<s>[INST] How do I show that $\cos^4x=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x )+\frac{3}{8}$ I know how to prove that 
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$ 
by substituting $\cos(2x)$ with $2\cos^2x-1$ according to the double angle identity
$$\cos(2x)=2\cos^2x-1$$
However, how do I do that for $\cos^4x$? 
Do I do it by writing $\cos^4x$ as $$\cos^2(x)\cdot \cos^2(x)$$ and thus get it by squaring the LHS of $$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
Im not sure how to proceed. Any ideas?
 [/INST] Since
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$ 
we have
$$\cos^2(2x)=\frac{1}{2}+\frac{1}{2}\cos(4x)$$ 
and thus, if we square first equation we get
 $$\cos^4x=\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\cos ^2(2x)=$$ 
$$ =\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\Big{(}\frac{1}{2}+\frac{1}{2}\cos(4x)\Big{)}$$
$$ =\frac{3}{8}+{1\over 2}\cos(2x)+\frac{1}{8}\cos(4x)$$
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