Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that

\begin{equation} \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)= \left\{ \begin{aligned} \sqrt{n} \space \space \text{for $n$ odd}\\ \\ \ 1 \space \space \text{for $n$ even}\\ \end{aligned} \right. \end{equation}

I found this identity here at$(35)$. At the moment I don't know where I should start from. Thanks!

share|cite|improve this question
Have you tried the identity $\tan(a+b)=...$ identity in reverse? – picakhu Oct 22 '12 at 15:27
@picakhu: no. This sounds interesting. The generalization seems hard though. – user 1618033 Oct 22 '12 at 15:27
up vote 8 down vote accepted

$$\tan nx=\frac{^nC_1t-^nC_3t^3+^nC_5t^5-\cdots }{^nC_0t^0-^nC_2t^2+^nC_4t^4-\cdots }$$ where $t=\tan x$

If $\tan nx=0, x=\frac {k\pi}n$ where $0\le k< n$, clearly, the roots of this $n$-degree equation are $\tan\frac{k\pi}n$

If $n$ is odd,


$^nC_n(-1)^{\frac{n-1}2}t^{n-1}+^nC_{n-2}(-1)^{\frac{n-3}2}t^{n-3}+\cdots-^nC_3t^2+^nC_1=0$ if we exclude $k=0$

So, $\prod_{k=1}^{n-1}\tan \left(\frac{k \pi}{n}\right)=n(-1)^{\frac{n-1}2}$ (applying Vieta's formula)

Now, $\tan \left(\frac{(n-k) \pi}{n}\right)=\tan \left(\pi-\frac{k \pi}{n}\right)=-\tan \left(\frac{k \pi}{n}\right)$

So, there are $\frac{n-1}2$ such pairs and $\lfloor \frac{n-1}2 \rfloor=\frac{n-1}2$ as $n$ is odd.

$\prod_{k=1}^{n-1}\tan \left(\frac{k \pi}{n}\right)$ $=(-1)^{\frac{n-1}2}\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan^2 \left(\frac{k \pi}{n}\right) $

$\implies (-1)^{\frac{n-1}2}\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan^2 \left(\frac{k \pi}{n}\right)=n(-1)^{\frac{n-1}2} $

$\implies \left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2=n$

$\implies \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=\sqrt n$ as all the angles lies in $(0,\frac \pi 2)$

If $n$ is even,

$ ^nC_1t^{n-1}-^nC_3t^{n-3}+^nC_5t^{n-5}-\cdots+^nC_{n-1}(-1)^{\frac n 2}t=0$ which has roots $\tan\frac{k\pi}n$ where $0\le k<n$ and $k\ne \frac n 2$ as $k=\frac n 2$ corresponds to $\tan \frac \pi 2(=\infty)$ which has occurred as the co-efficient of $t^n$ is $0$.

So, $ ^nC_1t^{n-2}-^nC_3t^{n-4}+^nC_5t^{n-6}-\cdots+^nC_{n-1}(-1)^{\frac n 2}=0$ if we exclude $k=0$ i.e., $(n-2)$ degree equation in $t$.

So, $$\prod_{\substack{k=1 \\ k\neq \frac{n}{2}}}^{n-1}\tan \left(\frac{k \pi}{n}\right)=-(-1)^{\frac n 2}$$

Now, $\tan \left(\frac{(n-k) \pi}{n}\right)=\tan \left(\pi-\frac{k \pi}{n}\right)=-\tan \left(\frac{k \pi}{n}\right)$

So, there are $\frac{n-2}2=(\frac n 2 -1)$ such pairs and $\lfloor \frac{n-1}2 \rfloor=\frac{n-2}2$ as $n$ is even.

$-(-1)^{\frac n 2}$ $={\displaystyle\prod_{\substack{k=1 \\ k \neq \frac{n}{2}}}^{n-1}} \tan \left(\frac{k \pi}{n}\right)$ $=(-1)^{\frac{n-2}2}\left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2$

$\implies \left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2=1$

$\implies \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=1$ as all the angles lies in $(0,\frac \pi 2)$

share|cite|improve this answer
thanks!(+1). However, the point $\prod_{k=1}^{n-1}\tan \left(\frac{k \pi}{n}\right)=n(-1)^{\frac{n-1}2}$ doesn't seem very clear. – user 1618033 Oct 24 '12 at 19:36
@Chris'ssister, could you please have a look into the modified answer? – lab bhattacharjee Oct 24 '12 at 19:37
oh, that's better now! Thanks. – user 1618033 Oct 24 '12 at 19:39
@Chris'ssister, welcome! hope I could make the point clear. – lab bhattacharjee Oct 24 '12 at 19:44
like many others here, you're very skillful at math and I'm glad that you share your solutions with us. Thanks. – user 1618033 Oct 24 '12 at 20:07

This answer only presents the answers already given, but in what I hope is a more accessible form.

