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Using that: $$ x^{n - 1} + x^{n - 2} + \cdots + x + 1 = \left(x - w\right)\left(x - w^{2}\right)\ldots\left(x - w^{n - 1}\right) $$

Prove that: $$ \sin\left(\pi \over n\right)\sin\left(2\pi \over n\right)\ldots \sin\left(\left[n-1\right]\pi \over n\right) = {n \over 2^{n - 1}} $$


marked as duplicate by Hans Lundmark, Scientifica, Arnaud D., José Carlos Santos, Namaste Oct 29 '18 at 13:47

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  • $\begingroup$ Vieta's formulae...? $\endgroup$ – Pedro Tamaroff Jan 10 '14 at 3:37

Notice that $$ w=e^{i\frac{2\pi}{n}}=z^2, $$ with $$ z=e^{i\frac{\pi}{n}} $$ and \begin{eqnarray} n&=&(1-w)\ldots(1-w^{n-1})=(|z|^2-z^2)\ldots(|z|^{2n-2}-z^{2n-2})\\ &=&(-1)^{n-1}z^{1+2+\ldots+(n-1)}(z-\bar{z})\ldots(z^{n-1}-\bar{z}^{n-1})\\ &=&(-1)^{n-1}z^{\frac{n(n-1)}{2}}2i\left(\sin\frac{\pi}{n}\right)\ldots2i\sin\left(\frac{(n-1)\pi}{n}\right)\\ &=&(-1)^{n-1}(2i)^{n-1}e^{i\frac{(n-1)\pi}{2}}\left(\sin\frac{\pi}{n}\right)\ldots\sin\left(\frac{(n-1)\pi}{n}\right)\\ &=&2^{n-1}\sin\left(\frac{\pi}{n}\right)\ldots\sin\left(\frac{(n-1)\pi}{n}\right). \end{eqnarray} Thus $$ \sin\left(\frac{\pi}{n}\right)\ldots\sin\left(\frac{(n-1)\pi}{n}\right)=\frac{n}{2^{n-1}}. $$

  • $\begingroup$ Thank you, how did you notice that $w=e^{i\frac{2\pi}{n}}=z^2$ ? $\endgroup$ – Victor Francisco Salazar Garci Jan 10 '14 at 4:54
  • $\begingroup$ By solving the equation $1+x+x^2+\ldots+x^{n-1}=0$, and the fact that $1+x+\ldots+x^{n-1}=\frac{1-x^n}{1-x}$ for $x\ne 1$. $\endgroup$ – Mercy King Jan 10 '14 at 4:57

Slightly different approach. Recall that $\sin z = \frac{e^{zi}-e^{-zi}}{2i}$, so substituting yields $$\prod_{j=1}^{n-1}\sin\frac{j\pi}{n}=\prod_{j=1}^{n-1}\left[\frac{e^{ji\pi/n}-e^{-ji\pi/n}}{2i}\right]$$ Factoring the numerators of the $\text{RHS}$ yields $$\prod_{j=1}^{n-1}\left[\frac{e^{ji\pi/n}-e^{-ji\pi/n}}{2i}\right]=\prod_{j=1}^{n-1}\left[\frac{e^{-ji\pi/n}\left(e^{2ji\pi/n}-1\right)}{2i}\right]\text{ } (*)$$

The factors in the numerator are similar to the factors of $x^n-1$, and it is well known that $x^n-1=(x-1)(x^{n-1}+\cdots + x+ x)$. It follows that the roots of $(x^{n-1}+\cdots + x+ x)$ are the $n$th roots of unity other than $1$. Hence we can write $$(x^{n-1}+\cdots + x+ x) = \prod_{j=1}^{n-1}(x-\omega^{j})$$ where $\omega$ is an $n$th root of unity.

$x=1$ yields $\prod_{j=1}^{n-1}(1-\omega^j)=n \text{ } (**)$.

Now we have to relate this back to the desired product of sines, so we reverse the factoring done in $(*)$. Observe that we have $1-e^{i\theta}=e^{i\theta/2}(e^{-i\theta/2}-e^{i\theta/2})=-e^{i\theta/2}(e^{i\theta/2}-e^{-i\theta/2})$, which yields $$\begin{align*}\left|1-e^{i\theta}\right|&=\left|-e^{i\theta/2}(e^{i\theta/2}-e^{-i\theta/2}\right|\\ &= \left|-e^{i\theta/2}\right|\left|e^{i\theta/2}-e^{-i\theta/2}\right|\\ &=\left|e^{i\theta/2}-e^{-i\theta/2}\right|\end{align*}$$

Now, using the fact that $\sin z =\frac{e^{iz}-e^{iz}}{2i}$ yields $$\begin{align*}\left|e^{i\theta/2}-e^{-i\theta/2}\right| &=\left|2i\sin\frac{\theta}{2}\right|\\ &=2\left|\sin\frac{\theta}{2}\right|\end{align*}$$

Since $\frac{\theta}{2}\in\left[0, \pi\right]$ we have $\sin\frac{\theta}{2}\ge 0\implies\left|1-e^{i\theta}\right|=\sin\frac{\theta}{2}$. In fact, for $\theta$ of the form $\frac{2k\pi}{n}$, where $k\in\left[1, n-1\right]$, we have $\omega^k=e^{2ki\pi/n}=e^{i\theta}$, so $$|1-\omega^k|=2\sin\frac{k\pi}{n}.$$

Taking the magnitude of both sides of the equation $\prod_{j=1}^{n-1}(1-\omega^j)=n$ yields $$\prod_{j=1}^{n-1}2\sin\frac{j\pi}{n}=n.$$ Dividing both sides by $2^{n-1}$ we have $$\prod_{j=1}^{n-1}\sin\frac{j\pi}{n}=\frac{n}{2^{n-1}}.$$


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