Sin x 的无穷乘积表达

注意到：

\[ \boxed{ \begin{aligned} \sin\left(3x\right) &= 3\sin\left(x\right)-4\sin^3\left(x\right) \\ \sin\left(5x\right) &= 5\sin\left(x\right)-20\sin^3\left(x\right) + 16\sin^5\left(x\right) \\ &\cdots \end{aligned} } \]

数学归纳容易证明

\[ \sin\left(\left(2n+1\right)x\right)=\sin x\cdot P\left(\sin^2\left(x\right)\right) \]

这里 \(P\left(x\right)\) 是关于 \(x\) 的 \(n\) 次多项式。

又因为

\[ \lim_{x\to 0}\dfrac{\sin\left(\left(2n+1\right)x\right)}{\sin\left(x\right)}=2n+1 \]

因此 \(\left[x^0\right]P\left(x\right)=2n+1\)。

同时，\(\sin\left(\left(2n+1\right)x\right)\) 的根为 \(\dfrac{k\pi}{2n+1},k\in\mathbb{Z}\)，因此 \(\sin^2\left(\dfrac{k\pi}{2n+1}\right),k=1,2,\ldots,n\) 为 \(P\left(x\right)\) 的 \(n\) 个根，即

\[ \begin{aligned} P\left(x\right) &= \left(2n+1\right)\prod_{k=1}^{n}\left(1-\dfrac{x}{\sin^{2}\left(\dfrac{k\pi}{2n+1}\right)}\right) \\ \dfrac{\sin\left(\left(2n+1\right)x\right)}{\sin\left(x\right)} &= \left(2n+1\right)\prod_{k=1}^{n}\left(1-\dfrac{\sin^2\left(x\right)}{\sin^{2}\left(\dfrac{k\pi}{2n+1}\right)}\right) \\ \dfrac{\sin\left(\left(2n+1\right)x\right)}{\left(2n+1\right)\sin\left(x\right)} &= \prod_{k=1}^{n}\left(1-\dfrac{\sin^2\left(x\right)}{\sin^{2}\left(\dfrac{k\pi}{2n+1}\right)}\right) \\ \dfrac{\sin\left(x\right)}{\left(2n+1\right)\sin\left(\dfrac{x}{2n+1}\right)} &= \prod_{k=1}^{n}\left(1-\dfrac{\sin^2\left(\dfrac{x}{2n+1}\right)}{\sin^{2}\left(\dfrac{k\pi}{2n+1}\right)}\right) \\ \end{aligned} \]

令 \(n\to\infty\)，有

\[ \dfrac{\sin\left(x\right)}{x} = \prod_{k=1}^{\infty}\left(1-\dfrac{x^2}{k^2\pi^2}\right) \]