Suppose $\theta=2\pi/5$. Apparently it is true that $1+ \cos^2 \theta + \cos^2(2\theta) + \cos^2(3 \theta) + \cos^2 (4\theta) = 5/2$, or equivalently, $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$. What is the easiest way to see this?
I see that $1 + \cos \theta + \dotsb + \cos 4\theta = 0$ by de Moivre.
This came up while calculating the character of the irreducible two-dimensional representation of the dihedral group of order $10$ (i.e. to prove it is irreducible).
Let $\zeta = e^{i \theta}$. Then you have
$$\cos^2 (2\theta) + \cos^2 (4\theta) = {1 \over 4} \left( e^{2\theta} + e^{-2\theta} \right)^2 + {1 \over 4} \left( e^{4\theta} + e^{-4\theta} \right)^2 $$
or, recalling the definition of $\zeta$,
$$ {1 \over 4} \left( \left( \zeta^2 + \zeta^{-2} \right)^2 + \left( \zeta^4 + \zeta^{-4} \right)^2 \right)$$
Now, if you expand those out you get
$$ {1 \over 4} \left( \zeta^4 + 2 + \zeta^{-4} + \zeta^8 + 2 + \zeta^{-8} \right) $$
or, recalling that $\zeta^5 = 1$, this is
$$ {1 \over 4} \left( 4 + \zeta^4 + \zeta + \zeta^2 + \zeta^3 \right). $$
Rearranging gives
$$ {1 \over 4} \left( 3 + (1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4) \right)$$
and by de Moivre we have $1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4$, so this is just $3/4$.
As $\pi=180^\circ$ $$S=\cos^272^\circ+\cos^2144^\circ$$
$$2S=1+\cos144^\circ+1+\cos288^\circ=2+\cos(180-36)^\circ+\cos(360-72)^\circ$$
As $\cos(180^\circ-A)=-\cos A,\cos(360^\circ-B)=+\cos B,$ $$2S-2=\cos72^\circ-\cos36^\circ$$
Now use Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
