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The question was also answered here: How can I deduce $\cos\pi z=\prod_{n=0}^{\infty}(1-4z^2/(2n+1)^2)$?
I found this formula on Wikipedia under Weierstrass factorization theorem,
$\cos (\pi z) = \underset{n \in \mathbb{Z}_{\text{odd}}}{\prod} (1-\frac{2 z}{n})e^{2z/n}$
However, I am not able to prove it from
$\sin (\pi z) = \pi z \underset{n \neq 0}{\prod} (1-\frac{z}{n})e^{z/n}$
The derivation of $\sin (\pi z)$ on page 175 of "Functions of One Complex Variable I" by Conway may be helpful.
Would someone please help me prove it? Thanks!
Edit: I would like to derive $\cos (\pi z)$ from $\sin (\pi z)$. Sorry for the confusion
Using the trigonometric identity $\sin (2\alpha) = 2 \sin \alpha \cos \alpha$ and the Weierstraß product of the sine, we obtain
\begin{align} \cos (\pi z) &= \frac{\sin (2\pi z)}{2\sin (\pi z)}\\ &= \frac{2\pi z \prod\limits_{n\neq 0}\bigl(1 - \frac{2z}{n}\bigr)e^{2z/n}}{2\pi z \prod\limits_{k\neq 0} \bigl(1 - \frac{z}{k}\bigr)e^{z/k}}\\ &= \frac{\prod\limits_{n\neq 0}\bigl(1 - \frac{2z}{n}\bigr)e^{2z/n}}{\prod\limits_{k\neq 0} \bigl(1 - \frac{z}{k}\bigr)e^{z/k}}\\ &= \frac{\prod\limits_{n\neq 0}\bigl(1 - \frac{2z}{n}\bigr)e^{2z/n}}{\prod\limits_{k\neq 0} \bigl(1 - \frac{2z}{2k}\bigr)e^{2z/2k}}\\ &= \frac{\prod\limits_{n \neq 0}\bigl(1 - \frac{2z}{n}\bigr)e^{2z/n}}{\prod\limits_{\substack{m\neq 0 \\ m \text{ even}}}\bigl(1 -\frac{2z}{m}\bigr)e^{2z/m}}\\ &= \prod_{n\text{ odd}}\biggl(1 - \frac{2z}{n}\biggr)e^{2z/n}. \end{align}