How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1?

Question

I can show for any given value of n that the equation

$$\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$$

is true and I can see that geometrically it is true. However, I can not seem to prove it out analytically. I have spent most of my time trying induction and converting the cosine to a sum of complex exponential functions

$$\frac{1}{2}\sum_{k=1}^n [\exp(\frac{i 2 \pi k}{n})+\exp(\frac{-i 2 \pi k}{n})] = 0$$

and using the conversion for finite geometric sequences

$$S_n = \sum_{k=1}^n r^k = \frac{r(1-r^n)}{(1-r)}$$

I have even tried this this suggestion I have seen on the net by pulling out a factor of $\exp(i \pi k)$ but I have still not gotten zero.

Please assist.

Your statement is false for $n=1$. This is because $\cos(2\pi)=1$, not $0$. — Sinister Cutlass, Nov 15 '15 at 18:53

score 9 · Accepted Answer · answered Nov 15 '15 at 18:58

9

We have $$\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{n}\right)=\textrm{Re}\left(\sum_{k=1}^{n}e^{2\pi ik/n}\right) $$ 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 $$e^{2\pi i}=\cos\left(2\pi\right)+i\sin\left(2\pi\right)=1 $$ so the claim follows.

answered Nov 15 '15 at 18:58

Marco Cantarini

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I can see that $\cos(2 \pi) = 1$ but not why $\sum_{k=1}^{n-1} \cos(\frac{2 \pi}{n}) = -1$. – AlephNaught Nov 15 '15 at 19:04
Sorry I don't understand your question... you have only to use the geometric series with $r=e^{2\pi i/n}$ and the last equality I wrote. – Marco Cantarini Nov 15 '15 at 19:11
I see, the value inside the parentheses evaluates to zero. Thanks. – AlephNaught Nov 15 '15 at 19:17

score 6 · Answer 2 · answered Nov 15 '15 at 19:03

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You are considering the real part of the sum of the roots of $x^n-1$. By Viète's theorem the sum of the roots of $x^n-1$ is the coefficient of $x^{n-1}$, i.e. zero.

answered Nov 15 '15 at 19:03

Jack D'Aurizio

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It is not clear to me that I am considering the real part of the sum of the roots $x^n-1$. – AlephNaught Nov 15 '15 at 19:10
@WilliamThomas: $$\cos\left(\frac{2\pi k}{n}\right) = \text{Re}\left(\exp\left(\frac{2\pi i k}{n}\right)\right).$$ – Jack D'Aurizio Nov 15 '15 at 19:12
@WilliamThomas $\cos(2\pi k/n)$ is the real part of $e^{2\pi ik/n}$, which is an $n$th root of unity. If $x$ is an $n$th root of unity, then $x^n=1$ or $x^n-1=0$. – Akiva Weinberger Nov 15 '15 at 19:20
Can you point me to a reference that shows the relationship to the nth root of unity comment? – AlephNaught Nov 15 '15 at 19:21
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@WilliamThomas: $$ \exp\left(\frac{2\pi i k}{n}\right)^n = \exp(2\pi i)^k = 1^k=1.$$ – Jack D'Aurizio Nov 15 '15 at 20:14

Mark Viola · Answer 3 · 2015-11-15T19:20:16.007

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One approach is to write

$$\cos\left(\frac{2\pi k}{n}\right)=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(k+1)}{n}\right)-\sin\left(\frac{2\pi(k-1)}{n}\right)\right)$$

Now, we have converted the sum into a telescoping sum, which we can evaluate directly as

$$\begin{align} \sum_{k=1}^n\cos\left(\frac{2\pi k}{n}\right)&=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(n+1)}{n}\right)+\sin\left(\frac{2\pi(n)}{n}\right)\right)\\\\ &-\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(2-1)}{n}\right)+\sin\left(\frac{2\pi(1-1)}{n}\right)\right)\\\\ &=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(n+1)}{n}\right)-\sin\left(\frac{2\pi(2-1)}{n}\right)\right)\\\\ &=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi}{n}\right)-\sin\left(\frac{2\pi}{n}\right)\right)\\\\ &=0 \end{align}$$

edited Nov 15 '15 at 19:20

answered Nov 15 '15 at 19:14

Mark Viola

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This is a nice answer, thank you. I would not have seen the identity shown on the first line. Can you point me to a reference of its derivation? – AlephNaught Nov 15 '15 at 19:27
You're welcome; my pleasure. And thank you! The identity is the addition angle formula for the sine function $\sin (x\pm y)=\sin x \cos y \pm \sin y \cos x$. – Mark Viola Nov 15 '15 at 19:30
I have not seen this addition angle formula rewritten in this form before. This is enlightening. – AlephNaught Nov 15 '15 at 19:35
Pleased I could help. – Mark Viola Nov 15 '15 at 19:36

How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1?

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