證明: sin(π/n)×sin(2π/n)×sin(3π/n)×...×sin((n-1)π/n) =n/2^(n-1)

正弦連乘積公式

$$\sin{\pi \over n} \cdot \sin{2\pi \over n} \cdots \sin{(n-1)\pi \over n} = \frac{n}{2^{n-1}}$$

證明：

$$\cos \theta +i\sin \theta=e^{i\theta} \Rightarrow \sin \theta = \cfrac{e^{i\theta}-e^{-i\theta}}{2i} \Rightarrow \sin \cfrac{k\pi}{ n} = \cfrac{e^{i{k\pi\over n}}-e^{-i{k\pi\over n}}}{2i} =\frac{e^{-i{k\pi \over n}}}{2i}\left(e^{i{2k\pi\over n}} -1\right)\\ \Rightarrow 
\sin\cfrac{\pi}{n} \cdot \sin\cfrac{2\pi}{n} \cdots \sin\cfrac{(n-1)\pi}{n} \\= \cfrac{e^{-i{\pi \over n}(1+2+\cdots +(n-1))}}{(2i)^{n-1}} \left( e^{i{2\pi\over n}}-1\right)\left( e^{i{4\pi\over n}}-1\right)\cdots \left( e^{i{2(n-1)\pi\over n}}-1\right)\\=\cfrac{e^{-i{\pi (n-1)\over 2}}}{(2i)^{n-1}} \left( e^{i{2\pi\over n}}-1\right)\left( e^{i{4\pi\over n}}-1\right)\cdots \left( e^{i{2(n-1)\pi\over n}}-1\right)\\= (2i)^{1-n}e^{-i{\pi (n-1)\over 2}} (\omega-1 ) ( \omega^2-1 )\cdots ( \omega^{n-1}-1 ),其中\omega = e^{i{2\pi\over n}}\\=2^{1-n}\cdot e^{i{\pi \over 2}(1-n)}\cdot e^{-i{\pi (n-1)\over 2}} (\omega-1 ) ( \omega^2-1 )\cdots ( \omega^{n-1}-1 ) (\because i=e^{i{\pi\over 2} }\Rightarrow i^{1-n}=e^{i{\pi\over 2}(1-n) })\\ =2^{1-n}e^{i(1-n)\pi}(\omega-1 ) ( \omega^2-1 )\cdots ( \omega^{n-1}-1 ) =2^{1-n}(-1)^{(1-n)}(\omega-1 ) ( \omega^2-1 )\cdots ( \omega^{n-1}-1 )\\ =(-2)^{1-n}(\omega-1 ) ( \omega^2-1 )\cdots ( \omega^{n-1}-1 )=(-2)^{1-n}(-1)^{1-n}(1-\omega ) (1- \omega^2)\cdots (1- \omega^{n-1}) \\ =2^{1-n}(1-\omega ) (1- \omega^2)\cdots (1- \omega^{n-1})\\ 令f(x)=x^n-1 =(x-1)(1+x+x^2+\cdots +x^{n-1})，\\則f(x)=0的根為1,\omega,\omega^2,\dots,\omega^{n-1},其中\omega = e^{i{2\pi\over n}}\\ \Rightarrow f(x)=(x-1)(1+x+x^2+\cdots +x^{n-1})= (x-1)(x-\omega) (x-\omega^2)\cdots(x-\omega^{n-1})\\ \Rightarrow 1+x+x^2+\cdots +x^{n-1}=(x-\omega) (x-\omega^2)\cdots(x-\omega^{n-1})\\ x=1代入上式\Rightarrow n=(1-\omega ) (1- \omega^2)\cdots (1- \omega^{n-1})\\ \Rightarrow \sin\cfrac{\pi}{n} \cdot \sin\cfrac{2\pi}{n} \cdots \sin\cfrac{(n-1)\pi}{n} =2^{1-n}(1-\omega ) (1- \omega^2)\cdots (1- \omega^{n-1}) =2^{1-n}\cdot n\\ =\frac{n}{2^{n-1}},故得證$$