<s>[INST] Deriving an expression for $\cos^4 x + \sin^4 x$ 
Derive the identity $\cos^4 x + \sin^4 x=\frac{1}{4} \cos (4x) +\frac{3}{4}$

I know $e^{i4x}=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\cos^4 x + \sin^4 x = \cos (4x) + 6 \cos^2 (x) \sin^2 (x)$.
So now I need to show that $\cos (4x) + 6 \cos^2 (x) \sin^2 (x)=\frac{1}{4} \cos (4x) +\frac{3}{4}$, which has stumped me.
 [/INST] $$\begin{align}
\cos^4x + \sin^4 x &= \left( \cos^2 x \right)^2 + \left( \sin^2 x \right)^2 \\[4pt]
&= \left( \frac{1 + \cos 2 x}{2}\right)^2 + \left( \frac{1-\cos 2x}{2} \right)^2 \\[4pt]
&= \frac{1}{2}\left( 1 + \cos^2 2 x \right) \\[4pt]
&= \frac{1}{2}\left( 1 + \frac{1+\cos 4 x}{2} \right) \\[4pt]
&= \frac{1}{4}\left(\; 3 + \cos 4 x \;\right)
\end{align}$$

From your particular stopping place, you could proceed thusly:
$$\begin{align}
\cos 4 x + 6 \sin^2 x \cos^2 x &= \cos 4 x + \frac32\cdot(2\sin x \cos x)^2 \\[4pt]
&= \cos 4 x + \frac32\cdot \sin^2 2 x \\[4pt]
&= \cos 4 x + \frac32\cdot \frac{1 - \cos 4 x}{2} \\[4pt]
&= \frac14\left(\;3 + \cos 4 x\;\right)
\end{align}$$
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