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}

With $\cos(4\theta)$ on the lhs, you can substitute $\theta = \frac{\pi}{8}$ in your first equation, which gives a polynomial equation for $\cos(\frac{\pi}{8})$.
$$0 = 8\cos^4\frac{\pi}{8}- 8\cos^2\frac{\pi}{8} + 1$$
This equation is biquadratic and can be solved with the quadratic formula to show that,
$$\cos\frac{\pi}{8} = \frac{1}{2}\sqrt{2+\sqrt{2}}$$
You can use related methods to show that,
\begin{align}
\cos\frac{\pi}{4}  &= \frac{1}{2}\sqrt{2} \\
\cos\frac{\pi}{8}  &= \frac{1}{2}\sqrt{2+\sqrt{2}} \\
\cos\frac{\pi}{16} &= \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}} \\
\cos\frac{\pi}{32} &= \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} \\
\cos\frac{\pi}{64} &= \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}
\end{align}
and the pattern goes on
