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If $x\notin \mathbb{R}$ then $$\sum_{k=0}^{n-1}\sin(2\pi kx)=\frac{\sin((n-1)\pi x)\sin(n\pi x)}{\sin(\pi x)}$$

This supposedly is a direct results from Dirichlet kernel. I was also told that this can simply found by using additions of sine functions, but I'm at a loss. Any help is appreciated.

Gregoire Rocheteau
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  • If you're familiar with geometric series, notice the above series is the imaginary part of $\sum_{k=0}^{n-1}e^{2\pi I kx}$. – Alex R. Aug 28 '20 at 17:58

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Hint :Consider $\displaystyle\sum_{k=0}^{n-1} \sin(2\pi kx)=\sum_{k=0}^{n-1} \Im\left({e^{2\pi kx}}\right)=\Im\left(\sum_{k=0}^{n-1} e^{2\pi kx}\right)=\Im\left(\frac{e^{2n\pi x}-1}{e^{2 \pi x}-1}\right)$, which is your result (why ?)

Bastien Tourand
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