Is there a proof that $\sin(\pi/n)+\sin(3\pi/n)+\ldots+\sin(\pi(2n-1)/n) = 0$?

Question

Is there a proof that $\sin(\pi/n)+\sin(3\pi/n)+\ldots+\sin(\pi(2n-1)/n) = 0$?

So I was trying to prove that MID$(n)$ for $$\int_{0}^{2\pi} \sin\theta \,d\theta$$ is equal to zero, and I ended up with the series above and did not know how to complete from there.

It's best not to fully rely on the title; you should write the full problem in the body of the question. — Christian E. Ramirez, Jan 31 '23 at 05:01
$(2n-1)$ is not a multiple of $4$ so is is not clear what numbers you are summing. — geetha290krm, Jan 31 '23 at 05:01
I take it MID(n) is the midpoint Riemann sum, and the 4 in your title is a typo which should be 3. — anon, Jan 31 '23 at 05:06
Yes it should be 3 instead of 4. Sorry for the mistake @geetha290krm — JoLiebert, Jan 31 '23 at 05:09
Compute $\sum_{k=0}^{n-1} e^{ik {2 \pi \over n}} $ and get imaginary. — copper.hat, Jan 31 '23 at 05:16
In other words, "by symmetry." To put a geometric spin on the previous two comments, the sum of sines in the title is the sum of the $y$-coordinates of certain points on the unit circle - these points are symmetric across the $y$-axis, and these points come in pairs whose $y$-coordinates cancel out, except for the point at $(-1,0)$ when $n$ is odd which contributes $0$ as its $y$-coordinate. — anon, Jan 31 '23 at 05:16
I think one can see the gist of Travis Willse's comment by considering the cases $n=5$ and $n=6$ first. — Li Kwok Keung, Jan 31 '23 at 05:23
https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro — lab bhattacharjee, Jan 31 '23 at 05:56

Mike · Accepted Answer · 2023-01-31T06:54:56.917

0

HINT: Note that the equation $$\sin((1+2k)\pi/n)=$$ $$=-\sin(-(1+2k)\pi/n))$$ $$=-\sin((2n-(2k+1))\pi/n)$$ holds for all positive integers $n$ and positive integers $k \le (n-1)/2$. So the $\sin((1+2k)\pi/n)$ cancels out with the $\sin((2n-(2k+1))\pi/n)$ for all positive intrgers $k<(n-1)/2$, leaving [for $n$ odd and $k=(n-1)/2$] the one term $\sin(n\pi/n)$, which is $0$....

edited Jan 31 '23 at 06:54

answered Jan 31 '23 at 06:00

Mike

20,434

score 0 · Answer 2 · answered Jan 31 '23 at 07:12

Let $$S=\sin(\pi/n)+\sin(3\pi/n)+\ldots+\sin(\pi(2n-1)/n)=\sum \sin([(2k-1)\pi/n]$$ $$\implies S=\Im\sum_{k=1}^{n} e^{i(2k-1)\pi/n}=\Im e^{-i\pi/n}\sum_{k=1}^{n} e^{2i\pi k/n} =\Im e^{i \pi/n}\left( \frac{e^{2i\pi k}-1}{e^{2i\pi/n}-1}\right)=0$$ As $e^{2i\pi k}=1, k \in I$

Is there a proof that $\sin(\pi/n)+\sin(3\pi/n)+\ldots+\sin(\pi(2n-1)/n) = 0$?

2 Answers2