )( )
2 6
sin 105 4 6 2tan 105
cos 105 2 6 2 6
4
6 2 2 6 6 2 2 6 2 8 4 32 3
2 6 42 6 2 6
+−
− − − − = = =− − −
− − + − − − − −= ⋅ = = = +
− −− +
�
�
�
11. Using the odd identity of tangent, followed by Exercise 5, yields
1 1 1 1cot
12 2 3tan tantan12 1212
1 1 2 3 2 32 3
4 32 3 2 3 2 3
π
π ππ
= = = =
− − +− −− −
+ += = ⋅ = = +
−− − +
368
13. First, we need to compute 11
cos12
π −
. Then, we use the definition of secant to
compute11
sec12
π −
. To this end, we have the following:
Step 1: Using the even identity for cosine and the addition formula
cos( ) cos cos sin sinA B A B A B− = + yields
1 0
11 11 11cos cos cos
12 12 12
11 11cos cos sin sin cos
12 12 12
π π ππ
π π ππ π
=− =
− = = −
= + = −
(1)
Using the addition formula cos( ) cos cos sin sinA B A B A B− = + (again) now yields
cos cos cos cos sin sin12 3 4 3 4 3 4
1 2 3 2 2 1 3 2 6
2 2 2 2 2 2 2 4
π π π π π π π = − = +
+ = + = + =
Hence, using this in (1) yields 11 2 6
cos12 4
π + − = −
.
Step 2: Using the definition of secant and the computation in Step 1 now yields
( )
11 1 1 4sec
1112 2 62 6cos12 4
4 2 64 2 62 6
2 62 6 2 6
π
π
− − = = =
+ +− −
− −− −= ⋅ = = −
−+ −
15. ( )( ) ( ) ( ) ( ) ( )
( )
1 1csc 255
sin 45 300 sin 45 cos 300 cos 45 sin 300
1
2 1 2 3
2 2 2 2
4 2 66 2
2 62 1 3
− = =− −
=
− −
−= ⋅ = −
−+
�
� � � � � �
17. Using the addition formula cos( ) cos cos sin sinA B A B A B− = + with 2A x= and
3B x= , followed by the even identity for cosine at the very end, yields
sin 2 sin 3 cos 2 cos 3 cos(2 3 ) cos( ) cosx x x x x x x x+ = − = − = .
369
19. Using the addition formula sin( ) sin cos cos sinA B A B A B− = − with A x= and
2B x= , followed by the odd identity for sine at the very end, yields
sin cos 2 cos sin 2 sin( 2 ) sin( ) sinx x x x x x x x− = − = − = −
21. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + with A xπ= − and
B x= yields sin cos( ) cos sin( ) sin( ) sin 0x x x x x xπ π π π− + − = + − = = .
23. Observe that
( ) ( )
( )
2 2
2 2 2 2
2 2
1
sin sin cos cos 2
sin 2sin sin sin cos 2cos cos cos 2
sin cos
A B A B
A A B B A A B B
A A
=
− + − − =
− + + − + − =
+�������
( )2 2
1
sin cosB B
=
+ +�������
[ ]2 sin sin cos cos 2A B A B− + −
[ ] [ ]2 sin sin cos cos 2 cos cos sin sin 2cos( )A B A B A B A B A B
=
− + = − + = − −
25. Observe that
( ) ( )
( ) ( )
( )
2 2
2 2 2 2
2 2
1
2 sin cos cos sin
2 sin 2sin cos cos cos 2cos sin sin
2 sin cos
A B A B
A A B B A A B B
A A
=
− + − + =
− + + − + +
− +�������
( )2 2
1
sin cosB B
=
− +�������
[ ]
[ ]
2 sin cos cos sin
2 sin cos cos sin 2sin( )
A B A B
A B A B A B
− + =
− + = − +
27. Using the addition formula tan tan
tan( )1 tan tan
A BA B
A B
−− =
+ with 49A = �
and 23B = �
yields tan 49 tan 23
tan(49 23 ) tan(26 )1 tan 49 tan 23
−= − =
+
� �� � �
� �
370
29. Observe that 1 1
cos( ) cos cos sin sin sin sin3 4
α β α β α β α β
+ = − = − − −
. We
need to compute both sinα and sin β . We proceed as follows:
Since the terminal side of α is in QIII, we
know that sin 0α < . As such, since we
are given that 1
cos3
α = − , the diagram is:
From the Pythagorean Theorem, we have
2 2 2( 1) 3z + − = , so that 2 2z = − .
Thus, 2 2sin
3α = − .
Since the terminal side of β is in QII, we
know that sin 0β > . As such, since we
are given that 1
cos4
β = − , the diagram is:
From the Pythagorean Theorem, we have
2 2 2( 1) 4z + − = , so that 15z = .
Thus, 15sin
4β = .
