Prove this identity: $\sin^4x = \frac{1}{8}(3 - 4\cos2x + \cos4x)$ The problem reads as follows.
Prove this identity: $$\sin^4x = \frac{1}{8}(3 - 4\cos2x + \cos4x)$$
I started with the right side and used double angles identities for $\cos2x$ and a sum and then then double angle identity for the $\cos4x...$ It all got messy and I hit a dead end. He doesn't give any hints and I'm pretty lost. 
I'm thinking that I will eventually need a product-sum identity to get the $\dfrac{1}{8}....$ But I'm just confused how to get there... 
Thanks in advance to anyone who can help!
 A: Or better yet, use Euler's formula!
$$\sin^4 x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^4={1\over 16}\left(e^{4ix}-4e^{2ix}+6-4e^{-2ix}+e^{-4ix}\right)={1\over 8}\left(\frac{e^{4ix}+e^{-4ix}}{2}-4\frac{e^{2ix}+e^{-2ix}}{2}+3\right)={1\over 8}\left(\cos{4x}-4\cos{2x}+3\right)$$ as desired.
A: Do you know power-reduction formulas?  
$$\sin^2(x) = \frac{1-\cos(2x)}{2}$$
$$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$
From here, it is fairly simple once we recognize that $\sin^4(x) = (\sin^2(x))^2$.
A: \begin{align}
\sin^4 x & = \left(\frac{1-\cos 2x}{2}\right)^2 \\
& = \frac{1-2\cos 2x + \cos^2 2x}{4} \\
& = \frac{1}{4}\left(1-2\cos 2x+\frac{1+\cos 4x}{2}\right) \\
& = \frac{1}{8}\left(3-4\cos 2x +\cos 4x\right)
\end{align}
Where identities $\sin^2 t=\frac{1-\cos (2t)}{2}$ and $\cos^2 t=\frac{1+\cos (2t)}{2}$ were used.
A: $\sin^4x = (1 - \cos^2x)^2 = \left(1 - \dfrac{1 + \cos2x}{2}\right)^2 = \dfrac{(1 - \cos2x)^2}{4} = \dfrac{1 - 2\cos2x + \cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\dfrac{1 + \cos4x}{2}}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{1 + \cos4x}{8} = \dfrac{3 - 4\cos2x + \cos4x}{8}$.
A: Everyone starts from LHS, I choose to start from RHS.
\begin{align}
\dfrac{1}{8}(3 - 4\cos2x + \cos4x)&=\dfrac{1}{8}(3 - 4\cos2x + 2\cos^2 2x-1)\\
&=\dfrac{1}{8}(2 - 4\cos2x + 2\cos^22x)\\
&=\dfrac{2}{8}(1 - 2\cos2x + \cos^22x)\\
&=\dfrac{1}{4}(1-\cos2x)^2\\
&=\dfrac{1}{4}(1-2\cos^2x+1)^2\\
&=\dfrac{1}{4}2^2(1-\cos^2x)^2\\
&=\sin^4x.
\end{align}
