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

I was not able (yet) to follow Chandrasekar's solution, but noticed this while trying to understand the argument (how it could possibly lead to the solution, or how exactly he arrives at $2x^2-x-1$ for $x=\cos\frac{\pi}{5}$, which to me seems non-obvious and even a fallacious deduction from his equations and prose -- apologies if I am just being dense)...perhaps it is what Chandrasekar meant all along, but in any case, it does seem to be the most elementary solution available.
Apply the double angle formula
$\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$
to $\theta=\frac{\pi}{5}$ and $\frac{2\pi}{5}$,
with $a=\cos\frac{\pi}{5}$ and $b=\frac{2\pi}{5}$
for convenience, recalling also that
$\cos(\pi\pm\theta)=-\cos\theta$:
$$
b=\cos\frac{2\pi}{5}=2\,\cos^2\frac{\pi}{5}-1=2a^2-1
$$
$$
-a=\cos\frac{4\pi}{5}=2\,\cos^2\frac{2\pi}{5}-1=2b^2-1
$$
Next, subtracting the equations
$$
\matrix{
2a^2=1+b\\
2b^2=1-a}
$$
we get
$$
\eqalign{
2\left(a^2-b^2\right)&=b+a\\
2\left(a+b\right)\left(a-b\right)&=b+a\\
2\left(a-b\right)&=1\\
a-b&=\frac12\,.
}
$$
Furthermore, multiplying, we get
$$
4(ab)^2=(1+b)(1-a)=1+(b-a)-ab=1+\left(-\tfrac12\right)-ab
$$
giving us the quadratic equation
$$4(ab)^2+(ab)-\tfrac12=0$$
$$8(ab)^2+2(ab)-1=0$$
$$\left(4ab-1\right)\left(2ab+1\right)=0$$
so that $ab=\frac14$ or $-\frac12$,
from which we can choose the former
since we know that $0&lta&ltb$.
