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Here is the problem I am stuck with: Is it true that for every positive integer $n > 1$, $$\sum\limits_{k=1}^n \cos \left(\frac {2 \pi k}{n} \right) =0= \sum \limits_{k=1}^n \sin \left(\frac {2 \pi k}{n} \right)$$ I'm imagining the unit circle and adding up the value of both trig functions separately but I cannot picture see how their sum add up to $0$.

EDIT I updated the question from earlier and realized that it was my fault because of a major typo. At least it was better to point it out late than never. So how would I go about solving this question as of now?

Originally the question was: $$\sum\limits_{k=1}^n \frac {\cos(2πk)}{n} =0= \sum\limits_{k=1}^n \frac {\sin(2πk)}{n}$$

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    The l.h.s. is $1+\frac12+\dots+\frac1n$, so it can't be $0$. – 2017-02-04
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    @Bernard Not that it matters that much as the question has flaws, but the left hand side is $\frac{1}{n} + \frac{1}{n} + \ldots + \frac{1}{n} = 1$ as the summation is wrt $k$ not $n$. – 2017-02-04
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    The question is correct. – 2017-02-04
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    So it's a contradiction then? – 2017-02-04
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    Not a contradiction. I mean the statement is false. – 2017-02-04
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    @Winther Yes I edited my question and noted your comment. – 2017-02-04
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    @Winther: You're right, I misread the formula. – 2017-02-04
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    Hint: let $\,z=\cos \left(\frac {2 \pi k}{n} \right)+ i \sin \left(\frac {2 \pi k}{n} \right)\,$, then $\;z^n=1$ and $\;\sum\limits_{k=1}^n \cos \left(\frac {2 \pi k}{n} \right) + i \sum \limits_{k=1}^n \sin \left(\frac {2 \pi k}{n} \right) = \sum\limits_{k=1}^n z^k\,$. – 2017-05-22
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    In the future, please do not edit questions so as to invalidate answers. (especially accepted answers, which will confuse future readers, even if the original question was not the question you had in mind) – 2017-05-22
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    Apologies from my end. I should have wrote it down correctly from the beginning, but mistakes happen you know. – 2017-05-22
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    No no, I meant that this should be a **new question** rather than an edit of an old one. This changes the question entirely and invalidates my answer, whereas posting a new question would've had no drawbacks. – 2017-05-22
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    Ah gotcha. I'll make sure this won't happen again. – 2017-05-22
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    Searching [with Approch0](https://approach0.xyz/search/?q=%24%5Csum%5Climits_%7Bk%3D1%7D%5En%20%5Ccos%20%5Cleft(%5Cfrac%20%7B2%20%5Cpi%20k%7D%7Bn%7D%20%5Cright)%20%3D0%24&p=1) returns several older copies of this question. For example, [How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any $n>1$?](https://math.stackexchange.com/q/1530598) and [Prove that $ \sum\limits_{n=1}^N \cos(2\pi n/N)= 0 $?](https://math.stackexchange.com/q/2188240) (You can try to search for posts about sum of sines as well.) – 2017-05-22
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    Why is this tagged ([tag:summation-by-parts])? Are you supposed to solve the problem [using summation by parts](https://en.wikipedia.org/wiki/Summation_by_parts)? – 2017-05-22
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    @MartinSleziak If it's improperly tagged, you're more than welcome to re-select the appropriate tags, or I can do it myself, doesn't matter. – 2017-05-23

2 Answers 2

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Old answer to old question:

Multiply both sides by $n$ to get

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

It's easy enough to see that for positive integers $k$, $\cos(2\pi k)=1$ and $\sin(2\pi k)=0$, thus,

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

$$\sum_{k=1}^n\cos(2\pi k)=n\ne0$$

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    I see it now. Can you explain the last line of your answer. Why does the summation equal $n$? – 2017-02-04
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    @Oliver821 Notice:$$\underbrace{1+1+\dots+1}_n=n$$ – 2017-02-04
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    Yes now I clearly see it. – 2017-02-04
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    It looks like you are saying that $n\cdot \cos(\frac{2\pi k}{n}) = \cos(2\pi k)$. – 2017-05-21
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    @AOrtiz The question was different when posted (and answered), look at the [edit history](https://math.stackexchange.com/posts/2129281/revisions). – 2017-05-21
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    @dxiv Hm, how do you manage to respond to comments below my *old* answer before me? O.O – 2017-05-22
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    @SimplyBeautifulArt It so happened that I was just writing a comment ;-) – 2017-05-22
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    @dxiv Interesting timing. – 2017-05-22
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It's a little bit easier with complex numbers. Set $\zeta = e^{2\pi i/n}$ and note that the real part of the sum $$ S = \sum_{k=1}^n\zeta^k $$ is precisely the sum $\sum_{k=1}^n\cos(\frac{2\pi k}{n})$ and the imaginary part of $S$ is precisely the sum $\sum_{k=1}^n\sin(\frac{2\pi k}{n})$. Therefore, if we show that $S = 0$, then we will have killed two birds with one stone.

However, $S$ is a geometric sum whose value is given explicitly by $$ S = \frac{\zeta(1 - \zeta^{n})}{1-\zeta}. $$ Since $\zeta^n = 1$, we indeed have $S = 0$, as desired.

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    Ah and does this have any relation with [Euler's Formula](https://en.wikipedia.org/wiki/Euler%27s_formula)? – 2017-05-22
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    @CartesianBread Yes, it comes directly from Euler's formula. – 2017-05-22
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    Plain and simple. Simply Beautiful Art: Again, much apologies for the confusion that occurred. – 2017-05-22
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    @CartesianBread No problem... – 2017-05-22