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.
 A: 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*}
A: You can also use that
$$2\sin(x)\Bigl[\cos 2x+\cos 4x + \cos 6x+\cos 8x\Bigr] = \sin 9x-\sin x = 2\cos 5x\sin 4x$$
so that inserting $x=\tfrac\pi5=\pi-4\tfrac\pi5$ yields the desired result.
A: Let $\zeta_n = e^{2\pi i/n} = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$ be a primitive $n^{\rm th}$ root of unity, so in particular, the roots of $z^5 - 1$ are $\zeta_5^k$ for $k = 0, 1, 2, 3, 4$.  Since the sum of the roots of a polynomial of degree $n$ is equal to the negative of the coefficient of the degree $n-1$ term, it follows that $$\sum_{k=0}^4 \zeta_5^k = 1 + \zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4 = 0.$$   Taking the real part of both sides of the equation immediately gives the desired identity.
A: 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.
A: Solve the equation $z^5=1$, for $z\in\mathbb{C}$. The solutions are $1$, $e^{i2\pi/5}$, $e^{i4\pi/5}$, $e^{i6\pi/2}$, $e^{i8\pi/2}$. Since those are the fifth roots of unity their sum is $0$. Use Euler's formula, $$e^{ix}=\cos x+ i\sin x,$$ and consider that since the sum of the roots is zero, then so must be its real part. Taking the real part of the roots and rearranging yields the required result.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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\begin{align}
&\color{#0000ff}{\large%
\sum_{n = 1}^{4}\cos\pars{2n\pi \over 5}}
=
\Re\sum_{n = 1}^{4}\expo{2n\pi\ic/5} = \Re\sum_{n = 1}^{4}\pars{\expo{2\pi\ic/5}}^{n}
=
\Re\braces{{\expo{2\pi\ic/5}\bracks{\pars{\expo{2\pi\ic/5}}^{4} - 1} \over \expo{2\pi\ic/5} - 1}}
\\[3mm]&=
\Re\bracks{{\expo{2\pi\ic/5}\pars{\expo{8\pi\ic/5} - 1} \over \expo{2\pi\ic/5} - 1}} 
=
\Re\bracks{{\overbrace{\expo{\pi\ic/5}\expo{4\pi\ic/5}}^{\ds{-1}}\
\pars{\expo{4\pi\ic/5} - \expo{-4\pi\ic/5}}
 \over
 \expo{\pi\ic/5} - \expo{-\pi\ic/5}}}
=
{-\sin\pars{4\pi/5} \over \sin\pars{\pi/5}}
\\[3mm]&=
{-\sin\pars{\pi - \pi/5} \over \sin\pars{\pi/5}}
=
{-\sin\pars{\pi/5} \over \sin\pars{\pi/5}} = \color{#0000ff}{\large -1}
\end{align}
A: The values represent the $x$-coördinates of four points of a regular pentagon centred at the origin, the missing fifth point being at $(1,0)$. The centre of mass of all five points is the origin, so in particular the mean of the $x$-coördinates is $0$. The sum of the four values is therefore $-1$.
A: 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 $\cos( 2a + 2a)$.    
