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)? 

Note that $$2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$$
therefore $$\cos\left(2\cdot \dfrac{2\pi}{5}\right) = \cos\left(3\cdot \dfrac{2\pi}{5}\right).$$
Put $\dfrac{2\pi}{5} = x$. Using the formulas
\begin{equation*}
\cos 2x = 2\cos^2 x - 1, \quad \cos 3x = 4\cos^3 x - 3\cos x,
\end{equation*}
we have
\begin{equation*}
4x^3 - 2x^2 -3x + 1 = 0 \Leftrightarrow (x - 1)(4x^2 + 2x - 1) = 0.
\end{equation*}
Because $\cos \dfrac{2\pi}{5}  \neq 1$, we get
\begin{equation*}
4x^2 + 2x - 1 = 0.
\end{equation*}
Solving the above quadratic equation for $x$ gives us $\cos \dfrac{2\pi}{5}  = \dfrac{-1 \pm \sqrt{5}}{4}$. Because $\cos \dfrac{2\pi}{5} > 0$, we take the positive sign, giving us $\cos \dfrac{2\pi}{5}  = \dfrac{-1 + \sqrt{5}}{4}$.
