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Let n, k be integers, $n>1$ and $k \perp n$ denote that k, n are coprime and let $S_n = \{1 \le k \le \lfloor n / 2 \rfloor : k \perp n \}.$ Then $$ n \left( \prod_{k \in S_{n}} \sin \left( k \frac {\pi}{n} \right) \right)^{-2} \in \mathbb{Z}. $$

I think this is surprising but I have no proof.

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up vote 2 down vote accepted

There are a number of results for products that are closely connected to your product. There are some inessential differences (inverse is not taken), and generally products are over all $k$ from $1$ to $n-1$ relatively prime to $n$, but that would be taken care of by your squaring. Here is a link to a fully available paper by Steven Galovich.

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Thank you very much for this nice and highly relevant paper. Yes, I think the above formula can be justified by Galovich's theorems. –  Jonas Alomo Jun 24 '11 at 21:48

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