Hence,
1 1 1 1 2 2 15 1 2 30 1 2 30cos( ) sin sin
3 4 3 4 3 4 12 12 12α β α β
+ + = − − − = − − − − = + =
α β
371
31. Observe that 3 1
sin( ) sin cos cos sin cos cos5 5
α β α β α β β α
− = − = − −
. We
need to compute both cosα and cos β . We proceed as follows:
Since the terminal side of α is in QIII, we
know that cos 0α < . As such, since we
are given that 3
sin5
α = − , the diagram is
as follows:
From the Pythagorean Theorem, we have
2 2 23 5z + = , so that 4z = − .
Thus, 4
cos5
α = − .
Since the terminal side of β is in QI, we
know that cos 0β > . As such, since we
are given that 1
sin5
β = , the diagram is as
follows:
From the Pythagorean Theorem, we have
2 2 21 5z + = , so that 24 2 6z = = .
Thus, 2 6
cos5
β = .
Hence,
3 1 3 2 6 4 1 6 6 4 6 6 4sin( ) cos cos
5 5 5 5 5 5 25 25 25α β β α
− − + − = − − = − − − = + =
αβ
372
33. Observe that sincetan tan
tan( )1 tan tan
α βα β
α β
++ =
−, we need to determine the values of
cosα and sin β so that we can then use this information in combination with the given
values of sinα and cos β in order to compute tan and tanα β . We proceed as
follows:
Since the terminal side of α is in QIII, we
know that cos 0α < . As such, since we
are given that 3
sin5
α = − , the diagram is:
From the Pythagorean Theorem, we have
2 2 2( 3) 5z + − = , so that 4z = − .
As such, 4
cos5
α = − , and so,
3sin 35tan
4cos 45
αα
α
−= = =
−
Since the terminal side of β is in QII, we
know that sin 0β > . As such, since we
are given that 1cos
4β = − , the diagram is:
From the Pythagorean Theorem, we have
2 2 2( 1) 4z + − = , so that 15z = .
As such, 15
sin4
β = , and so
15sin 4tan 15
1cos4
ββ
β= = = −
−
Hence,
( )
3 3 4 1515
tan tan 4 4tan( )31 tan tan 4 3 15
1 154 4
3 4 15 4 3 15 12 16 15 9 15 12(15) 192 25 15
16 9(15) 1194 3 15 4 3 15
α βα β
α β
−−
++ = = =
− +− −
− − − − + −= ⋅ = =
− −+ −
35. Observe that
sin( ) sin( ) (sin cos sin cosA B A B A B B A+ + − = + ) (sin cos sin cosA B B A+ − )
2sin cosA B=
So, the given equation is an identity.
α
β
373
37. The equation sin cos2 2
x xπ π
− = +
is a conditional since it is false for 2
xπ
= , for
instance.
39. Observe that ( ) ( ) ( ) ( )2
4 4 4 2sin sin cos sin cos sin cosx x x x xπ π π+ = + = + . So, the
given equation is an identity.
41. Observe that
( ) ( )( )2 2
22 2
1 cos 1 cos1 cos cos sin sin1 cos 2
2 2 2
2 2cos1 cos sin
2
x xx x x xx
xx x
− − −− −−= =
−= = − =
So, the given equation is an identity.
43. Observe that
sin 2 sin( ) sin cos sin cos 2sin cosx x x x x x x x x= + = + = .
So, the given equation is an identity.
45. The equation sin( ) sin sinA B A B+ = + is a conditional since it is false for
2A B
π= = , for instance.
47. Observe that
( )tan tan 0 tan
tan( ) tan1 tan tan 1 0 tan
π β βπ β β
π β β
+ ++ = = =
− −.
So, the given equation is an identity.
49. Observe that
1 1cot(3 )
tan(3 ) tanx
x xπ
π+ = =
+.
So, the given equation is an identity.
51. The given equation is a conditional since it is false when 4
x π= , for instance.
53. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + with A x= and
3B
π= yields cos sin sin cos sin
3 3 3y x x x
π π π = + = +
.
So, the graph is the graph of siny x= shifted to the left 3
π units, as seen below.
374
55. Using the addition formula cos( ) cos cos sin sinA B A B A B− = + with A x= and
4B
π= yields
sin sin cos cos cos cos sin sin cos4 4 4 4 4
y x x x x xπ π π π π
= + = + = −
.
So, the graph is the graph of cosy x= shifted to the right 4
π units, as seen below.
57. Using the addition formula sin( ) sin cos cos sinA B A B A B+ = + with A x= and
3B x= yields
[ ] ( )sin cos3 cos sin 3 sin cos3 cos sin 3 sin 3 sin 4y x x x x x x x x x x x= − − = − + = − + = −
So, the general shape of the graph is that of siny x= , but with period is 2
4 2
π π= and
then reflected over the x-axis, as seen below.
59. Observe that ( )4
1 tantan
1 tan
xx
x
π+= −
−.
