Using the infinite product of $\sin(\pi z)$, one can find the Hadamard product for $e^z-1$:
$$e^z-1 =2ie^{z/2}\sin(-iz/2)= 2i e^{z/2} (-iz/2) \prod_n \left(1+\frac{z^2}{4\pi n^2}\right)\\= e^{z/2} z \prod_n \left(1+\frac{z^2}{4\pi n^2}\right).$$
I don't see a way to find the product for $\cos\pi z$. A naive attempt is letting $\{a_n\}\subset{\Bbb C}$ be all the zeros of $\cos(\pi z)$ and showing the possible convergence of
$$
\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)
$$
Is there an alternative way to find the Hadamard product in the title for $\cos\pi z$?
Best Answer
Hint: Use $\sin(2z)=2\sin(z)\cos(z)$ so that $$\cos(z)=\frac{\sin(2z)}{2\sin(z)}.$$ If you're careful about how you write it, you will see that all of the 'even terms' cancel nicely. I do not have time right now, but if you haven't been able to solve it within a few hours, I will return and post my solution.