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.

Why not use trig?
for $$\cos\left(\frac{2\pi}{5}\right)+ \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) =-1.$$
First of notice that if $a = \frac{2\pi}{5}$ then we have
$$\cos(a) + \cos(2a) + \cos(3a)+ \cos(4a) =-1.$$
Knowing the identity that $$\cos(2a) = 2\cos^2(a) -1$$ we can directly derive:
$$\cos(3a) = 4 \cos^3(a)-3\cos(a)$$   and 
$$\cos(4a) = 8 \cos^4(a) - 8\cos^2(a) +1$$
Also we know that $\cos\left(\frac{2\pi}{5}\right) = \frac{1}{5^{1/2} + 1}$ which is $\color{red}{1/2 1/ \phi}$ the golden ratio 
Yes there is a lot arithmetic, but once you have gotten to the arithmetic the problem is solved...just use care in expanding the powers.  I worked it out and proved the problem, using the above scheme.    Look once you know the key trig identity for $\cos(2a)$ which is $\cos( a+a)$ you can very simply derive the other identities for $\cos(3a) = \cos(2a + a)$ and $\cos(4a) = \cos(3a + a)$ or $\cos( 2a + 2a)$.    
