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 $

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*}

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.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}