So, the graph is as follows:
61. Observe that ( )6
1 3 tantan
3 tan
xx
x
π+= +
−.
So, the graph is as follows:
63. ( ) ( ) ( )2 3 4 52 2
4 4 4 2 2 2! 3! 4! 5!cos cos cos sin sin cos sin 1 ...x x x xx x x x x xπ π π− = + = + = + − − + + − − + +
375
65. Observe that ( ) 2 1 2 12 1
2 1 2 1
tan tantan
1 tan tan 1
m m
m m
θ θθ θ
θ θ
− −− = =
+ +.
67. ( ) ( ) ( ) ( ) ( )cos cos cos sin sinE A kz ct A kz ct kz ct= − = +
If z
nλ
= , an integer, then z nλ= , so that 2
2kz kn kn nk
πλ π
= = =
. Hence,
sin( ) 0kz = . So, the formula for E simplifies to ( ) ( ) ( )cos cos cosE A kz ct A kz ct= − = .
69. tan( ) tan tanA B A B+ ≠ + . Should have used tan tan
tan( )1 tan tan
A BA B
A B
++ =
−.
71. False. Observe that
( ) ( ) ( ) ( ) ( ) ( )cos 15 cos 45 30 cos 45 cos 30 sin 45 sin 30
2 3 2 1 6 2
2 2 2 2 4
= − = +
+ = + =
� � � � � � �
while
( ) ( ) 2 3 2 3cos 45 cos 30
2 2 2
−− = − =
� � .
73. False. Observe that ( )4
1 tantan
1 tan
xx
x
π+= +
−.
75. Approaches may vary slightly here, but they will result in equivalent answers as
long as the identities are applied correctly. The final form of the simplification depends
on how you initially choose to group the input quantities. For instance, the following is
a legitimate simplification:
( ) ( )( )[ ] [ ]
sin sin sin( )cos sin cos( )
sin cos sin cos cos sin cos cos sin sin
sin cos cos sin cos cos cos cos sin sin sin sin
A B C A B C A B C C A B
A B B A C C A B A B
A B C B A C A B C A B C
+ + = + + = + + +
= + + −
= + + −
77. Observe that sin( ) sin cos sin cosA B A B B A− = − . We seek values of A and B such
that the right-side equals sin sinA B− . Certainly, if we choose values of A and B such
that cos cos 1B A= = , then this occurs. And this occurs precisely when A and B are
integer multiples of 2π (and they need not be the same multiple of 2π !). So, the
solutions are 2 , 2B m A nπ π= = , where n and m are integers.
376
79. a. The following is the graph of
1
sin1 1 cos1cos sin
1 1y x x
− = −
, along
with the graph of cosy x= (dotted
graph):
b. The following is the graph of
2
sin 0.1 1 cos 0.1cos sin
0.1 0.1y x x
− = −
,
along with the graph of cosy x= (dotted
graph):
c. The following is the graph of
3
sin 0.01 1 cos 0.01cos sin
0.01 0.01y x x
− = −
,
along with the graph of cosy x= (dotted
graph):
Refer back to Exercise 63 for a
development of this difference quotient
formula. Looking at the sequence of graphs
in parts a – c, it is clear that this sequence of
graphs better approximates the graph of
cosy x= as 0h → .
377
81. Observe that
( )
( )
( )
sin 2( ) sin 2 sin 2 cos(2 ) sin(2 ) cos2 sin 2
sin 2 cos(2 ) 1 sin(2 ) cos 2
sin 2 cos(2 ) 1 sin(2 ) cos 2
sin(2 ) 1 cos(2 )cos 2 sin 2
x h x x h h x x
h h
x h h x
h
x h h x
h h
h hx x
h h
+ − + −=
− +=
−= +
− = −
a. The following is the graph of
1
sin(2) 1 cos(2)cos 2 sin 2
1 1y x x
− = −
,
along with the graph of 2cos 2y x=
(dotted graph):
b. The following is the graph of
2
sin(0.2) 1 cos(0.2)cos 2 sin 2
0.1 0.1y x x
− = −
,
along with the graph of 2cos 2y x= (dotted
graph):
c. The following is the graph of
3
sin(0.02) 1 cos(0.02)cos2 sin 2
0.01 0.01y x x
− = −
,
along with the graph of 2cos 2y x=
(dotted graph):
Looking at the sequence of graphs in parts a
– c, it is clear that this sequence of graphs
better approximates the graph of
2cos 2y x= as 0h → .
378
83. Observe that ( ) ( )
( )
sin cos( ) sin sin( )cos sin cos( ) 1 sin( )cossin( ) sin sin cos( ) sin( ) cos sin
sin cos( ) 1 sin( )cos cos( ) 1 sin( ) sin( ) 1 cos( )sin cos cos sin
x h x h x x h h xx h x x h h x x
h h h h
x h h x h h h hx x x x
h h h h h h
− + − ++ − + −= = =
− − − = + = + = −
85. sin( ) 0
sin cos sin cos 0
sin cos sin cos
sin sin
cos cos
tan tan
x y
x y y x
x y y x
x y
x y
x y
+ =
+ =
= −
= −
= −
87.
