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I would like to prove the following, $$\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$$ This is equivalent to showing that if we have a regular n-gon, all of the vectors in $\mathbb{R}^2$ pointing to every vertex cancel eachother out to $\textbf{0}$. It seems obvious enough and i've tried a number of examples for small $n$. I am not sure how to start a proof though.

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    $\begingroup$ $$\sum_{k=0}^{n-1}\exp\left(\frac{2π\mathrm ik}n\right)=0.$$ $\endgroup$
    – Ѕᴀᴀᴅ
    Commented Jun 24, 2018 at 2:29
  • $\begingroup$ @Alex Francisco this doesn't seem obvious to me, I will try and show this now. Thanks. $\endgroup$
    – user445909
    Commented Jun 24, 2018 at 2:33
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    $\begingroup$ That's a geometric progression. $\endgroup$
    – Ѕᴀᴀᴅ
    Commented Jun 24, 2018 at 2:34
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    $\begingroup$ Well, it's a standard manipulation to calculate trigonometric sums using complex numbers. $\endgroup$
    – Ѕᴀᴀᴅ
    Commented Jun 24, 2018 at 2:54
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    $\begingroup$ Not much harm done, Leekboi. New users have scant resources to locate a duplicate. More experienced users OTOH... Anyway, I recalled having answered this. I do have misgivings for using a question I myself answered as a dupe target, but this time I didn't find alternatives. One may exist though :-) $\endgroup$
    – Jyrki Lahtonen
    Commented Jun 24, 2018 at 11:42

2 Answers 2

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(Too long for a comment.)  The algebraic proof using complex numbers is the most direct, yet...

This is equivalent to showing that if we have a regular n-gon, all of the vectors in $\mathbb{R}^2$ pointing to every vertex cancel each other out to $\textbf{0}$.

...this is too good a geometric insight to let go to waste. The following sketches how a proof could be built around this intuition (though it doesn't formalize it in full detail).

Let the respective vectors be $\,\mathbf{u_1}+\mathbf{u_2}+\ldots+\mathbf{u_n} = \mathbf{u}\,$, and assuming $\,n \gt 1\,$.

If $\,n\,$ is even then the regular $n$-gon has central symmetry i.e. $\,\mathbf{u_{j+n/2}}=-\mathbf{u_j}\,$ and therefore the $\,n\,$ vectors cancel out pairwise, so $\mathbf{u}=\mathbf{0}$.

Otherwise if $\,n\,$ is odd, let $\,T\,$ be the rotation of $\,\mathbb{R}^2\,$ by $\,\pi / n\,$. It is geometrically obvious that the endpoints of $\,\mathbf{u_1}, T(\mathbf{u_1}), \mathbf{u_2}, T(\mathbf{u_2}), \ldots \mathbf{u_n}, T(\mathbf{u_n})\,$ are the vertices of a regular $2n$-gon, thus by the previous observation $\,\mathbf{u_1} + T(\mathbf{u_1}) + \mathbf{u_2} + T(\mathbf{u_2}) + \ldots + \mathbf{u_n} + T(\mathbf{u_n})=\mathbf{0}\,$. On the other hand, $\,T\,$ is a linear transformation, so the latter implies $\,\mathbf{u} + T(\mathbf{u})=\mathbf{0}\,$, which in turn implies $\,\mathbf{u}=\mathbf{0}\,$.

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    $\begingroup$ Regarding the last part, could you please explain how $\textbf{u} + T(\textbf{u}) = \textbf{0} \Rightarrow \textbf{u} = \textbf{0}$? Does this not just imply $T(\textbf{u}) = -\textbf{u}$? $\endgroup$
    – user445909
    Commented Jun 24, 2018 at 5:12
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    $\begingroup$ @Leekboi $\,T\,$ is a rotation. $\,T(\mathbf{u})=-\mathbf{u}\,$ can only happen for rotations by $\,\pi\,$, which is not the case here because $\,n \gt 1\,$. $\endgroup$
    – dxiv
    Commented Jun 24, 2018 at 5:14
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    $\begingroup$ Haha, I was just about to say that. Thanks for the clarification. Very cool proof, thank you. $\endgroup$
    – user445909
    Commented Jun 24, 2018 at 5:15
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I assume $n \ge 2$.

Set

$\omega = \exp \left ( \dfrac{2\pi i}{n} \right ); \tag 1$

then

$\omega^n = \left ( \exp \left ( \dfrac{2\pi i}{n} \right ) \right )^n = \exp \left ( \dfrac{2 \pi i n}{n} \right ) = \exp ( 2 \pi i ) = 1; \tag 2$

we have

$( \omega - 1) \displaystyle \sum_0^{n - 1} \omega^k = \omega^n - 1 = 0; \tag 3$

since

$\omega \ne 1, \tag 4$

(3) yields

$ \displaystyle \sum_0^{n - 1} \omega^k = 0; \tag 5$

from (1)

$\omega^k = \exp \left ( \dfrac{2\pi i k}{n} \right ) = \cos \left ( \dfrac{2 \pi k}{n} \right ) + i \sin \left ( \dfrac{2 \pi k}{n} \right ) ; \tag 6$

substituting (6) into (5) and separating out the real and imaginary parts gives us

$\displaystyle \sum_0^{n - 1} \cos \left ( \dfrac{2 \pi k}{n} \right ) + i \sum_0^{n - 1} \sin \left ( \dfrac{2 \pi k}{n} \right ) = 0, \tag 7$

from which we immediately have

$\displaystyle \sum_0^{n - 1} \cos \left ( \dfrac{2 \pi k}{n} \right ) = \sum_0^{n - 1} \sin \left ( \dfrac{2 \pi k}{n} \right ) = 0. \tag 8$

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