<s>[INST] Prove the trigonometric identity $\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$ I need to show the steps to prove this identity:
$$\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$$
I know that $\cos{2x}=\cos^2{x}-\sin^2{x}$. From there I do not know what to do.
The solution should look like:
$$\sin^4{x}=sin^4{x}$$
I need to prove the right side equals the left side.
 [/INST] Let $ z = \cos(\theta) +i \sin(\theta)$
We can then find through De Moive's formula and the fact that $\cos -\theta = \cos \theta$ and $\sin -\theta = - \sin \theta$:
$ z - \frac1z \equiv 2i\sin\theta$
Thus: ($i^4=1$)
$ (z-\frac1z)^4 \equiv 16\sin^4\theta$ 
Expanding the RHS:
$ (z-\frac1z)^4 \equiv z^4 - 4z^2 + 6 - 4z^{-2} + z^{-4}$
Then subsituting $z$ back into the equation. (The $\sin$s will cancel out)
$ (z-\frac1z)^4 \equiv 2\cos4\theta - 8\cos2\theta + 6$
Finally dividing by 16 to get $\sin^4\theta$:
$\sin^4\theta = \frac {\cos4\theta - 4\cos2\theta + 3} 8$
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