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?

As an alternative we can use that
$$\cos x=\Re(e^{ix})$$$$\implies \cos 4x=\Re(e^{i4x})=\Re[(\cos x+i\sin x)^4]=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$
that is
$$\cos 4x=\cos^4 x-6\cos^2x \sin^2x+\sin^4x$$
$$\cos 4x=\cos^4 x-6\cos^2x (1-\cos^2 x)+(1-\cos^2 x)^2$$
$$\cos 4x=\cos^4 x-6\cos^2x+6\cos^4x+1-2\cos^2x+\cos^4 x$$
$$8\cos^4 x=\cos 4x+8\cos^2x-1$$
$$8\cos^4 x=\cos 4x+8\left(\frac{1}{2}+\frac{1}{2}\cos(2x)\right)-1$$
$$8\cos^4 x=\cos 4x+4\cos(2x)+3$$
$$\cos^4 x=\frac18\cos 4x+\frac12\cos(2x)+\frac38$$