( )
tan( ) 2
tan tan2
1 tan tan
tan tan 2 2 tan tan
tan 1 2 tan 2 tan
2 tantan
1 2 tan
x y
x y
x y
x y x y
x y y
yx
y
+ =
+=
−
+ = −
+ = −
−=
+
Section 6.3 Solutions --------------------------------------------------------------------------------
1. Since 1
sin5
x = and cos 0x < , the
terminal side of x must be in QII.
The diagram is as follows:
From the Pythagorean Theorem, we have
( )2
2 21 5z + = , so that 2z = − .
As such, 2
cos5
x = − . Therefore,
1 2 4sin 2 2sin cos 2
55 5x x x
= = − = −
3. Since 5
cos13
x = and sin 0x < , the
terminal side of x must be in QIV.
The diagram is as follows:
From the Pythagorean Theorem, we have
2 2 25 13z + = , so that 12z = − .
As such, 12
sin13
x = − , so that
12
sin 1213tan5cos 5
13
xx
x
−= = = − .
Therefore,
22
12 2422 tan 1205 5tan 2
1191 tan 119121
255
xx
x
− − = = = =
− −− −
5
379
5. Since 12 12
tan5 5
x−
= =−
and
3
2x
ππ < < , the diagram is as follows:
From the Pythagorean Theorem, we have
2 2 2( 5) ( 12) z− + − = , so that 13z = .
As such, 12
sin13
x = − and 5
cos13
x = − .
Therefore,
12 5sin 2 2sin cos 2
13 13
120
169
x x x
= = − −
=
7. Since sec 5x = , it follows that
15
cos x= and so,
1cos
5x = . Since
also sin 0x > , the terminal side of x must
be in QI. The diagram is as follows:
From the Pythagorean Theorem, we have
( )2
2 21 5z + = , so that 2z = .
As such, 2
sin5
x = , so that
2
sin 5tan 2
1cos
5
xx
x= = = .
Therefore,
2 2
2 tan 2(2) 4 4tan 2
1 tan 1 2 3 3
xx
x= = = = −
− − −.
5
380
9. Since csc 2 5x = − , it follows that
12 5
sin x= − and so, 1 5
sin102 5
x = − = − .
Since also cos 0x < , the terminal side of x
must be in QIII. The diagram is as
follows:
From the Pythagorean Theorem, we have
( ) ( )2 22 5 10z + − = , so that 95z = − .
As such, 95
cos10
x = − . Therefore,
5 95sin 2 2sin cos 2
10 10
10 19 19
100 10
x x x
= = − −
= =
11. Since csc 0x < , it follows that
sin 0x < . Since also12
cos13
x = − , the
terminal side of x must be in QIII. The
diagram is as follows:
From the Pythagorean Theorem, we have
( )22 212 13z + − = , so that 5z = − .
As such, 5
sin13
x = − , so that
5
sin 513tan12cos 12
13
xx
x
−= = =
−
.
Therefore, 2
2
2
1 1 1 tancot 2
2 tantan 2 2 tan
1 tan
51
119 6 11912
5 144 5 1202
12
xx
xx x
x
−= = =
−
− = = ⋅ =
.
13. Using 2
2 tantan 2
1 tan
AA
A=
− with 15A = �
yields
( )2
12 tan15 sin 30 32tan 2 15 tan 30
1 tan 15 cos30 33
2
= ⋅ = = = =−
� �� �
� �
5−
381
15. Using sin 2 2sin cosA A A= with 8
Aπ
= yields
1 1 1 2 2sin cos 2sin cos sin
8 8 2 8 8 2 4 2 2 4
π π π π π = = = =
17. Using 2 2cos2 cos sinA A A= − with 2A x= yields
2 2cos 2 sin 2 cos(2 2 ) cos 4x x x x− = ⋅ =
19. ( )
( )( ) ( )
( )( )
5 5 112 6 25 5
12 62 5 5 3212 6
2 tan sin 3tan 2 tan
1 tan cos 3
π ππ π
π π= ⋅ = = = = −
− −
21. ( ) ( ) ( )2 37 7 712 12 6 2
1 2sin cos 2 cosπ π π− = ⋅ = = −
23. ( ) ( ) ( ) ( )2 37 7 7 712 12 6 6 2
2cos 1 cos 2 cos cosπ π π π− − = − ⋅ = − = = −
25. Observe that
1 1 1 1 1 1csc 2 csc sec
sin 2 2sin cos 2 sin cos 2A A A
A A A A A
= = = =
.
27. Observe that
( )( ) 2 2 2 2sin cos cos sin sin cos cos sin cos 2x x x x x x x x x − + = − = − − = − .
