<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] $\cos^2 x \sin^2 x = (1 - \sin^2 x) \sin^ 2 x = \sin^2 x - \sin^ 4 x$ and by the same argument $\cos^2 x \sin^2 x = \cos^2 x - \cos^ 4 x$. Take the average of these two identities to obtain
$$
\cos^2 x \sin^2 x = \frac{1}{2} - \frac{\cos^4x + \sin^4 x}{2} \, .
$$
Now substitute this and simplify.
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