How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator

The complex roots of $x^5-1$ are:
$$
\begin{align}
x_1&=1\\
x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\
x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\
x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\
x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5
\end{align}
$$
using Vieta's formulas you get
$$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2\pi}5-\cos\frac\pi5\right)=0,$$
which yields your first equation. 

From now on, let $\varphi=\frac{\pi}5$ (for brevity).
We know that
$\cos2\varphi-\cos\varphi+\frac12=(2\cos^2\varphi-1)-\cos\varphi+\frac12=2\cos^2\varphi-\cos\varphi-\frac12=0$, i.e.
$$2\cos^2\varphi-\cos\varphi=\frac12.$$
Now from $\cos2\varphi=\cos\varphi-\frac12$ you get
$$\cos\varphi\cos2\varphi=\cos^2\varphi-\frac{\cos\varphi}2=\frac{2\cos^2\varphi-\cos\varphi}2=\frac14.$$
(Or, as suggested in Chandrasekhar's answer, from $2\cos^2\varphi-\cos\varphi=\frac12$ you can find the value of $\cos\varphi$ by solving the quadratic equation and taking the positive root. Once you know $\cos\varphi$, you can compute $\cos2\varphi$ and many other things. If you try it this way, you can check your result e.g. here.)
