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Mathematics
Mathematics
Problem
sin
3
θ
cos
θ
\sin 3\theta\,\cos \theta
sin
3
θ
cos
θ
is equal to a)
1
2
[
sin
(
4
θ
)
−
sin
(
2
θ
)
]
\dfrac{1}{2}\,[\sin(4\theta)-\sin(2\theta)]
2
1
[
sin
(
4
θ
)
−
sin
(
2
θ
)]
b)
1
2
[
cos
(
4
θ
)
+
cos
(
2
θ
)
]
\dfrac{1}{2}\,[\cos(4\theta)+\cos(2\theta)]
2
1
[
cos
(
4
θ
)
+
cos
(
2
θ
)]
c)
1
2
[
sin
(
4
θ
)
+
sin
(
2
θ
)
]
\dfrac{1}{2}\,[\sin(4\theta)+\sin(2\theta)]
2
1
[
sin
(
4
θ
)
+
sin
(
2
θ
)]
d)
1
2
sin
(
4
θ
)
\dfrac{1}{2}\,\sin(4\theta)
2
1
sin
(
4
θ
)
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