29. Note that starting with the right-side is easier this time. Indeed, observe that
( ) ( )2 2 2 2
2 2 22
1 cos sin 1 sin cos1 cos 2
2 2 2
cos cos 2coscos
2 2
x x x xx
x x xx
+ − − ++= =
+= = =
31. Observe that
( )( )4 4 2 2 2 2
1
cos sin cos sin cos sin cos 2x x x x x x x
=
− = − + =�������
33. Observe that
[ ]
[ ]
( )
2 2 2 2
2
2
2
2
2
2 2 2
2 2
2 2
8sin cos 2 4sin cos
2 2sin cos
2 sin 2
2 1 cos 2
2 2cos 2
1 1 2cos 2
cos 2 sin 2 1 2cos 2
1 cos 2 sin 2
1 cos 2 sin 2
1 cos(2 2 ) 1 cos 4
x x x x
x x
x
x
x
x
x x x
x x
x x
x x
= ⋅
=
=
= −
= −
= + −
= + + −
= − +
= − −
= − ⋅ = −
382
35. Note that starting with the right-side is easier this time. Indeed, observe that
[ ]
2 2 22 2
22
2sin 2sin 2 sin2sin csc 2
sin 2 2sin cos
x x xx x
x x x
− − −− = = =
24 sin x
2
22
1 1sec
2cos 2cosx
xx= − = −
37. Note that starting with the right-side is easier this time. Indeed, observe that
( ) ( )
( )( )( )
( )
2 2 2
2 2
2 2
2 3
2 2 2
sin 4cos 1 sin 3cos cos 1
sin 3cos 1 cos
sin 3cos sin
3sin cos sin
sin cos sin 2sin cos
sin cos 2 (2sin cos ) cos
sin cos 2 sin 2 cos
sin( 2 ) sin 3
x x x x x
x x x
x x x
x x x
x x x x x
x x x x x
x x x x
x x x
− = + −
= − −
= −
= −
= − +
= +
= +
= + =
39. Note that starting with the right-side is easier this time. Indeed, observe that
( )
( )
( )
2
3 2
2 2
cos
2 2
2sin cos 4sin cos 2sin cos 1 2sin
2sin cos 1 sin sin
sin 2 cos sin
sin 2 cos 2
12sin 2 cos 2
2
1sin 4
2
x
x x x x x x x
x x x x
x x x
x x
x x
x
=
− = −
= − −
= −
=
=
=
�����
383
41. Before graphing, observe that
( ) ( ) ( )2 2 2 2 2 2
2
sin 2 2sin cos 2sin cos
1 cos 2 1 cos sin cos sin cos sin
2sin cos coscot
2sin sin
x x x x xy
x x x x x x x
x x xx
x x
= = =− − − + − −
= = =
So, a snapshot of the graph is as follows:
43. Before graphing, observe that
2 2cos sincos sin
cot tan sin cossin coscos sincot tan
sin cos
x xx x
x x x xx xyx xx x
x x
++
+= = =
−−
2 2cos sin
sin cos
x x
x x
−
1
2 2
2 2
cos 2
cos sin 1sec 2
cos sin cos 2x
x xx
x x x
=
=
+= = =
−
�������
�������
So, the period of this function is 2
2
ππ= . The graph is as follows:
384
45. Before graphing, observe that 12
sin(2 )cos(2 ) sin(4 )x x x= . The graph is
as follows:
47. Before graphing, observe that
12
tan cot 11 1
1 1sec csc
cos sin
1 sin cos
1 sin(2 )
x x
x x
x x
x x
x
− = −
⋅
= −
= −
The graph is as follows:
49. Before graphing, observe that
sin 2 2sin cos3cos 2
cos
x x xx
x− =
cos x3cos 2
2sin 3cos 2
x
x x
−
= −
The graph is as follows:
51. Use 1 cos
sin2 2
A A− =
with
30A = � to obtain
( ) 30 1 cos30sin 15 sin
2 2
31
2 32
2 4
2 3
2
−= =
−−
= =
−=
� ��
.
53. Use 1 cos
cos2 2
A A+ = −
with
11
6A
π= to obtain
11 11 31 cos 1
11 2 3 2 36 6 2cos cos12 2 2 2 4 2
π ππ
+ + + +
= = − = − = − = −
.
385
55. Use 1 cos
cos2 2
A A+ =
with 150A = �
to obtain
( )3
1150 1 cos150 2 3 2 32cos 75 cos
2 2 2 4 2
− + − −= = = = =
� ��
.
57. Use 1 cos
tan2 1 cos
A A
A
− =
+ with 135A = �
to obtain
( ) 135 1 cos135tan 67.5 tan
2 1 cos135
−= =
+
� ��
�.
