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Mathematics
Mathematics
Problem
Let
θ
=
3
π
2
+
α
\theta = \dfrac{3\pi}{2} + \alpha
θ
=
2
3
π
+
α
, where
0
<
α
<
π
2
0<\alpha<\dfrac{\pi}{2}
0
<
α
<
2
π
. Evaluate
sin
(
2
θ
)
\sin(2\theta)
sin
(
2
θ
)
in terms of
α
\alpha
α
. a)
sin
(
2
α
)
\sin(2\alpha)
sin
(
2
α
)
b)
−
cos
(
2
α
)
-\cos(2\alpha)
−
cos
(
2
α
)
c)
−
sin
(
2
α
)
-\sin(2\alpha)
−
sin
(
2
α
)
d)
cos
(
2
α
)
\cos(2\alpha)
cos
(
2
α
)
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