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How to prove this equation: $$\sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n}\right) \cdots \sin\left(\frac{(n-1)\pi}{n}\right)=\frac{2n}{2^n}$$ There's a hint: Consider the product of $n-1$ non-zero roots of equation $(z+1)^n=1$, but I fail to understand it.

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To begin with, those roots will be given by $z+1=e^{ik/n}$ for $k=1,2,\ldots,n-1$. Does that help? Possibly useful is that these come in complex conjugate pairs (except for one, if $n$ is even). – Harald Hanche-Olsen May 2 '13 at 13:56
up vote 5 down vote accepted

To explain the given hint:

the roots of $x^n=1$ are $$x_k=e^{2ik\pi/n}\quad k=0,\ldots,n-1$$ so the product of roots except $0$ of $(z+1)^n=1$ is $$\prod_{k=1}^{n-1} (x_k-1)=\prod_{k=1}^{n-1}e^{2ik\pi/n}-1=\prod_{k=1}^{n-1}e^{ik\pi/n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=\prod_{k=1}^{n-1}2ie^{ik\pi/n}\prod_{k=1}^{n-1}\sin({k\pi/n})$$

hence to find the desired result you have to calculate (which isn't difficult) $$\prod_{k=1}^{n-1}(x_k-1)\quad\text{and}\quad \prod_{k=1}^{n-1}2ie^{ik\pi/n}$$

Hint for the first product you can use the Vieta's formulas and I trust you that you understand this second hint.

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+1 for the carefully calibrated hint. – Did May 2 '13 at 14:30

We know the roots of $z^n-1=0$ are $e^{\frac{2k\pi i}n}$ where $0\le k<n$

$$\implies z^n-1=\prod_{0\le k<n}(z-e^{\frac{2k\pi i}n})$$

$$(z-1)\sum_{0< r<n}z^r=(z-1)\prod_{0<k<n}(z-e^{\frac{2k\pi i}n})$$

So, $e^{\frac{2k\pi i}n}, (1\le k\le n-1)$ are roots of the equation $\sum_{1\le r\le n-1}z^r=0$

i.e., $\sum_{1\le r\le n-1}z^r=\prod_{0<k<n}(z-e^{\frac{2k\pi i}n})$

$$\text{Putting }z=1 \text{ in the above identity } ,\prod_{1\le k\le n-1}(1-e^{\frac{2k\pi i}n})=n$$

$$\text{ Taking modulus, }\prod_{1\le k\le n-1}\left|1-e^{\frac{2k\pi i}n}\right|=n--->(1)$$

$$\text{ Now, }1-e^{\frac{2k\pi i}n}=1-\left(\cos \frac{2k\pi}n+i\sin\frac{2k\pi}n\right)=2\sin^2\frac{k\pi}n-i2\sin\frac{k\pi}n\cos\frac{k\pi}n$$ $$=-2i\sin\frac{2k\pi}n\left(\cos\frac{k\pi}n-i\sin\frac{k\pi}n\right)$$

$$\prod_{1\le k\le n-1}(1-e^{\frac{2k\pi i}n})=\prod_{1\le k\le n-1}\left(-2i \sin\frac{2k\pi}n\left(\cos\frac{k\pi}n-i\sin\frac{k\pi}n\right)\right)$$

$$\text{ Taking modulus, }\prod_{1\le k\le n-1}\left|1-e^{\frac{2k\pi i}n}\right|=\prod_{1\le k\le n-1}\left|-2i \sin\frac{2k\pi}n\left(\cos\frac{k\pi}n-i\sin\frac{k\pi}n\right)\right|=2^{n-1}\prod_{1\le k\le n-1}\sin\frac{k\pi}n--->(2)$$ as $\sin\frac{k\pi}n>0$ for $1\le k \le n-1$

Compare $(1)$ and $(2)$

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