<s>[INST] Confusing double angle identity How would I solve the following double angle identity. 
$$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$$
So far my work is 
$$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$$
But how would I proceed. 
 [/INST] \begin{align*}
\cos^4(x) &= \left(\frac{e^{ix}+e^{-ix}}{2}\right)^4\\
 &= \frac{e^{4xi} + 4e^{2xi} + 6 + 4e^{-2xi} + e^{-4xi}}{16}\\
 &= \frac{3}{8} + \frac{1}{2} \frac{e^{2xi} + e^{-2xi}}{2} + \frac{1}{8} \frac{e^{4xi}+e^{-4xi}}{2}\\
 &= \frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)
\end{align*}
</s>