<s>[INST] Simplify: $\sin \frac{2\pi}{n} +\sin \frac{4\pi}{n} +\ldots +\sin \frac{2\pi(n-1)}{n}$. Can you help me solve this problem?

Simplify: $\sin \dfrac{2\pi}{n} +\sin \dfrac{4\pi}{n} +\ldots +\sin \dfrac{2\pi(n-1)}{n}$.

 [/INST] Take the terms in opposite pairs, and note the change of sign,
$$\sin \dfrac{2k\pi}{n}+\sin \dfrac{2\pi(n-k)}{n}=\sin \dfrac{2k\pi}{n} +\sin(2\pi-\dfrac{2k\pi}{n})=\sin \dfrac{2k\pi}{n} -\sin\dfrac{2k\pi}{n}=0.$$
In case that $n$ is even, the central term remains, but $$\sin \dfrac{2n\pi}{2n}=0.$$
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