$$ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $$
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.
On
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.
On
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.
On
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.
On
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*}
On
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.
On
Here's a couple other trigonometric approaches:
$$ \cos\left(\frac{2\pi}{5}\right) \ + \ \cos\left(\frac{4\pi}{5}\right) \ + \ \cos\left(\frac{6\pi}{5}\right) \ + \ \cos\left(\frac{8\pi}{5}\right) $$
$$ = \ \left[ \cos\left(\frac{2\pi}{5}\right) \ + \ \cos\left(\frac{8\pi}{5}\right) \right] \ + \ \left[\cos\left(\frac{4\pi}{5}\right) \ + \ \cos\left(\frac{6\pi}{5}\right) \right] $$
$$ = \ ( 2 \cdot \cos \left[ \frac{\frac{8\pi}{5} + \frac{2\pi}{5}}{2} \right] \cdot \cos \left[ \frac{\frac{8\pi}{5} - \frac{2\pi}{5}}{2} \right] ) \ + \ ( 2 \cdot \cos \left[ \frac{\frac{6\pi}{5} + \frac{4\pi}{5}}{2} \right] \cdot \cos \left[ \frac{\frac{6\pi}{5} - \frac{4\pi}{5}}{2} \right] ) $$
$$ = \ ( 2 \cdot \cos \pi \cdot \cos \frac{3\pi}{5} ) \ + \ ( 2 \cdot \cos \pi \cdot \cos \frac{\pi}{5} ) \ = \ -2 \cdot ( \cos \frac{3\pi}{5} + \cos \frac{\pi}{5} ) $$
$$ = \ -2 \cdot ( 2 \cdot \cos \left[ \frac{\frac{3\pi}{5} + \frac{\pi}{5}}{2} \right] \cdot \cos \left[ \frac{\frac{3\pi}{5} - \frac{\pi}{5}}{2} \right] ) \ = \ -4 \ \cdot \ \cos \frac{2\pi}{5} \ \cdot \ \cos \frac{\pi}{5} $$
$$ = \ -4 \ \cdot \ ( 2 \cos^2 \frac{\pi}{5} \ - \ 1) \ \cdot \ \cos \frac{\pi}{5} \ = \ 4 \ \cos \frac{\pi}{5} \ - \ 8 \ \cos^3 \frac{\pi}{5} \ , $$
to which we shall return. We also have
$$ \cos\left(\frac{2\pi}{5}\right) \ + \ \cos\left(\frac{4\pi}{5}\right) \ + \ \cos\left(\frac{6\pi}{5}\right) \ + \ \cos\left(\frac{8\pi}{5}\right) $$
$$ = \ \cos\left(\frac{2\pi}{5}\right) \ + \ \cos\left(\frac{4\pi}{5}\right) \ + \ \cos\left(- \frac{4\pi}{5}\right) \ + \ \cos\left(- \frac{2\pi}{5}\right) $$
$$ = \ 2 \cdot \left[ \cos\left(\frac{2\pi}{5}\right) \ + \ \cos\left(\frac{4\pi}{5}\right) \right] \ = \ 2 \cdot \left[ \cos\left(\frac{2\pi}{5}\right) \ + \ ( 2 \cos^2 \left(\frac{2\pi}{5} \right) \ - \ 1 ) \right] $$
$$ = \ 4 \cos^2 \left(\frac{2\pi}{5} \right) \ + \ 2 \cos\left(\frac{2\pi}{5}\right) \ - \ 2 \ . $$
A bit unfortunately, we do not get any obvious simplification; instead, we need to bring in the values for $ \ \cos\left(\frac{\pi}{5}\right) \ \ $ and $ \cos\left(\frac{2\pi}{5}\right) \ , \ $ which are known exactly. [One discussion can be found here . ] We thus find
$$ 4 \ \cos \frac{\pi}{5} \ - \ 8 \ \cos^3 \frac{\pi}{5} \ = \ 4 \ \left( \frac{1 \ + \ \sqrt{5}}{4} \right) \ - \ 8 \ \left( \frac{1}{4} \ + \ \frac{\sqrt{5}}{4} \right)^3 $$
$$ = \ ( 1 \ + \ \sqrt{5} ) \ - \ 8 \ \left( \frac{1 \ + \ 3 \sqrt{5} \ + \ 15 \ + \ 5 \sqrt{5}}{64} \right) \ = \ ( 1 \ + \ \sqrt{5} ) \ - \ ( 2 \ + \ \sqrt{5} ) \ = \ -1 $$
or
$$ 4 \cos^2 \left(\frac{2\pi}{5} \right) \ + \ 2 \cos\left(\frac{2\pi}{5}\right) \ - \ 2 \ = \ 4 \ \left( \frac{ \sqrt{5} \ - \ 1}{4} \right)^2 \ + \ 2 \ \left( \frac{ \sqrt{5} \ - \ 1}{4} \right) \ - \ 2 $$
$$ = \ \left( \frac{ 5 \ - \ 2 \sqrt{5} \ + \ 1}{4} \right) \ + \ \left( \frac{ \sqrt{5} \ - \ 1}{2} \right) \ - \ 2 \ = \ \frac{3}{2} \ - \ \frac{1}{2} \ - \ 2 \ = \ -1 \ . $$
Naturally, this is related to the methods discussed by the other solvers through geometry in the complex plane.
$$ \\ $$
It would be a fair question as to why I bothered writing all of this out. I also want to show that this produces some other interesting identities through the "golden triangle/pentagram". Since $ \ \cos \frac{\pi}{5} \ = \ \frac{\phi}{2} \ $ , we can write
$$ 4 \left( \frac{\phi}{2} \right) \ - \ 8 \left( \frac{\phi}{2} \right) \ = \ 2 \phi \ - \ \phi^3 \ = \ -1 \ , $$
which bears some resemblance to one of its defining equations, $ \ \phi^2 \ - \ 1 \ = \ \phi \ . \ $ Writing $ \ \cos \frac{2\pi}{5} \ = \ \frac{\phi^2}{2} \ - \ 1 \ $ (from the "double-angle" formula), we also have
$$ 4 ( \frac{\phi^2}{2} \ - \ 1 )^2 \ + \ 2 (\frac{\phi^2}{2} \ - \ 1 ) \ - \ 2 \ = \ \phi^4 \ - \ 3 \phi^2 \ = \ -1 \ . \ $$ (Of course, since we could also write $ \ \cos \frac{2\pi}{5} \ = \ \frac{\phi \ - \ 1}{2} \ , $ this second identity is closely related to the previous one. All of this is in fact underlain by those fifth-roots of unity discussed in many of the other answers here. [This is part of what makes "5" one of John Baez' favorite numbers...]
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}