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.

If $z^{5}-1=0$, then $$(z^{4}+z^{3}+z^{2}+z+1)(z-1)=0,$$ because $z^{5}-1=(z^{4}+z^{3}+z^{2}+z+1)(z-1)$. This implies that $$
z^{4}+z^{3}+z^{2}+z+1=0$$ or $z-1=0$. Since
\begin{equation*}
z=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}=e^{i2\pi /5}\ne 1
\end{equation*}
is a root of $z^5-1=0 $ and
\begin{equation*}
z^{k}=\cos \frac{2k\pi }{5}+i\sin \frac{2k\pi }{5}=e^{i2k\pi /5},
\end{equation*}
for $k\in \left\{ 1,2,3,4\right\} $, we have
\begin{eqnarray*}
0 &=&z^{4}+z^{3}+z^{2}+z+1=\left( \cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}%
\right) +\left( \cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\right) +\cdots +1
\\
&=&\left( \cos \frac{8\pi }{5}+\cos \frac{6\pi }{5}+\cdots +1\right)
+i\left( \sin \frac{8\pi }{5}+\cos \frac{6\pi }{5}+\cdots +\sin \frac{2\pi }{%
5}\right) .
\end{eqnarray*}
So
\begin{equation*}
\cos \frac{8\pi }{5}+\cos \frac{6\pi }{5}+\cdots +1=0.
\end{equation*}
