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I was working on a problem and I need this proof to proceed. I'm fairly confident that this is true, because each point: $2\pi/n*k$ is spread out evenly, but I don't know how to go about proofing this.

Can someone direct me to proofing or disproofing this?

I tried using this equality with no success $$\sum_{k = 0}^{n-1}\sin(2\pi/n*k)= \sum_{k = 0}^{n-1}\sin(2\pi/n*(n-k))$$

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    have you tried the formula: $e^{ix} = \cos x + i\sin x$ – 2017-01-08
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    No, but I don't see how I would use it. I used that in my original problem to break it up all the way to here, so it might be going backwards. – 2017-01-08
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    The point is that $\sum_{k=0}^{n-1} \exp(2\pi i k/n)$ is a geometric progression. Its real and imaginary parts are the sums with $\cos$ and $\sin$. – 2017-01-08
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    Note also that your equation is **not** true for $n=1$. – 2017-01-08
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    @Goldname: use the [formula for the sum of a geometric series](https://en.wikipedia.org/wiki/Geometric_series#Formula) and [Euler's Formula](https://en.wikipedia.org/wiki/Euler's_formula). – 2017-01-08

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Hint: What is the sum of all the $n$th roots of unity.

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    that is what I am trying to proof. – 2017-01-08
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    @Goldname: Use a well-known formula called Viete's theorem for $n$th degree polynomial. Look it up wikipedia first about this topic. – 2017-01-08
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    @DeepSea to proof that all roots of unity add to 0 or for my current question? – 2017-01-08
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Hint: The formula for the sum of a geometric series says $$ \sum_{k=0}^{n-1}e^{2\pi ik/n}=\frac{e^{2\pi in/n}-1}{e^{2\pi i/n}-1} $$

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    I am missing something here - why $\sum_{k=0}^{n-1}e^{2\pi ik/n}=\frac{e^{2\pi in/n}-1}{e^{2\pi i/n}-1}$ instead of $\frac{1-e^{2\pi in/n}}{1-e^{2\pi i/n}}$ as with standard sum of geometric series? – 2018-03-14
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    There's a difference? – 2018-03-14
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    I understand there isn't, but I can't understand why - even though it is a naive question can you please clarify why they are equivalent? – 2018-03-14
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    Is $\frac{-1}{-1}=\frac11$? Is $1+r+r^2+\cdots+r^{n-1}=r^{n-1}+r^{n-2}+\cdots+1$? Now consider $\frac{1-r^n}{1-r}=\frac{r^n-1}{r-1}$. Different authors are free to use whichever of these equal forms they wish. Neither is *the* "standard" sum. When $r\lt1$, then $\frac{1-r^n}{1-r}$ is used so that numerator and denominator are positive and when we let $n\to\infty$, we get $\frac1{1-r}$. However, when $r\gt1$, $\frac{r^n-1}{r-1}$ is used since then numerator and denominator are positive. – 2018-03-14
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I think you're missing the fact that both of the sums equal $0$, for $n>1$. For any $k$, $\sin(2\pi k/n) = -\sin(2\pi (n-k)/n)$, so the first half of the sum cancels the second half. Similarly for $\cos$.

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    This is a good observation. But if there are an odd number of terms, won't the middle term survive the cancellation? – 2017-01-08
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    The middle term is $\sin(\pi)=0$. – 2017-01-08