<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] $$\cos^4x+\sin^4x\\=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\\=1-\frac12\sin^22x=1-\frac12\left(\frac12(1-\cos4x)\right)=\frac34+\frac14\cos4x$$
As:
$$\sin^2x+\cos^2x=1;\\(a+b)^2=a^2+b^2+2ab;\\\sin2x=2\sin x\cos x;\\\cos2x=\cos^2x-\sin^2x=1-2\sin^2x$$

Also:
$$\cos^4x+\sin^4x=\left(\frac{e^{ix}+e^{-ix}}2\right)^4+\left(\frac{e^{ix}-e^{-ix}}2\right)^4\\=\frac1{16}({\small e^{-4 ix}+4 e^{-2 ix}+4 e^{2 ix}+e^{4 ix}+6+e^{-4 ix}-4 e^{-2 ix}-4 e^{2 ix}+e^{4i x}+6})\\=\frac34+\frac14\left(\frac{e^{4ix}+e^{-4ix}}2\right)=\frac34+\frac14\cos4x$$
</s>