Trig and de Moivre's theorem

Question

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}

Answer 1

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

Answer 2

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 $8u^4-8u^2 + 1$

$\cos \frac {\pi}{8} = \sqrt {\frac 12 + \frac {\sqrt{2}}{4}}\\ \cos \frac {3\pi}{8} = \sqrt {\frac 12 - \frac {\sqrt{2}}{4}}\\ \cos \frac {5\pi}{8} = -\sqrt {\frac 12 - \frac {\sqrt{2}}{4}}\\ \cos \frac {7\pi}{8} = -\sqrt {\frac 12 + \frac {\sqrt{2}}{4}}$