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Prove that, for a natural number $n$, $$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$$

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 with the sines and tangents roots may be helpful.

Blue
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DVD
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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
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    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

2 Answers2

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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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Nathaniel Bubis
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The products for $\sin(x)$ and $\cos(x)$ can be evaluated in the same manner. Taking the ratio gives the product for $\tan(x)$. $$ \begin{align} \prod_{k=1}^n\sin\left(\frac{k\pi}{2n+1}\right) &=\prod_{k=1}^n\left|\frac{e^{i\frac{k\pi}{2n+1}}-e^{-i\frac{k\pi}{2n+1}}}{2i}\right|\tag{1a}\\ &=\frac1{2^n}\prod_{k=1}^n\left|1-e^{-i\frac{2k\pi}{2n+1}}\right|\tag{1b}\\ &=\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|1-e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{1c}\\ &=\lim_{z\to1}\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|z-e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{1d}\\ &=\lim_{z\to1}\frac1{2^n}\left|\frac{z^{2n+1}-1}{z-1}\right|^{1/2}\tag{1e}\\[6pt] &=\frac{\sqrt{2n+1\vphantom{-}}}{2^n}\tag{1f} \end{align} $$ Explanation:
$\text{(1a):}$ $\sin(x)\gt0$ over the product
$\phantom{\text{(1a):}}$ write $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(1b):}$ $|i|=\left|e^{i\frac{k\pi}{2n+1}}\right|=1$
$\text{(1c):}$ $|1-z|=|1-\bar z|$
$\text{(1d):}$ write $1=\lim\limits_{z\to1}z$
$\text{(1e):}$ $\prod\limits_{\substack{k=-n\\k\ne0}}^n\left(z-e^{-i\frac{2k\pi}{2n+1}}\right)=\frac{z^{2n+1}-1}{z-1}$
$\text{(1f):}$ evaluate the limit $$ \begin{align} \prod_{k=1}^n\cos\left(\frac{k\pi}{2n+1}\right) &=\prod_{k=1}^n\left|\frac{e^{i\frac{k\pi}{2n+1}}+e^{-i\frac{k\pi}{2n+1}}}2\right|\tag{2a}\\ &=\frac1{2^n}\prod_{k=1}^n\left|1+e^{-i\frac{2k\pi}{2n+1}}\right|\tag{2b}\\ &=\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|1+e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{2c}\\ &=\lim_{z\to1}\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|z+e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{2d}\\ &=\lim_{z\to1}\frac1{2^n}\left|\frac{z^{2n+1}+1}{z+1}\right|^{1/2}\tag{2e}\\[6pt] &=\frac1{2^n}\tag{2f} \end{align} $$ Explanation:
$\text{(2a):}$ $\cos(x)\gt0$ over the product
$\phantom{\text{(2a):}}$ write $\cos(x)=\frac{e^{ix}+e^{-ix}}2$
$\text{(2b):}$ $\left|e^{i\frac{k\pi}{2n+1}}\right|=1$
$\text{(2c):}$ $|1+z|=|1+\bar z|$
$\text{(2d):}$ write $1=\lim\limits_{z\to1}z$
$\text{(2e):}$ $\prod\limits_{\substack{k=-n\\k\ne0}}^n\left(z+e^{-i\frac{2k\pi}{2n+1}}\right)=\frac{z^{2n+1}+1}{z+1}$
$\text{(2f):}$ evaluate the limit

Thus, taking the ratio of $(1)$ and $(2)$ gives $$ \prod_{k=1}^n\tan\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1\vphantom{-}}\tag3 $$

robjohn
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