# :Duplicate: How to prove that $\cos (\pi * z) = \prod_{n \in \mathbb{Z}_{\text{odd}}} (1-\frac{2 z}{n})e^{2z/n}$ [duplicate]

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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

## marked as duplicate by Rahul, Daniel W. Farlow, Jack, JonMark Perry, RohanDec 10 '16 at 9:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• $$\cos(x)=\prod_{n =1}^\infty(1-\frac{4x^2}{\pi^2 (2n-1)^2}) = \lim_{N \to \infty}\prod_{n = 1}^N(1-\frac{4x^2}{\pi^2 (2n-1)^2})$$ $$=\lim_{N \to \infty}\prod_{n = 1}^N(1-\frac{2x}{\pi (2n-1)})(1+\frac{2x}{\pi (2n-1)}) =\lim_{N \to \infty}\prod_{n=-N+1}^N(1-\frac{2x}{\pi (2n-1)})$$ $$= \lim_{N \to \infty}\prod_{n=-N+1}^N(1-\frac{2x}{\pi (2n-1)})e^{2x / (\pi(2n-1))} = \prod_{n =-\infty}^\infty(1-\frac{2x}{\pi (2n-1)})e^{2x / (\pi(2n-1))}$$ – reuns Oct 26 '16 at 5:50
• You're missing a factor of $\pi z$ in your $\sin$ product. The easiest way to obtain the product for the cosine is to use $$\cos (\pi z) = \frac{\sin (2\pi z)}{2\sin (\pi z)}$$ and expand the sines in the numerator and denominator into their Weierstraß products. The terms for even $n$ cancel. – Daniel Fischer Oct 26 '16 at 17:24
• Thank you for the correction. The even term can cancel, but how about the exponential part? I am a little confused. – Shizkuyi Oct 26 '16 at 17:29

## 1 Answer

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}