How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from 
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)? 

for any $\theta \in \mathbb{R}$ the transformation $\psi: x \rightarrow 2x^2 -1$ sends $cos \theta$ to $cos 2\theta$ . hence $\psi^2 $ sends $cos \theta$ to $cos4\theta$
if $\alpha = \frac{2\pi}5$ then $cos 4\alpha = cos \alpha$ so that $cos \alpha$ is a fixed point for $\psi$ and if $c=cos \alpha$ we have
$$
\psi^2(c) = c
$$
i.e.
$$
2(2c^2-1)^2-1= c
$$
or 
$$
(c-1)(8c^3-1) = 0
$$
i.e.
$$
(c-1)(2c-1)(4c^2+2c-1) = 0
$$
if we disregard the roots corresponding to angles with rational cosines we have:
$$
4c^2+2c-1 =0,
$$
giving
$$
c = \frac14\left(-1 \pm \sqrt{5}\right)
$$
the negative root corresponds to the angle $\frac{4\pi}5$
