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Here's the identity:

For $0 < m < N$, we have $$ \sum_{k=0}^{N-1}\cos \left( \frac{2m \pi}{N}k\right) = 0 $$

I know I can solve this using a variety of methods, i.e. anything described here: How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?.

But what I'm looking for is a 1-2 liner type thing. I feel the argument for my progression is just so nice, there must be a super nice way to prove this? I was able to do it for the case where $\cos$ is replaced by $\sin$... But maybe I'm being too greedy.

(More background if you're curious, but I don't think it's necessary to know: This is basically a follow up question to what I asked yesterday; basically we're finishing up the second equation: Clean proof for trigonometry identity? I know what the answer is, but I feel like there should be like a $1$-$2$ liner to compute this)

mrtaurho
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sweetpotato
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    This answer to the linked question is a short one-liner. – dxiv Sep 18 '21 at 22:12
  • yep i think you're right! facepalm so actaully, that answer caught my eye too, but i didnt know what the commenter to the answer meant by 'extract the real and imaginary' parts...but i think now i do, seems like the answer below is doing exactly that! – sweetpotato Sep 18 '21 at 22:32
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    @sweetpotato Yes, exactly. Extracting real and imaginary parts just means considering $a$ and $b$ separetely in $z=a+bi$. An interesting thing is now to note that $z=0$ if and only if $a=b=0$, which is what I used below. – mrtaurho Sep 18 '21 at 22:34
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    gotcha. actually even better, your way proves both the sin and cos relations at the same time lol...so no need for separate arguments like i had before – sweetpotato Sep 18 '21 at 22:36
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    @sweetpotato Exactly. By the way, if you want to respond to someone in particular use "@[username]" (which can be auto-completed by pressing TAB after you have typed @). Then the user is notified that you made a comment referring to them. – mrtaurho Sep 18 '21 at 22:38
  • @mrtaurho ahh lol i totally forgot thats a feature. good to know, thx again – sweetpotato Sep 18 '21 at 22:39

1 Answers1

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Combine that

$$ \zeta=\exp\left(\frac{2\pi i}N\right) $$

is a non-trivial $N$-th root of unity, hence satisfies

$$ \frac{\zeta^N-1}{\zeta-1}=\zeta^{N-1}+\cdots+1=0\,, $$

with the fact that

$$ \exp\left(\frac{2\pi i}Nk\right)=\cos\left(\frac{2\pi}Nk\right)+i\sin\left(\frac{2\pi }Nk\right) $$

to get the result for $m=1$. For arbitary $0<m<N$ the same works since the summands corresponding to $m$ are just powers of $\zeta^m\ne1$, i.e. still powers of a non-trivial $N$-th roots of unity.

mrtaurho
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