Answers and Explanations 173
Answers
101–200
Continue, using the identity 1 – cos
2 
x = sin
2 
x to simplify the denominator:
=
+
= +
=
+
=
+
sin ( cos )
sin
cos
sin
sin
cos
sin
csc
cot
x
x
x
x
x
x
x
x
x
x
1
1
1
2
132.
4
3
3
cos
c os
θ
θ
−
Use the identity cos(A + B) = cos A cos B – sin A sin B:
cos( )
cos(
)
cos cos
sin sin
3
2
2
2
θ
θ θ
θ
θ
θ
θ
=
+
=
−
Then use the identities cos( )
cos
2
2
1
2
θ
θ
=
− and sin( )
sin cos
2
2
θ
θ
θ
=
and simplify:
cos cos
sin sin
cos
cos
s in ( sin cos )
cos
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
2
2
2
1
2
2
2
3
−
=
−
(
) −
=
−c cos
sin cos
cos
cos
cos
cos
cos
cos
c
θ
θ
θ
θ
θ
θ
θ
θ
θ
−
=
−
−
−
(
)
=
−
−
2
2
21
2
2
2
2
3
3
o os
cos
cos
cos
θ
θ
θ
θ
+
=
−
2
4
3
3
3
133.
π
6
,
5
6
π
Solve for sin x:
2
1 0
1
2
sin
sin
x
x
− =
=
The solutions are x = π
6
and 5
6
π in the interval [0, 2π].
134.
0, π, 2π
Make one side of the equation zero, use the identity tan
sin
cos
x
x
x
=
, and then factor:
sin
tan
sin
tan
sin
sin
cos
sin
cos
x
x
x
x
x
x
x
x
x
=
−
=
−
=
−
(
) =
0
0
1
1
0
Setting the first factor equal to zero gives you sin x = 0, which has the solutions x = 0, π,
and 2π. Setting the second factor equal to zero gives you 1
1
0
−
=
cos x
so that cos x = 1,
which has the solutions x = 0 and 2π.