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
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
 A: \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*}
A: $$\eqalign{1 - 2 \sin^2 x \cos^2 x &= 1 - (1-\cos^2 x) \cos^2 x - \sin^2 x (1 - \sin^2 x)\cr
&= 1 - \cos^2 x + \cos^4 x - \sin^2 x + \sin^4 x\cr
&= \cos^4 x + \sin^4 x}$$
A: Well, you can always use the Euler formulas
$$
\cos{x} = \frac{e^{ix} + e^{-ix}}{2}, \quad \sin{x} = \frac{e^{ix} - e^{-ix}}{2i}
$$
and expand the powers on the left side if you are ok with that solution.
A: 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 $$\mathbf {\cos^4 x + \sin^4 x = \dfrac {3 + \cos 4x}{4}}$$
A:                    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}$