Since 2
cos135 cos 452
= − = −� � , the above further simplifies to
( )
2 2 212 2
tan 67.52
12
+− − = =
+ −
�
2 2
2
−
2 2 2 2 4 4 2 23 2 2
4 22 2 2 2
+ + + += ⋅ = = +
−− +
59. Observe that
9 9 1 1sec sec
9 98 8cos 4cos8
2
π π
π π
− = = =
(1).
Also, note that 9 2
cos cos 2 cos4 4 4 2
π π ππ
= + = =
. Hence, using the formula
1 coscos
2 2
A A+ = −
with
9
4A
π= gives
9 21 cos9 12 244 2cos
2 2 2 2
ππ
+ + + = − = − = −
.
Using this in (1) then yields: 9 1 2sec
8 2 2 2 2
2
π − = = − + +
−
386
61. First, note that using 1 costan
2 1 cos
A A
A
− =
+ with 13
4A
π= , we obtain
( )( )
13 1 1 1cot
13 138 131 costan 4 4tan82 131 cos
4
π
π π π
π
= = =
− +
(1)
Next, observe that ( ) ( ) ( ) 213 5 5cos cos 2 cos4 4 4 2
π π ππ= + = = − . Using this in (1)
subsequently yields:
13 1 1cot
8 2 221
22
21
2
π = =
+− −
+ −
2 2
2
−
1 1 1 1
2 2 2 2 2 2 4 4 2 2 3 2 2
4 22 2 2 2 2 2
−= = = =
+ + + + + +⋅
−− − +
Note: Recall that there are two other formulae equivalent to 1 cos
tan2 1 cos
A A
A
− =
+ , and
either one could be used in place of this one to obtain an equivalent (but perhaps
different looking) result. For instance, using sin
tan2 1 cos
A A
A
=
+ with 13
4A
π= yields
( )( )
13 1 1 1cot
13 13 138 sintan 4 4tan82 131 cos
4
π
π π π
π
= = =
+
(2)
Since
( ) ( ) ( ) 213 5 5cos cos 2 cos4 4 4 2
π π ππ= + = = −
( ) ( ) ( ) 213 5 5sin sin 2 sin4 4 4 2
π π ππ= + = = −
we can use these in (2) to obtain
13 1 2 2 2 2 2 2 2 2cot 1 2
8 22 2 2 2
2
21
2
π − − − = = = ⋅ = = −
−− − −
−
.
Even though the forms of the two answers are different, they are equivalent.
387
63. First, note that using the formula 1 cos
cos2 2
A A+ = −
with
5
4A
π= yields
5 21 cos5 15 2 244 2cos cos8 2 2 2 2
ππ
π
+ − − = = − = − = −
.
Then, we have
5 1 1 2sec
58 2 2 2 2cos8 2
π
π
− = = =
− − −
.
65. Observe that
( ) ( )( ) ( ) ( )
( )
2702 270 270
2 2
1 1 1 1cot 135 cot 1
1 0tan tan 1 cos 270
1 01 cos 270
− = − = = − = = =−− −+
+
�
� �
�
�
�
.
67. Since 5
cos13
x = and sin 0x < , the terminal side of x lies in QIV. Hence, the
terminal side of 2
x lies in QII, and so sin 0
2
x >
. Therefore, we have
51
1 cos 13 5 8 2 2 1313sin2 2 2 26 26 1313
x x−
− − = = = = = =
.
69. Since 12
tan5
x = and 3
2x
ππ < < , we
know that sin 0x < and cos 0x < . Also,
note that since 3
2 2 4
xπ π< < , sin 0
2
x >
.
Consider the following diagram:
We see from the diagram that
5cos
13x = − . So, we have
51
1 cos 13sin
2 2 2
3 3 13
1313
x x
− − − = =
= =
388
71. Since sec 5x = , we know that
1cos
5x = . This, together with sin 0x > ,
implies that the terminal side of x is in QI.
Hence, the terminal side of 2
x is also in
QI. Thus, tan 02
x >
.
Consider the following diagram:
We see from the diagram that 2sin
5x = .
So, we have
( )
2
sin 25tan
12 1 cos 5 115
2 5 1 5 1
5 1 2
x x
x
= = =
+ + +
− −= =
−
Alternatively, we could have used one of
the other two equivalent formulae for
tan2
x
. Specifically, the following
computation is also valid:
11
1 cos 5 15tan
12 1 cos 5 115
5 1 5 1 6 2 5
45 1 5 1
3 5
2
x x
x
−− −
= = = + + +
− − −= ⋅ =
+ −
−=
73. Since csc 3x = , we know that 1sin
3x = .
This, together with cos 0x < , implies that
the terminal side of x is in QII. Hence, the
terminal side of 2
x is in QI. So, sin 02
x >
.
Consider the following diagram:
We see from the diagram that
2 2cos
3x = − . So, we have
2 21
31 cossin
2 2 2
3 2 2
6
x x
− −
− = =
+=
5
2 2−
389
75. Since csc 0x < , it follows that
sin 0x < . This, together with 1cos
4x = − ,
implies that the terminal side of x is in
QIII. Hence, the terminal side of 2
x is in
QII. Thus, cot 02
x <
.
