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)? 

Let $\omega$ be a primitive fifth root of unity. The quadratic residues in $\mathbb{Z}/(5\mathbb{Z})^*$ are $1$ and $4$, hence by setting $q=\omega^1-\omega^2-\omega^3+\omega^4$ such Gauss sum fulfills $q^2=5$. On the other hand
$$\omega^1-\omega^2-\omega^3+\omega^4= 2\cos\frac{2\pi}{5}-2\cos\frac{4\pi}{5} \tag{A}$$
is real an positive, so $q=\sqrt{5}$, and 
$$\omega^1+\omega^2+\omega^3+\omega^4=2\cos\frac{2\pi}{5}+2\cos\frac{4\pi}{5}=-1,\tag{B}$$
so by summing $(A)$ and $(B)$ we get $4\cos\frac{2\pi}{5}=\sqrt{5}-1$ as wanted.
