Why is $\cos^2(2\pi/5) + \cos^2(4\pi/5)=3/4$?

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 $3/4$.