Proof of a summation

Question

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

The l.h.s. is $1+\frac12+\dots+\frac1n$, so it can't be $0$. — Bernard, Feb 04 '17 at 20:27
@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 $n$. — Winther, Feb 04 '17 at 20:31
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 $;\sum\limits_{k=1}^n \cos \left(\frac {2 \pi k}{n} \right) + i \sum \limits_{k=1}^n \sin \left(\frac {2 \pi k}{n} \right) = \sum\limits_{k=1}^n z^k,$. — dxiv, May 22 '17 at 00:03
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) — Simply Beautiful Art, May 22 '17 at 00:11
Apologies from my end. I should have wrote it down correctly from the beginning, but mistakes happen you know. — John W. Smith, May 22 '17 at 00:12
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. — Simply Beautiful Art, May 22 '17 at 00:14
Searching with Approch0 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 $n>1$? and Prove that $ \sum\limits_{n=1}^N \cos(2\pi n/N)= 0 $? (You can try to search for posts about sum of sines as well.) — Martin Sleziak, May 22 '17 at 15:31
Why is this tagged ([tag:summation-by-parts])? Are you supposed to solve the problem using summation by parts? — Martin Sleziak, May 22 '17 at 15:34
@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. — John W. Smith, May 23 '17 at 04:19

Simply Beautiful Art · Accepted Answer · 2017-05-22T00:14:30.097

3

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,

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

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

edited May 22 '17 at 00:14

answered Feb 04 '17 at 20:24

Simply Beautiful Art

74,685

I see it now. Can you explain the last line of your answer. Why does the summation equal $n$? – John W. Smith Feb 04 '17 at 20:36
1

@Oliver821 Notice:$$\underbrace{1+1+\dots+1}_n=n$$ – Simply Beautiful Art Feb 04 '17 at 20:37
Yes now I clearly see it. – John W. Smith Feb 04 '17 at 20:39
It looks like you are saying that $n\cdot \cos(\frac{2\pi k}{n}) = \cos(2\pi k)$. – Alex Ortiz May 21 '17 at 23:55
2

@AOrtiz The question was different when posted (and answered), look at the edit history. – dxiv May 21 '17 at 23:58
@dxiv Hm, how do you manage to respond to comments below my old answer before me? O.O – Simply Beautiful Art May 22 '17 at 00:00
@SimplyBeautifulArt It so happened that I was just writing a comment ;-) – dxiv May 22 '17 at 00:03
@dxiv Interesting timing. – Simply Beautiful Art May 22 '17 at 00:10

score 1 · Answer 2 · answered May 22 '17 at 00:04

1

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.

answered May 22 '17 at 00:04

Alex Ortiz

24,844

Ah and does this have any relation with Euler's Formula? – John W. Smith May 22 '17 at 00:16
1

@CartesianBread Yes, it comes directly from Euler's formula. – Simply Beautiful Art May 22 '17 at 00:17
Plain and simple. Simply Beautiful Art: Again, much apologies for the confusion that occurred. – John W. Smith May 22 '17 at 00:19
@CartesianBread No problem... – Simply Beautiful Art May 22 '17 at 00:25

Proof of a summation

2 Answers2

Old answer to old question: