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

Duplicate

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.

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

• $$\cos(x)=\prod_{n =1}^\infty(1-\frac{4x^2}{\pi^2 (2n-1)^2}) = \lim_{N \to \infty}\prod_{n = 1}^N(1-\frac{4x^2}{\pi^2 (2n-1)^2})$$ $$=\lim_{N \to \infty}\prod_{n = 1}^N(1-\frac{2x}{\pi (2n-1)})(1+\frac{2x}{\pi (2n-1)}) =\lim_{N \to \infty}\prod_{n=-N+1}^N(1-\frac{2x}{\pi (2n-1)})$$ $$= \lim_{N \to \infty}\prod_{n=-N+1}^N(1-\frac{2x}{\pi (2n-1)})e^{2x / (\pi(2n-1))} = \prod_{n =-\infty}^\infty(1-\frac{2x}{\pi (2n-1)})e^{2x / (\pi(2n-1))}$$ – 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
Using the trigonometric identity $\sin (2\alpha) = 2 \sin \alpha \cos \alpha$ and the Weierstraß product of the sine, we obtain