<s>[INST] Why is $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$? 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).
 [/INST] 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 $\cos(36^\circ) - \cos(72^\circ) = 1/2$
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