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)? 

By my post, $$
\cos \frac{\pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}
$$
By $\cos (\pi-\dfrac{\pi}{5} )=-\cos \dfrac{\pi}{5}, $ we have
$$
-\cos \frac{4 \pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}
$$
By double-angle formula, $$
-\left(2 \cos ^{2} \frac{2 \pi}{5}-1\right)-\cos \frac{2 \pi}{5}=\frac{1}{2} \Leftrightarrow 
4 \cos ^{2} \frac{2 \pi}{5}+2 \cos \frac{2 \pi}{5}-1=0
$$
Using quadratic formula gives $$
\boxed{\cos \frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}}
$$
By the way, $$\cos \frac{\pi}{5}= \cos \frac{2\pi}{5}+\frac{1}{2}= \frac{1+\sqrt{5}}{4}
$$
