<s>[INST] Prove $\cos^2x\sin^4x = \frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))$ I need help with a problem some may consider odd but here it is.
I have the following trig identity I been working on and I managed to get it to.
$$\cos^2x\sin^4x=\frac{3}{16}-\frac{\cos(2x)}{4}+\frac{3\cos(2x)}{16}+\frac{1}{8}(1+\cos(4x))+\frac{1}{32}(\cos(6x)+\cos(2x))$$
However I am not sure how to make this simplify to $\frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))=\cos^2x\sin^4x$ 
and thus I am stuck. 
 [/INST] The equation you have written is not correct in general.
What you can do is note that there exists some $\theta \in [0, \pi)$ such that $\sin^2 x = \sin \theta$.
Then $1 - \sin^2 x = 1 - \sin \theta \Rightarrow \cos^2 x = 1 - \sin \theta$.
Thus we know that $\cos^2 x \sin^4 x = (1 - \sin \theta) \sin^2 \theta$.
Let $1 - 2 \sin \theta = T$. Then we know:
$\cos^2 x \sin^4 x = (1 - \sin \theta) \sin^2 \theta = \frac{1}{8} (1 + T)(1 - T)^2$.
Similarly,
$\cos(2x) = T,$
$\cos(4x) = 2T^2 - 1,$
$\cos(8x) = 8T^4 - 8T^2 + 1$.
The sum-to-product formula tells us:
$\cos (2x) + \cos (6x) = 2 \cos(4x) \cos (8x) = 32T^6 - 48T^4 + 20T^2 - 2$.
Perhaps you can go somewhere from here.
</s>