Show that $\cos4x=8\sin^4x-8\sin^2x+1$ using De Moivre's theorem 
Use De Moivre's theorem to show that $$\cos4x=8\sin^4x-8\sin^2x+1.$$
Hence show the one of the roots of the equation $8z^4-8z^2+1=0$ is $\sin\frac{\pi}{8}$ and express the other roots in polar form.
Deduce that $\sin \frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$ and find an exact expression for $\sin \frac{11}{8}\pi$.

My attempt,
$\cos4x=Re(\cos4x+i\sin4x)$
$=Re(\cos x+i\sin x)^4$
$=Re(c^4+4c^3is+6c^2i^2s^2+4ci^3s^3+i^4s^4)$
$=c^4-6c^2s^2+s^4$
So, $\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x$
$=(1-\sin^2x)^2-6(1-\sin^2x)\sin^2x+\sin^4x$
$1-2\sin^2x+\sin^4x-6\sin^2x+6\sin^4x+\sin^4x$
$=8\sin^4x-8\sin^2x+1$
How to proceed then?

let $\cos x =c $, $\sin x = s$
$\cos4x = (1-s^2)^2-6(1-ss)\cdot s^2+s^4$
$= 1 -2s^2 +s^4 - 6s^2 +6s^4 +s^4$
$= 8s^4 -8s^2 +1$
$\cos 4x = 0  = 8s^4 -8s^2 +1$
$\cos \frac{\pi}{2} =0$
$x= \frac{\pi}{8}$
so $\sin \frac{\pi}{8}$ is a solution
Look at the roots of $8s^4-8s^2 +1$ and use the quadratic equation to solve
we see that there is only one solution in range $(0,0.5)$ which is $\frac{\sqrt{(2-\sqrt{(2)})}}{2}$ so this has to be $\sin \frac{\pi}{8}$.
