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.

$\cos^4x+\sin^4 x\\=(\cos^2x+\sin^2x)^2-2(\sin x\cos x)^2\\=1-2(\frac{1}{2}\sin(2x))^2\\=1-\frac{1}{2}\sin^2(2x)\\=1-\frac{1}{2}\frac{1-\cos(4x)}{2}\\=\frac{1}{4}\cos(4x)+\frac{3}{4}$
