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How to evaluate this? $$\prod_{k=1}^m \tan \frac{k\pi}{2m+1}$$

My work

I couldn't figure out a method to solve this product. I thought that this identity could help. $$\frac{e^{i\theta}-1}{e^{i\theta}+1}=i\tan \frac{\theta}{2}$$

By supposing $z=e^{\frac{2k\pi}{2m+1}i}$, then, $$i\tan \frac{k\pi}{2m+1}=\frac{z-1}{z+1}$$ So, the product $$\displaystyle\prod_{k=1}^m \tan\frac{k\pi}{2m+1}=\displaystyle\prod_{k=1}^m \frac{z-1}{i(z+1)}=\displaystyle\prod_{k=1}^m \frac{e^{\frac{2k\pi}{2m+1}i}-1}{i(e^{\frac{2k\pi}{2m+1}i}+1)}$$

which is getting more complicated.

Answer is $\sqrt{2m+1}$. Any help is appreciated.

user
  • 26,272

2 Answers2

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So you can start by breaking the product to get: $$\prod_{k=1}^{2m}\tan{(\frac{k\pi}{2m+1})}=\prod_{k=1}^{2m}\sin{(\frac{k\pi}{2m+1})}\prod_{k=1}^{2m}\frac{1}{\cos{(\frac{k\pi}{2m+1}})}$$ Now : $$\prod_{k=1}^{2m}\cos{(\frac{k\pi}{2m+1})}=\frac{(-1)^{m}}{2^{2m}}$$ Proof: $$\cos{(\frac{k\pi}{2m+1})}=\exp{(\frac{ki\pi}{2m+1})}\frac{(1+\exp{(\frac{-2ik\pi}{2m+1}}))}{2}$$ If we have polynomial $z^{2m+1}-1$ then its $2m+1$ roots of unity are $\exp{(\frac{-2ik\pi}{2m+1}}):1\le k \le 2m+1$ Thus : $$z^{2m+1}-1=-\prod_{k=1}^{2m+1}(-z+\exp{(\frac{-2ik\pi}{2m+1}}))$$ $$\therefore \frac{(-1)^{2m+1}-1}{2}=-1=-\prod_{k=1}^{2m}(1+\exp{(\frac{-2ik\pi}{2m+1}}))$$ $$\therefore \prod_{k=1}^{2m}\cos{(\frac{k\pi}{2m+1})}=\frac{1}{2^{2m}}\exp(\frac{i(2m)(2m+1)\pi}{2(2m+1)})=\frac{(-1)^{m}}{2^{2m}}$$ Now we have to evaluate : $$\prod_{k=1}^{2m}\sin{(\frac{k\pi}{2m+1})}$$ Using similar process like one above we get: $$\prod_{k=1}^{2m}\sin{(\frac{k\pi}{2m+1})}=(-1)^{m}\lim_{z\to1}\frac{z^{2m+1}-1}{(z-1)(2i)^{2m+1}}=\frac{(2m+1)(-1)^{m}}{(2i)^{2m}}$$ Okay we have these two products but how do they relate to the original problem. Well $$\tan{(\frac{(2m+1-k)\pi}{2m+1})}=\tan{(\pi-\frac{k\pi}{2m+1})}=-\tan{(\frac{k\pi}{2m+1})}$$ thus: $$\prod_{k=1}^{2m}\tan{(\frac{k\pi}{2m+1})}=\prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})}\prod_{k=m+1}^{2m}\tan{(\frac{k\pi}{2m+1})}=(-1)^{m}(\prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})})^{2}$$ $$\prod_{k=1}^{2m}\tan{(\frac{k\pi}{2m+1})}=\frac{(2m+1)}{(2i)^{2m}}2^{2m}=(-1)^{m}(2m+1)$$ $$\therefore (-1)^{m}(\prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})})^{2}=(-1)^{m}(2m+1)$$ $$\therefore \prod_{k=1}^{m}\tan{(\frac{k\pi}{2m+1})}=\sqrt{2m+1} $$

mathlover123
  • 528
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Let

$$f(x) = x^{n-1} + x^{n-2} + x^{n-3} \>\cdots \> +\>x +1 = \prod_{k=1}^{n-1}(x - e^{i\frac{2\pi k}n}) $$ and set $n=2m+1$ to evaluate

$$\frac{f(1)}{f(-1)}= 2m+1 = \prod_{k=1}^{2m}\frac{1 - e^{i\frac{2\pi k}{2m+1}}}{1 + e^{i\frac{2\pi k}{2m+1}}} = i^{2m} \prod_{k=1}^{2m}\tan \frac{k\pi}{2m+1} = \prod_{k=1}^{m}\tan^2\frac{k\pi}{2m+1} $$ Thus

$$\prod_{k=1}^m \tan \frac{k\pi}{2m+1}=\sqrt{2m+1}$$

Quanto
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