Consider the following diagram:
We see from the diagram that
15sin
4x = − . So, we have
1 1cot
2 1 costan
2 1 cos
1 1 3
5511 4
4 341
14
15
5
x
x x
x
= = =
− −
+
−= = = −
− − −
+ −
= −
77. Since 24
cot5
x = − and 2
xπ
π< < , we
know that sin 0x > and cos 0x < . Also,
note that since 02 2
x π< < , cos 0
2
x >
.
Consider the following diagram:
We see from the diagram that
24cos
601x = − . So, we have
241
601cos
2 2
x−
=
15−
601
390
79. Since sec 0x > , it follows that
cos 0x > . This, together with
sin 0.3x = − , implies that the terminal side
of x is in QIV. Hence, the terminal side of
2
x is in QII. Thus, tan 0
2
x <
.
Consider the following diagram:
We see from the diagram that
cos 0.91x = . So, we have
1 cos 1 0.91tan
2 1 cos 1 0.91
x x
x
− − = − = −
+ + .
81. Since sec 2.5x = , we know that
1cos 0.4
2.5x = = . This, together with
tan 0x > , implies that the terminal side of
x is in QI. Hence, the terminal side of 2
x
is in QI. Thus, cot 02
x >
.
Consider the following diagram:
We have
1 1 7cot
2 31 0.4tan
2 1 0.4
x
x
= = =
− +
83. Using the formula 1 cos
cos2 2
A A+ =
with
5
6A
π= yields
51 cos 5
56 6cos cos2 2 12
ππ
π
+ = =
.
0.91
5.25
391
85. Using the formula sin
tan2 1 cos
A A
A
=
+ with 150A = �
yields
( )sin150 150
tan tan 751 cos150 2
= =
+
� ��
�.
87. ( )5 54 4 5
854
1 costan tan
1 cos 2
π ππ
π
− = − = −
+
89. Use the known Pythagorean identity 2 2sin cos 1θ θ+ = with
2
xθ = to conclude that
the given equation is in fact an identity.
91. Observe that
( ) ( ) ( )2 2 22sin cos sin 2 sin sin( )x x x x x− = − ⋅ = − = − .
93. Observe that
22
2 1 cos 1 costan tan
2 2 1 cos 1 cos
x x x x
x x
− − = = ± = + +
.
95. Observe that
( )( )( )
( ) ( )
( )( )
( )( ) ( )( )( )
( )
22 22
2
sin 1 sin 1 costan cot
sin2 2 1 cos 1 cos sin
1 cos
1 cos 1 cossin 1 cos
sin 1 cos sin 1 cos
1 cos 1 cos 1 cos
sin 1 cos
1 cos
A A A A A
AA A A
A
A AA A
A A A A
A A A
A A
A
+ + = + = +
+ + +
− + ++ += =
+ +
− + + +=
+
+=
( ) ( )
( ) ( )
1 cos 1 cos
sin 1 cos
A A
A A
− + +
+
22csc
sinA
A= =
97. Observe that
( ) ( )
2
22
2 2
1 1 2csc
2 1 cos1 cossin2 2
2 1 cos 2 1 cos2 1 cos
1 cos 1 cos 1 cos sin
A
A AA
A AA
A A A A
= = =
− −
+ ++= ⋅ = =
− + −
99. ( )
( )2
2
2 1 coscsc csc 2 2cos csc 2 2cos
2 sin
AAA A A A
A
+ = ± = ± + = ± +
.
392
101. Before graphing, observe that 2
2 1 cos 1 cos4cos 4 4 2 2cos
2 2 2
x x xy x
+ + = = = = +
.
Observe that the period is 2π , the amplitude is 2, there is no phase shift, and the vertical
shift is up 2 units. So, following the approach in Chapter 6, we see that the graph is as
follows:
103. Before graphing, observe that
( ) ( )
( )2
2
1 cos 1 cos1 cos
1 tan 1 1 cos2 1 cos
1 cos1 tan 1
2 1 cos
x xx x
xxy
x x
x
+ − −−
− − ++ = = =−
+ + +
( ) ( )
( )
1 cos 1 cos
1 cos
x x
x
+ + −
+
2coscos
2
xx= = .
So, the graph is as follows:
393
105. Before graphing, observe that
( )( )22
2
1 cos 24sin 1 4 1 2 2cos 1 1 2cos
2
x
xy x x − ⋅
= − = − = − − = −
.
The graph is as shown below:
107. Before graphing, observe that
( )22
1 cos 2tan tan
1 cos 2
xx
y xx
−= = =
+.
The graph is as shown below:
109. Observe that
( )21500sin 60 (200) 750 1 cos120 3.75sin120
2
3 40,000 1 3500 750 1 3.75
2 2 2 2
250 3 22,500,000 37,500 3
22,565,385 lbs.
