$\cos\left(\frac{\pi}{5}\right)$ using De Moivre's Theorem This is an Exercise 3.2.5 from Beardon's Algebra and Geometry:

Show that $\cos\left(\frac{\pi}{5}\right)=\frac{\lambda}{2}$, where $\lambda$ = $\frac{1+\sqrt{5}}{2}$ (the Golden Ratio).
[Hint: As $\cos(5\theta) = 1$, where $\theta = \frac{2\pi}{5}$, we see from De Moivre's theorem that $P(\cos\theta) = 0$ for some polynomial $P$ of degree five. Now observe that $P(z)=(1-z)Q(z)^2$ for some quadratic polynomial $Q$.]

So, by De Moivre's theorem:
$$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5=\cos(2\pi)+i\sin(2\pi)=\cos(2\pi)=1$$
And so:
$$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1=0$$
Thus, $\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1$ is our polynomial $P$ of degree five. But how can I get from here to $Q$ and from $Q$ to $\cos\left(\frac{\pi}{5}\right)$?

Start with something simpler, an expression for $e^{i\frac{\pi}5}$:
$$\left(\cos\frac{\pi}5+i\sin\frac{\pi}5\right)^5=-1$$
Write explicitly the terms of the expansion. We know that we can ignore therms in even powers of $i$. Using $z=\cos\frac{\pi}5$ and $\sin^2\frac{\pi}5=1-z^2$ one gets:
$$z^5-10z^3(1-z^2)+5z(1-z^2)^2+1=0$$
Hopefully you can continue from here. Just note that your final polynomial first term in the expansion might be $1+z$, not $1-z$.
