# Prove the trigonometric identity $(35)$

Prove that

\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.

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

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 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. – Chris's wise sister Oct 22 '12 at 15:27

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

$^nC_n(-1)^{\frac{n-1}2}t^n+^nC_{n-2}(-1)^{\frac{n-3}2}t^{n-2}+\cdots-^nC_3t^3+^nC_1t=0$

$^nC_n(-1)^{\frac{n-1}2}t^{n-1}+^nC_{n-2}(-1)^{\frac{n-3}2}t^{n-3}+\cdots-^nC_3t^2+^nC_1=0$ 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, $\prod_{k=1}^{n-1}_{(k\ne \frac n 2)}\tan \left(\frac{k \pi}{n}\right)=-(-1)^{\frac n 2}$

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.

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

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

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

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 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. – Chris's wise sister 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. – Chris's wise sister 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. – Chris's wise sister 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 $$\prod_{k=1}^{(n-1)/2}\tan\left(\dfrac{k\pi}{n}\right)=\sqrt{n}\tag{5}$$

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 Thanks for your work! (+1) – Chris's wise sister Oct 24 '12 at 19:38

Partial solution:

Note that

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

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

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

hence

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

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

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For $n$ even, using

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

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