求证：$\cos{\dfrac{\pi}{2n+1}}\cos{\dfrac{2\pi}{2n+1}}\cdots\cos{\dfrac{2n\pi}{2n+1}}=\dfrac{(-1)^n}{2^{2n}}$，其中 $n\in \mathbb{N}$．
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【解析】
根据题意，有$$\left|\cos{\dfrac{\pi}{2n+1}}\cos{\dfrac{2\pi}{2n+1}}\cdots\cos{\dfrac{2n\pi}{2n+1}}\right|
=\dfrac{1}{2^{2n}}\cdot\left|\dfrac{\sin{\dfrac{2\pi}{2n+1}}\sin{\dfrac{4\pi}{2n+1}}\cdots\sin{\dfrac{4n\pi}{2n+1}}}{\sin{\dfrac{\pi}{2n+1}}\sin{\dfrac{2\pi}{2n+1}}\cdots\sin{\dfrac{2n\pi}{2n+1}}}\right|,$$记 $[m]$ 表示正整数 $m$ 除以 $2n+1$ 所得的余数，则有$$\left\{[1],[2],\cdots,[2n]\right\}=\left\{[2],[4],\cdots,[4n]\right\},$$由以上两式可得$$\left|\cos{\dfrac{\pi}{2n+1}}\cos{\dfrac{2\pi}{2n+1}}\cdots\cos{\dfrac{2n\pi}{2n+1}}\right|=\dfrac{1}{2^{2n}}.$$
=\dfrac{1}{2^{2n}}\cdot\left|\dfrac{\sin{\dfrac{2\pi}{2n+1}}\sin{\dfrac{4\pi}{2n+1}}\cdots\sin{\dfrac{4n\pi}{2n+1}}}{\sin{\dfrac{\pi}{2n+1}}\sin{\dfrac{2\pi}{2n+1}}\cdots\sin{\dfrac{2n\pi}{2n+1}}}\right|,$$记 $[m]$ 表示正整数 $m$ 除以 $2n+1$ 所得的余数，则有$$\left\{[1],[2],\cdots,[2n]\right\}=\left\{[2],[4],\cdots,[4n]\right\},$$由以上两式可得$$\left|\cos{\dfrac{\pi}{2n+1}}\cos{\dfrac{2\pi}{2n+1}}\cdots\cos{\dfrac{2n\pi}{2n+1}}\right|=\dfrac{1}{2^{2n}}.$$
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