Infinite Product of 1-1/n^4

Question

When I look at wolframalpha, I get $$\prod_{n=2}^\infty \left(1-\frac{1}{n^4}\right) = \frac{\sinh(4\pi)}{4\pi}.$$

My only guess where this comes from would be the euler's sine product formula$$\sin(z) = z\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2\pi^{2}}\right).$$ But, this gives $\sinh(π)=-i\sin(i\pi)= i\pi\prod\left(1+\frac{1}{n^2}\right).$ I've fiddled around and cannot manage to get an $n^{4}$ showing up anywhere.

Answer 1

Hint: You already have $\prod \left(1+\frac{1}{n^2}\right)$. What do you need to multiply by to get what you are after? Can you create that with what you already know?

Answer 2

Hint: Observe you have \begin{align} \prod^\infty_{n=2} \left(1-\frac{1}{n^2} \right) \prod^\infty_{n=2} \left(1+\frac{1}{n^2} \right) \end{align}