<s>[INST] Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
 [/INST] $$\eqalign{1 - 2 \sin^2 x \cos^2 x &= 1 - (1-\cos^2 x) \cos^2 x - \sin^2 x (1 - \sin^2 x)\cr
&= 1 - \cos^2 x + \cos^4 x - \sin^2 x + \sin^4 x\cr
&= \cos^4 x + \sin^4 x}$$
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