Trig and de Moivre's theorem 
A) Use de Moivre's theorem to prove that $\cos^4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
B) Therefore deduce that $\cos(\pi/8) = \left(\frac{2 + \sqrt{2}}{4}\right)^{1/2}$
C) and write down an expression for $\cos(3\pi/8)$.

I have proved the first part of the question but I am not sure where to go from there.
My attempt
de Moivre's theorem states that $(\cos x +i\sin x)^n = \cos nx + i\sin nx$
using this and $\cos ^2θ + \sin ^2θ =1$, I did the following:
\begin{align}
\cos 4θ + i\sin 4θ &= (\cos θ +i\sin θ)^4 + \cos 4θ + 4i\cos ^3θ\sin θ - 6\cos ^2θ\sin ^2θ-4i\cos θ\sin 3θ+\sin ^4θ\\
\cos 4θ &= \cos ^4θ -6\cos ^2θ\sin ^2θ +\sin ^4θ\\
\cos 4θ &= \cos ^4θ - 6\cos ^2θ(1-\cos ^2θ) + (1-\cos ^2θ)^2\\
\cos 4θ &= \cos ^4θ - 6\cos ^2θ + 6\cos ^4θ +1 - 2\cos ^2θ + \cos ^θ\\
\cos 4θ &= 8\cos ^4θ -8\cos ^2θ +1
\end{align}

As noted, the firt proposition should be
$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
What you show is fine.
What is $\cos \frac {\pi}{8}$?
$\cos 4\frac {\pi}{8} = 8\cos^4\frac {\pi}{8} - 8\cos^2\frac {\pi}{8} +1\\
\cos \frac {\pi}{2} = 0 =8\cos^4\frac {\pi}{8} - 8\cos^2\frac {\pi}{8} +1$
Let $u = \cos \frac{\pi}{8}$
$8u^4 - 8u^2 + 1 = 0$
Apply the quadratic formula.
$u^2 = \frac {8\pm\sqrt{64-32}}{16} = \frac 12 \pm \frac {\sqrt{2}}{4}$
$u = \pm \sqrt {\frac 12 \pm \frac {\sqrt{2}}{4}}$
But this gives 4 answers... how do we make sense of that?
All values of $\cos(\frac {(2n+1)\pi}{8})$ will solve $8u^4-8u^2 + 1$
$\cos \frac {\pi}{8} = \sqrt {\frac 12 + \frac {\sqrt{2}}{4}}\\
\cos \frac {3\pi}{8} = \sqrt {\frac 12 - \frac {\sqrt{2}}{4}}\\
\cos \frac {5\pi}{8} = -\sqrt {\frac 12 - \frac {\sqrt{2}}{4}}\\
\cos \frac {7\pi}{8} = -\sqrt {\frac 12 + \frac {\sqrt{2}}{4}}$
