Periodicity of Sine from infinite Product Formula: $z\prod_{n=1}^{\infty}\left( 1 - \frac{z^2}{n^2\pi^2}\right)$

Question

Let $\sin z$ be defined by the following infinite product:

$$ \sin (z) = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2\pi^2} \right) $$

How can one derive that $\sin(z + 2\pi) = \sin(z)$?

Answer 1

One may write $$ \begin{align} &(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{(z+2\pi)^2}{n^2\pi^2}\right) \\\\=\:&(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{z+2\pi}{n\pi}\right)\prod_{n=1}^{N} \left(1+\frac{z+2\pi}{n\pi}\right) \\\\=\:&(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{z}{n\pi}-\frac{2}{n}\right)\prod_{n=1}^{N} \left(1+\frac{z}{n\pi}+\frac{2}{n}\right) \\\\=\:&(z+2\pi)\prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right) \prod_{n=1}^{N} \left(1-\frac{2}{n\left(1-\frac{z}{n\pi} \right)}\right)\left(1+\frac{2}{n\left(1+\frac{z}{n\pi} \right)}\right) \\\\=\:&(z+2\pi)\prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right) \cdot\prod_{n=1}^{N} \frac{((n-2) \pi -z) ((n+2) \pi +z)}{(n \pi -z) (n \pi +z)} \end{align} $$ then factors telescope in the latter product giving $$ \begin{align} &(z+2\pi) \prod_{n=1}^{N} \left(1-\frac{(z+2\pi)^2}{n^2\pi^2}\right) \\\\=\:&(z+2\pi)\cdot \prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right)\cdot \frac{z(\pi+N\pi +z) ((2+N)\pi+z)}{(z+2\pi)((N-1)\pi-z)(N\pi-z)} \\\\=\:&z\cdot \prod_{n=1}^{N} \left(1-\frac{z^2}{n^2\pi^2}\right)\cdot \frac{(\pi+N\pi+z)((2+N)\pi+z)}{((N-1)\pi-z) (N\pi-z)} \end{align} $$ then letting $N \to \infty$ gives the desired identity.