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.
 A: If you know $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix}+e^{-ix}}{2}$, then just expand using the binomial theorem:
$$\sin^4 x = \frac{e^{4ix} - 4e^{3ix}e^{-ix}+6e^{2ix}e^{-2ix}-4e^{ix}e^{-3ix}+e^{-4ix}}{(2i)^4}$$
$$=\frac{e^{4ix}+e^{-4ix} -4e^{2ix}-4e^{-2ix} + 6}{16}$$
$$=\frac{\frac{e^{4ix}+e^{-4ix}}{2} -4\frac{e^{2ix}+e^{-2ix}}{2} + 3}{8}$$
$$=\boxed{\frac{\cos 4x -4\cos 2x + 3}{8}}$$
as desired.
A: Hint: $$\cos{2x}=\cos^2{x}-\sin^2{x}=1-2\sin^2{x}\\
\implies \sin^2{x}=\frac{1-\cos{2x}}{2}$$
A: Hint: $(\cos^2 x-\cos2x)(\cos^2 x-\cos 2x)=(\sin^2 x)^2$
A: You could also use the fact that $\text{Re} ((\cos x + i \sin x)^4) = \text{Re} (\cos 4x + i \sin 4x)$.
A: 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 $\sin^4\theta$:
$\sin^4\theta = \frac {\cos4\theta - 4\cos2\theta + 3} 8$
