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!

$\sin^4x = (1 - \cos^2x)^2 = \left(1 - \dfrac{1 + \cos2x}{2}\right)^2 = \dfrac{(1 - \cos2x)^2}{4} = \dfrac{1 - 2\cos2x + \cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\dfrac{1 + \cos4x}{2}}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{1 + \cos4x}{8} = \dfrac{3 - 4\cos2x + \cos4x}{8}$.
