<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] Notice that 
\begin{eqnarray}
\cos(2x)&=& \cos^2 x - \sin^2 x \\
&=& 2 \cos^2 x - 1.\\
\end{eqnarray}
Then
\begin{equation}
\cos^2 x = \dfrac{1}{2}(1+\cos(2x)).
\end{equation}
Hence,
\begin{eqnarray}
\cos^4 x &=& (\cos^2 x)^2\\
&=& \left[\dfrac{1}{2}(1 + \cos(2x))\right]^2\\
&=& \dfrac{1}{4}(1 +2 \cos(2x)+ \cos^2(2x))\\
&=& \dfrac{1}{4} +\dfrac{1}{2} \cos(2x) + \dfrac{1}{4}\dfrac{1}{2}(1+\cos(4x))\\
&=& 3/8 + 1/2 \cos(2x) +1/8 \cos(4x)
\end{eqnarray}
</s>