<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] \begin{align*}
\frac{3+\cos4x}{4}&=\frac{3+2\cos^22x-1}{4}=\frac{(\cos^2x-\sin^2x)^2+1}{2}\\
&=\dfrac{\sin^4x+\cos^4x+(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}{2}\\
&=\sin^4x+\cos^4x
\end{align*}
</s>