<s>[INST] Showing Trigonometric Identity Prove that:
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$
Attempt:
\begin{align*}
L.H.S & = \cos^2\theta\sin^4\theta\\
& = \cos^2\theta\sin^2\theta\sin^2\theta\\
& = \frac{1+\cos2\theta}{2}.\frac{1-\cos2\theta}{2}.\frac{1-\cos2\theta}{2}\\
& = \frac{1}{8} (1-\cos^22\theta)(1-\cos2\theta)
\end{align*}
Now, what should I do?
 [/INST] With $z=e^{i\theta}$,
$$\left(\frac{z+z^{-1}}2\right)^2\left(\frac{z-z^{-1}}{2i}\right)^4=\frac1{32}\left(\frac{z^6+z^{-6}-z^2-z^{-2}+4-2z^4-2z^{-4}}2\right).$$
For convenience, multiply by $64z^6$, set $t:=z^2$ and rewrite
$$(t^2-1)^2(t-1)^2=t^6+1-t^4-t^2+4t^3-2t^5-2t.$$
By direct expansion of the LHS you get the equality.
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