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

If $\displaystyle x=\frac\pi5,5x=\pi\implies3x=\pi-2x\implies\cos3x=\cos(\pi-2x)\  \   \  \ (1)$
But as $\displaystyle\cos(\pi-2x)=-\cos2x$, (1)$\displaystyle\implies\cos3x+\cos2x=0\ \ \ \ (2)$
If $\displaystyle\cos x=c(2)$ becomes $4c^3+2c^2-3c-1=0\  \ \ \ (3)$
Again  from $(1)\implies3x=2n\pi\pm(\pi-2x)$ 
$\displaystyle'-'\implies x=(2n-1)\pi\implies\cos x=-1$
$\displaystyle'+'\implies x=\frac{(2n+1)\pi}5$ where $n=0,1,2,3,4$
Observe that $\displaystyle\cos\frac{(2\cdot2+1)\pi}5=\cos\pi=-1$(which converges with $'-'$ case )
and $\displaystyle\cos\frac{(2\cdot0+1)\pi}5=\cos\frac{(2\cdot4+1)\pi}5$ as $\displaystyle\cos(2\pi-y)=\cos y$
Similarly,  $\displaystyle\cos\frac{(2\cdot1+1)\pi}5=\cos\frac{(2\cdot3+1)\pi}5$
So, the quadratic equation whose roots are $\displaystyle\cos\frac{\pi}5=\cos\frac{9\pi}5,\cos\frac{3\pi}5=\cos\frac{7\pi}5$ will be $\displaystyle\frac{4c^3+2c^2-3c-1}{c+1}=0\iff4c^2-2c-1=0\  \ \ \ (4)$
Now use the fact $\displaystyle c^2=\cos^2x=\frac{1+\cos x}2$
