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Prove that for a natural number $n$,

$$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}.$$

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This follows from a continued fraction identity for which, I think, there is a lengthy proof. But, I thought, that there may be a direct geometric or another proof. Constructing a polynomial w/the sines and tangents roots may be helpful. – DVD May 15 '13 at 22:20
That's very informative, thanks - in the future, why not include that sort of thing to begin with? (If you don't want a lengthy symbolic proof, tell people so that they don't put lots of time and effort into something that wasn't what you're after). – Zev Chonoles May 15 '13 at 22:42
up vote 2 down vote accepted

The proof for this should be identical to the one for: $$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$

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