F = + − +
= + − − +
= + +
≈
� � �
111. Observe that 2
2 1sin cos 2sin cos sin
2 2 2 2 2 2
aA bh a a a
θ θ θ θθ
= = = ⋅ ⋅ =
394
113. Consider the following diagram:
Observe that
2
2 tan 4 1tan 2 tan
1 tan
BB B
B y y= = =
−(1) (2)
Substituting (2) into (1) yields the following equation that we solve for y:
( )( ) ( )
2
2 2
1 2 2 2
2 21 11
2 2 2 2
2 4 4 4 2 4
1 11
2 4 4 4 2 2 2 ft.
y y y
y
y yy
y
y y y yy y
y y y y y
−= ⇒ = ⇒ = ⇒ =
− −−
⇒ = − ⇒ = ⇒ = ⇒ =
115. Should use 2 2
sin3
x = − since we are assuming that sin 0x < .
117. If 3
2x
ππ < < , then
3
2 2 4
xπ π< < , so that sin
2
x
is positive, not negative.
119. False. Let 4
Aπ
= . Observe that
( )
sin 2 sin 14 2
sin 4 sin 04
π π
ππ
⋅ = =
⋅ = =
Since 1 1 0+ ≠ , the statement is false.
121. False. Note that 2
2 tantan 2
1 tan
xx
x=
−.
Take any ,4 2
xπ π
∈
. For such x,
tan 1x > , so that 21 tan 0x− < . So, for
these values of x, tan 2 0x < .
123. False. Use A π= , for instance. Indeed, observe that sin sin 22 2
π π + =
, while
sin 0π = .
125. False. Use 5
4x
π= , for instance. Observe that
5tan 1 0
4
π = >
, while
5tan 0
8
π <
since the terminal side of
5
8
πis in QII.
395
127. Observe that
( )
( )
( ) ( ) ( )( )
2
2
2
2
2 2
1 cos1 cos 1 cos 1 costan
2 1 cos 1 cos 1 cos 1 cos
1 cos 1 cos 1 cos 1 cos
sin sin sin 1 cos
1 cos sin sin
sin 1 cos sin 1 cos 1 cos
AA A A A
A A A A
A A A A
A A A A
A A A
A A A A A
−− − − = ± = ± ⋅ = ±
+ + − −
− − − += ± = ± = ± ⋅
+
−= ± = ± = ±
+ + +
Case 1: The terminal side of 2
A is in QI or QIII.
Here, we use sin
tan2 1 cos
A A
A
=
+ . Further, for such angles A, both sin 0A > and
1 cos 0A+ ≥ . Hence, sin
01 cos
A
A>
+, so that we have
sin sintan
2 1 cos 1 cos
A A A
A A
= =
+ + .
Case 2: The terminal side of 2
A is in QII or QIV.
Here, we use sin
tan2 1 cos
A A
A
= −
+ . Further, for such angles A, both sin 0A < and
1 cos 0A+ ≥ . Hence, sin
01 cos
A
A<
+, so that we have
sin sin sintan
2 1 cos 1 cos 1 cos
A A A A
A A A
= − = − − =
+ + + .
So, in either case, we conclude that sin
tan2 1 cos
A A
A
=
+ .
One cannot evaluate the identity at A π= since the denominator on the right-side would
be 0 (hence, the fraction is not well-defined) and 2
π is not in the domain of tangent.
129. One cannot verify the identity for this value of x since it does not belong to the
domains of the functions involved.
131. Observe that
( )( ) ( ) ( )
( )
( )( )
2
4
4 224
2
2
1 cos1 1 1cot
tan 1 cos1 costan
1 cos
A
A
A AA
A
A
+= = = = ±
−−±
+
.
133. Observe that ( )2tan 0x > when 3
2 2 2 20 or x xπ ππ< < < < . So, the desired values of
x in [ ]0, 2π for which is true are 0 x π< < .
396
135. Consider the following graphs:
Observe that 1
y (dotted graph) is a good
approximation of 2y on [ 1,1]− .
137. Consider the graphs of the following
functions: 3 5
1
2 2
2 3! 5!
x x
xy
= − +
2 sin2
xy
=
Observe that 1
y (dotted graph) is a good
approximation of 2y on [ 1,1]− .
139. 2
2 22 2
2
2sincos 2sin cos 2sin cos sin2
cos cos2 cos2cos sincos sin
cos
tan 2xx x x x x x
x x xx xx x
x
x−−
= ⋅ = = =
141.
( ) ( )
2 3 2 3 2
2 2
3sin cos sin sin cos sin 2sin cos
sin cos sin 2sin cos cos
sin cos 2 sin 2 cos
sin( 2 )
sin 3
x x x x x x x x
x x x x x x
x x x x
x x
x
− = − +
= − +
= +
= +
=