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

Question

How to show that

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

using the well-know representation of sine as infinite product that is

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

I have tried this problem for quite considerable amount of time. Indeed I also found a nice way to prove the former without using the latter. However I'm wondering if we can prove the result using the latter infinite product. Your help would be highly appreciated.

Thanks in advance.

Answer 1

Since $\sinh z = \frac{1}{i}\sin(iz)$, we have

$$\begin{align*} \sinh z &= \frac{1}{i}\sin(iz) \\ &= \frac{1}{i}\times (iz)\prod\limits_{n=1}^{\infty}\left(1 - \frac{(iz)^2}{n^2\pi^2}\right) \quad (\text{using the product formula for }\sin(\cdot))\\ &= z\prod\limits_{n=1}^{\infty}\left(1 -\frac{-z^2}{n^2\pi^2}\right)\\ &= z\prod\limits_{n=1}^{\infty}\left(1 +\frac{z^2}{n^2\pi^2}\right), \\ \end{align*} $$ as required.

Answer 2

As suggested in the comments, noting that $$\text{sinh}(z) = \frac{i}{1}\text{sin}(iz)$$ We have

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