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 $a=\cos\frac{\pi}5$ and $b=\cos\frac{2\pi}5=2a^2-1$ (by the double-angle identity for cosine), we want to show that
$$
0=(x-a)(x+b)=x^2-(a-b)x-ab=x^2-\tfrac12x-\tfrac14,
$$
i.e., that $a$ and $-b$ are roots of $4x^2-2x-1$.
Note also that $4x^2+2x-1=4(x+a)(x-b)$ has roots $-a$ and $b$.
What is special about the numbers $\{a,b,-b,-a\}$?
They are the $x$-coordinates (real parts) of the nonreal $10$th roots of unity,
$$
x+iy=e^{\pm\pi i\cdot\frac{k}5}
\qquad
\text{for}
\qquad
k\not\equiv0\pmod5.
$$
But these satisfy the equation $(x+iy)^5=(-1)^k=\pm1$.
Taking the imaginary part, we have
$$
\eqalign{
0 &= \Im\left[(x+iy)^5\right] \\
  &= \Im\left[x^5+5x^4(iy)+10x^3(iy)^2+10x^2(iy)^3+5x(iy)^4+(iy)^5\right] \\
  &= \Im\left[iy\left(5x^4+10x^2(iy)^2+(iy)^4\right)\right] \\
  &= y\left[5x^4-10x^2y^2+y^4\right] \\
\implies
0 &= \left[5x^4-10x^2(1-x^2)+(1-x^2)^2\right] \\
  &= 16x^4-12x^2+1 \\
  &= 16x^4-8x^2+1 ~-~4x^2 \\
  &= \left( 4x^2-1 \right) - \left( 2x \right)^2 \\
  &= \left( 4x^2+2x-1\right)\left( 4x^2-2x-1\right) \,.
}
$$
So far, we have shown this has roots $\{\pm a,\pm b\}$.
It only remains to show that
$\{-a,b\}$ are the roots of the first factor
and that the desired pair $\{a,-b\}$
splits the second quadratic factor into linear terms
(as we wrote at the outset).
Perhaps there is a clever way to infer the correct linear order
of this set of roots by noticing that $a>\cos\frac\pi4>b$ so that $a^2>\frac12>b^2$ and combining this with our identity $b=2a^2-1$ above.
I propose in stead to use that $a>b$ but then to notice
that our quadratic factors are in fact parabolas,
with roots that are very easy to order on the $x$-axis.
If we let $t=2x$ (which preserves order), then we have
$$
t^2\pm t-1 = t\,(t\pm1)-1
$$
which both pass through $(0,-1)$
and alternately pass through $(\mp1,-1)$,
shown respectively in red and blue below;
the result follows.

It's only a small extra step to note that the roots of our factor,
$t^2-t-1$, are $t=\frac{1\pm\sqrt5}{2}$, or $x=\frac{1\pm\sqrt5}{4}$.
