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)? 

Notice that $$\sin 3\theta=3\sin \theta-4\sin^3 \theta$$ Now substitute $\theta=\frac{\pi}{10}$ in the above equation, we get 
$$\sin \frac{3\pi}{10}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies \cos\left( \frac{\pi}{2}-\frac{3\pi}{10}\right)=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies \cos\frac{\pi}{5}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies 1-2\sin^2\frac{\pi}{10}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies 4\sin^3\frac{\pi}{10}-2\sin^2 \frac{\pi}{10}-3\sin \frac{\pi}{10}+1=0$$ It is obvious that the sum of all the coefficients of above cubic equation is $0$ hence  above cubic equation has one real root $1$. Now it can be easily factorized as follows $$\left(\sin \frac{\pi}{10}-1\right)\left(4\sin^2 \frac{\pi}{10}+2\sin \frac{\pi}{10}-1\right)=0$$ 
$$\color{red}{ \text{if}\quad \sin\frac{\pi}{10}-1=0 \implies \sin\frac{\pi}{10}=1}$$ but  $\frac{\pi}{10}<\frac{\pi}{2} \implies \sin\frac{\pi}{10}<1$ Hence, above value is unacceptable 
$$\color{blue}{ \text{if}\quad 4\sin^2\frac{\pi}{10}+2\sin\frac{\pi}{10}-1=0 \implies \sin\frac{\pi}{10}=\frac{-2\pm \sqrt{(-2)^2-4(4)(-1)}}{2(4)}}$$ $$\sin\frac{\pi}{10}=\frac{-1\pm \sqrt{5}}{4}$$ but  $0<\frac{\pi}{10}<\frac{\pi}{2}\implies 0<\sin\frac{\pi}{10}<1$  Thus we get 
$$\sin\frac{\pi}{10}=\frac{-1+\sqrt{5}}{4}$$
$$\implies \cos\left(\frac{\pi}{2}-\frac{\pi}{10}\right)=\frac{-1+\sqrt{5}}{4}$$
$$\implies \color{blue}{\cos\frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}}$$
