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I am stuck trying to prove

$$\prod_{k=1}^{n-1} \sin {\pi k \over n} = {n \over 2^{n-1}}$$

and I'd appreciate help.

What I have done so far:

$z^n - 1 = \prod_{k=1}^n (z - \xi^k)$ where $\xi = e^{2 \pi i k\over n}$. Dividing both sides by $z-1$ we get

$$ {z^n - 1 \over z - 1} = \prod_{k=1}^{n-1} (z - \xi^k)$$

Taking the limit $z \to 1$ on both sides, $$ n = \prod_{k=1}^{n-1} (1 - \xi^k)$$

So I'm getting close to what I want to prove but unfortunately, this is where I am stuck!

How to proceed form here?

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You might try using the identity $$\sin\left(\frac{\pi k}{n}\right)=\frac{e^{i\pi k/n}-e^{-i\pi k/n}}{2i}= -\frac{e^{-\pi i k/n}}{2i}\left(1-e^{2i\pi k/n}\right)$$ to convert the desired product to something that resembles the computation you have done so far.

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  • $\begingroup$ Thank you, I tried that. Now from $$ -2 i \sin z = e^{-iz} - e^{iz} = e^{-iz}(1 - e^{2iz})$$ I have $$ n = \prod_{k=1}^{n-1} {-2i \sin {\pi k \over n} \over e^{-i 2\pi k \over n}} = 2^{n-1} \prod_{k=1}^{n-1} -i (\sin {\pi k \over n}) e^{i{2 \pi k \over n}}$$ from which the denominator becomes $1$ in the product as $1 + 2 + \cdot + n - 1 = (n-1)n /2$. But what to do about the $i$ and the factor of $-1$? $\endgroup$ – Anna Apr 27 '15 at 5:01
  • $\begingroup$ You can pull them out to get $(-i)^{n-1}$. To get rid of this, you also have to also pull out the product of the $e^{i\pi k/n}$'s. $\endgroup$ – Rolf Hoyer Apr 27 '15 at 5:04
  • $\begingroup$ I also noticed that you introduced a factor of $2$ into the exponents that shouldn't be there. $\endgroup$ – Rolf Hoyer Apr 27 '15 at 5:09
  • $\begingroup$ Right, thank you. I introduced the factor of $2$ by mistake after doing the calculation for the second time. I will try now and see if I can eliminate the factor of $-i$... $\endgroup$ – Anna Apr 27 '15 at 5:16
  • $\begingroup$ I managed. Thank you. In retrospect it was rather easy. $\endgroup$ – Anna Apr 27 '15 at 5:26
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Let $$ P=\prod_{k=1}^{n-1}\sin\frac{k\pi}{n},\quad a:=e^{\frac{\pi i}{n}}=\cos\frac{\pi}{n}+i\sin\frac{\pi}{n},\quad k=0,1,\ldots,n-1. $$ Notice that the roots of $z^n-1$ are: $$ a^0=1,\, a^2,\,a^4,\, \ldots,\, a^{2(n-1)}, $$ therefore we have $$ \sum_{l=0}^{n-1}z^l=\frac{z^n-1}{z-1}=\prod_{k=1}^{n-1}(z-a^{2k}) \quad \forall z\ne 1. $$ By letting $z\to1$ we get: $$\tag{1} n=\prod_{k=1}^{n-1}(1-a^{2k})=\prod_{k=1}^{n-1}a^k(a^{-k}-a^k)=\left(\prod_{k=1}^{n-1}a^k\right)\cdot\left(\prod_{k=1}^{n-1}(a^{-k}-a^k)\right) $$ But $$\tag{2} \prod_{k=1}^{n-1}(a^{-k}-a^k)=\prod_{k=1}^{n-1}\left[-2i\sin\frac{k\pi}{n}\right]=P\prod_{k=1}^{n-1}(-2i)=(-2i)^{n-1}P, $$ and $$\tag{3} \prod_{k=1}^{n-1}a^k=a^{1+2+\ldots+(n-1)}=a^{n(n-1)/2}=e^{i(n-1)\pi/2}=i^{n-1} $$ Combining (10, (2), and (3), we get: $$ n=i^{n-1}(-2i)^{n-1}P=(-2)^{n-1}i^{2(n-1)}P=2^{n-1}P. $$ Hence $$ P=\frac{n}{2^{n-1}}. $$

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