Even n

A key identity is $\tan(x)\tan(\pi/2-x)=1$. For $n$ even, this immediately verifies that $$ \prod_{k=1}^{n/2-1}\tan\left(\frac{k\pi}{n}\right)=1\tag{1} $$ Odd n

Since $\cos(nx)(1+i\tan(nx))=e^{inx}=(\cos(x)(1+i\tan(x)))^n$. Considering the ratio of the imaginary part to the real part, for odd $n$, we get $$ \tan(nx)=\frac{\displaystyle\sum_{k=0}^{(n-1)/2}(-1)^{k}\binom{n}{2k+1}\tan^{2k+1}(x)}{\displaystyle\sum_{k=0}^{(n-1)/2}(-1)^{k}\binom{n}{2k}\tan^{2k}(x)}\tag{2} $$ Therefore, $$ \sum_{k=0}^{(n-1)/2}(-1)^{k}\binom{n}{2k+1}x^{2k}=0\tag{3} $$ has roots at $x\in\left\{\tan\left(\dfrac{k\pi}{n}\right):1\le k\le n-1\right\}$. Since $(3)$ has even degree, the product of the roots is the ratio of the constant coefficient to the lead coefficient: $$ \prod_{k=1}^{n-1}\tan\left(\dfrac{k\pi}{n}\right)=(-1)^{(n-1)/2}n\tag{4} $$ Another key identity is $\tan(\pi-x)=-\tan(x)$. Combined with $(4)$, this yields $$ \prod_{k=1}^{(n-1)/2}\tan\left(\dfrac{k\pi}{n}\right)=\sqrt{n}\tag{5} $$

share|cite|improve this answer
Thanks for your work! (+1) – user 1618033 Oct 24 '12 at 19:38

I forgot that I had posted an answer to this question, and answered a duplicate question recently. Since this answer to the odd case is significantly different from the other answers, I have moved it here.

Note that $$ \tan^2(\theta/2)=-\left(\frac{e^{i\theta}-1}{e^{i\theta}+1}\right)^2 $$ Therefore, for odd $n$, $$ \begin{align} \prod_{k=1}^{(n-1)/2}\tan^2(k\pi/n) &=\prod_{k=1}^{(n-1)/2}(-1)\left(\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\right)^2\\ &=\prod_{k=1}^{(n-1)/2}\left(\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\right)\left(\frac{e^{-2\pi ik/n}-1}{e^{-2\pi ik/n}+1}\right)\\ &=\prod_{k=1}^{(n-1)/2}\left(\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\right)\left(\frac{e^{2\pi i(n-k)/n}-1}{e^{2\pi i(n-k)/n}+1}\right)\\ &=\prod_{k=1}^{n-1}\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\\ &=\prod_{k=1}^{n-1}\frac{1-e^{2\pi ik/n}}{1+e^{2\pi ik/n}}\\ &=\lim_{z\to1}\prod_{k=1}^{n-1}\frac{z-e^{2\pi ik/n}}{z+e^{2\pi ik/n}}\\ &=\lim_{z\to1}\frac{z^n-1}{z-1}\frac{z+1}{z^n+1}\\[12pt] &=n \end{align} $$ Since tangent is positive in the first quadrant, $$ \prod_{k=1}^{(n-1)/2}\tan(k\pi/n)=\sqrt{n} $$

share|cite|improve this answer
Nice answer (+1) – user 1618033 Aug 24 '13 at 16:57

Partial solution:

Note that

$$\tan(\frac{\pi}{2}-x) \tan(x)=1$$

This solves the problem for $n$ even: if $n=2m$, then

$$P:=\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=\prod_{k=1}^{m-1}\tan \left(\frac{k \pi}{2m}\right) =\prod_{k=1}^{m-1}\tan \left(\frac{(m-k) \pi}{2m}\right) $$


$$P^2=\prod_{k=1}^{m-1}\tan \left(\frac{k \pi}{2m}\right)\prod_{k=1}^{m-1}\tan \left(\frac{(m-k) \pi}{2m}\right)=1$$

and since $P$ is positive, you are done...

share|cite|improve this answer

For $n$ even, using

this equation

it is obvious since $\cos(\frac {\pi}{2}) = 0$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.