قرار دهید:
$$A:=sin\frac{\pi}{2n+1}.sin\frac{2\pi}{2n+1}...sin\frac{n\pi}{2n+1}$$
$$B:=cos\frac{\pi}{2n+1}.cos\frac{2\pi}{2n+1}...cos\frac{n\pi}{2n+1}$$
$$C:=sin\frac{\pi}{2n+1}.sin\frac{3\pi}{2n+1}...sin\frac{(2n-1)\pi}{2n+1}$$
$$D:=sin\frac{2\pi}{2n+1}.sin\frac{4\pi}{2n+1}...sin\frac{2n\pi}{2n+1}$$
$$E:=sin\frac{(n+1)\pi}{2n+1}.sin\frac{(n+2)\pi}{2n+1}...sin\frac{2n\pi}{2n+1}$$
حالا اگر $1 \leq k \leq n$ با توجه به اینکه $(2k-1)+2(n-k+1)=2n+1$ و $sin(\pi-x)=sinx$ داریم $C=D$ و با استدلالی مشابه $A=E$. از طرفی دیگر:
$$2^nAB=(2sin\frac{\pi}{2n+1}.cos\frac{\pi}{2n+1}).(2sin\frac{2\pi}{2n+1}.cos\frac{2\pi}{2n+1})...$$
$$.(2sin\frac{n\pi}{2n+1}.cos\frac{n\pi}{2n+1})$$
$$=sin\frac{2\pi}{2n+1}.sin\frac{4\pi}{2n+1}...sin\frac{2n\pi}{2n+1}$$
$$=C=D$$
$$ \Rightarrow (2^nAB)^2=C^2=CC=CD=AE=A^2$$
$$ \Rightarrow (2^nB)^2=1$$
$$ \Rightarrow 2^nB=1$$
$$ \Rightarrow B= \frac{1}{2^n} $$
$\Box$
