<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] HINT: use that $$\cos(6\theta)=32\, \left( \cos \left( \theta \right)  \right) ^{6}-48\, \left( \cos
 \left( \theta \right)  \right) ^{4}+18\, \left( \cos \left( \theta
 \right)  \right) ^{2}-1
$$
$$\cos(2\theta)=2\, \left( \cos \left( \theta \right)  \right) ^{2}-1$$
$$\cos(4\theta)=8\, \left( \cos \left( \theta \right)  \right) ^{4}-8\, \left( \cos
 \left( \theta \right)  \right) ^{2}+1
$$
</s>