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

To check, you can see if the LHS equals the RHS by using
$$\cos^2 x = \dfrac {\cos 2x + 1}{2} \\ \sin^2 x = \dfrac {1- \cos 2x}{2}$$
Squaring both sides
$$\cos^4 x = \dfrac 14(\cos^2 2x + 2 \cos 2x +1) \\ \sin^4 x = \dfrac 14(1-2 \cos 2x + \cos^2 2x)$$
Adding we get
$$\cos^4 x + \sin^4 x = \left(\dfrac 14 {\cos^2 2x} + \dfrac 14 \right) +\left(\dfrac 14 + \dfrac 14 {\cos^2 2x}\right) \\ \cos^4 x + \sin^4 x = \dfrac 12 ({1+\cos^2 2x}) $$
Since $2 \cos^2 2x - 1 = \cos 4x$, $\cos^2 2x = \dfrac 12 ({\cos 4x +1})$ so
$$\cos^4 x + \sin^4 x = \dfrac 12 (1 + \dfrac 12 (\cos 4x + 1)) \\ \cos^4 x + \sin^4 x = \dfrac 12 + \dfrac 14 (\cos 4x + 1)$$ or $$\mathbf {\cos^4 x + \sin^4 x = \dfrac {3 + \cos 4x}{4}}$$
