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

                   Answer: 

After Euler formula We know that:
$Cos(x) ^4 = \frac{cos(4x)}{8}+\frac{cos(2x)}{2}+\frac{3}{8}$
And
$sin(x) ^4 =\frac{cos(4x)}{8}-\frac{cos(2x)}{2}+\frac{3}{8}$
So,
$ Cos(x) ^4 +sin(x) ^4=\frac{cos(4x)}{4}+\frac{3}{4} =\frac{cos(4x)+3}{4}$
