# How to prove <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> k

How to prove $\sum _{k=1}^{n}\mathrm{cos}\left(\frac{2\pi k}{n}\right)=0$ for any n>1?
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cefflid6y
We have
$\sum _{k=1}^{n}\mathrm{cos}\left(\frac{2\pi k}{n}\right)=\text{Re}\left(\sum _{k=1}^{n}{e}^{2\pi ik/n}\right)$
and so
$\sum _{k=1}^{n}{e}^{2\pi ik/n}=\frac{{e}^{2\pi i/n}\left(1-{e}^{2\pi i}\right)}{1-{e}^{2\pi i/n}}$
and notice that
${e}^{2\pi i}=\mathrm{cos}\left(2\pi \right)+i\mathrm{sin}\left(2\pi \right)=1$
so the claim follows.
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fythynwyrk0
One approach is to write
$\mathrm{cos}\left(\frac{2\pi k}{n}\right)=\frac{1}{2\mathrm{sin}\left(\frac{2\pi }{n}\right)}\left(\mathrm{sin}\left(\frac{2\pi \left(k+1\right)}{n}\right)-\mathrm{sin}\left(\frac{2\pi \left(k-1\right)}{n}\right)\right)$
Now, we have converted the sum into a telescoping sum, which we can evaluate directly as
$\begin{array}{rl}\sum _{k=1}^{n}\mathrm{cos}\left(\frac{2\pi k}{n}\right)& =\frac{1}{2\mathrm{sin}\left(\frac{2\pi }{n}\right)}\left(\mathrm{sin}\left(\frac{2\pi \left(n+1\right)}{n}\right)+\mathrm{sin}\left(\frac{2\pi \left(n\right)}{n}\right)\right)\\ \\ & -\frac{1}{2\mathrm{sin}\left(\frac{2\pi }{n}\right)}\left(\mathrm{sin}\left(\frac{2\pi \left(2-1\right)}{n}\right)+\mathrm{sin}\left(\frac{2\pi \left(1-1\right)}{n}\right)\right)\\ \\ & =\frac{1}{2\mathrm{sin}\left(\frac{2\pi }{n}\right)}\left(\mathrm{sin}\left(\frac{2\pi \left(n+1\right)}{n}\right)-\mathrm{sin}\left(\frac{2\pi \left(2-1\right)}{n}\right)\right)\\ \\ & =\frac{1}{2\mathrm{sin}\left(\frac{2\pi }{n}\right)}\left(\mathrm{sin}\left(\frac{2\pi }{n}\right)-\mathrm{sin}\left(\frac{2\pi }{n}\right)\right)\\ \\ & =0\end{array}$