\(\sin{x}\)的无穷乘积式
观察以下两个式子:
\[\boxed{\begin{aligned}
&\sin{3x}=3\sin{x}-4\sin^3{x}\\
&\sin{5x}=5\sin{x}-20\sin^3{x}+16\sin^5{x}\\
&\cdots
\end{aligned}}
\]
可以发现
\[\sin{(2n+1)x}=\sin{x}\cdot\mathrm{P}(\sin^2{x})
\]
其中\(\mathrm{P}(x)\)是关于\(x\)的\(n\)次多项式.
因为\(\lim_{x\to0}\dfrac{\sin{(2n+1)x}}{\sin{x}}=2n+1\)，所以\(\mathrm{P}(x)\)的常数项为\(2n+1\).
同时，\(\sin{(2n+1)x}\)的根为\(\dfrac{k\pi}{2n+1},k\in\mathbb{Z}\)，所以\(\sin^2{\dfrac{k\pi}{2n+1}},k=1,2,\cdots,n\)恰为\(\mathrm{P}(x)\)的\(n\)个根.
\[\begin{aligned}
&\mathrm{P}(x)=(2n+1)\left(1-\frac{x}{\sin^2{\dfrac{\pi}{2n+1}}}\right)\left(1-\frac{x}{\sin^2{\dfrac{2\pi}{2n+1}}}\right)\cdots\left(1-\frac{x}{\sin^2{\dfrac{n\pi}{2n+1}}}\right)\\
&\mathrm{P}(x)=(2n+1)\prod_{k=1}^{n}\left(1-\frac{x}{\sin^2{\cfrac{k\pi}{2n+1}}}\right)\\
&\frac{\sin{(2n+1)x}}{\sin{x}}=(2n+1)\prod_{k=1}^{n}\left(1-\frac{\sin^2{x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)\\
&\frac{\sin{x}}{(2n+1)\sin{\cfrac{1}{2n+1}x}}=\prod_{k=1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)\\
&\frac{\sin{x}}{(2n+1)\sin{\cfrac{1}{2n+1}x}\displaystyle\prod_{k=1}^{m}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)}=\prod_{k=m+1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)\\
\end{aligned}
\]
左边对\(n\)取极限，使\(n\to\infty\).
\[\frac{\sin{x}}{x \displaystyle \prod_{k=1}^{m}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}
\]
对于右边，有这一不等式
\[\boxed{\begin{aligned}
&\frac{2}{\pi}x<\sin{x}<x,x\in\left(0,\frac{\pi}{2}\right)\\
&\sin^2{\frac{1}{2n+1}x}<\frac{x^2}{(2n+1)^2}\\
&\sin^2{\frac{k\pi}{2n+1}}>\frac{4k^2}{(2n+1)^2}\\
&\frac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\cfrac{k\pi}{2n+1}}}<\frac{x^2}{4k^2}\\
\end{aligned}}
\]
于是就有
\[\begin{aligned}
&1>\prod_{k=m+1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)>\prod_{k=m+1}^{n}\left(1-\cfrac{x^2}{4k^2}\right)>\prod_{k=m+1}^{\infty}\left(1-\cfrac{x^2}{4k^2}\right)\\
&1>\prod_{k=m+1}^{n}\left(1-\cfrac{\sin^2{\cfrac{1}{2n+1}x}}{\sin^2{\dfrac{k\pi}{2n+1}}}\right)>\prod_{k=m+1}^{\infty}\left(1-\cfrac{x^2}{4k^2}\right)
\end{aligned}
\]
这个式子是否成立与\(n\)的取值无关，所以代入\(n\to\infty\)的极限可得
\[\boxed{1>\frac{\sin{x}}{x \displaystyle \prod_{k=1}^{m}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}>\prod_{k=m+1}^{\infty}\left(1-\cfrac{x^2}{4k^2}\right)}
\]
再在两边对\(m\)取极限，使\(m\to\infty\).两边极限都是\(1\)，所以根据极限夹逼准则，可得
\[\boxed{\frac{\sin{x}}{x }= \prod_{k=1}^{\infty}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}
\]
\[\boxed{\sin{x}=x \prod_{k=1}^{\infty}\left(1-\cfrac{x^2}{k^2\pi^2}\right)}
\]
\(\cfrac{\sin{x}}{x }\)也被称为\(\mathrm{sinc}\,x\).