<s>[INST] Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it is a correct identity, although I can't prove it.
I've tried to use the formula 
$$ \cos(z) = \frac{e^ {iz} - e^ {iz}}{2} $$ but without any satisfying result.
This exercise is from chapter on series. 
EDIT: I corrected a mistake in the formula I wanted to use.
 [/INST] Let $\omega = \exp\left(\frac{2\pi i}{5}\right) = \cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right)$. Then $\omega$ is a fifth root of unity ($\omega^5 = 1$). Then $$1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.$$ By taking the real part of both sides (after applying De Moivre's Theorem), we obtain $$1 + \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) = 0.$$ Subtracting one from both sides gives the identity you're after.
</s>