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I have already shown $\cos \pi z=\prod\limits_{n=1}^{\infty}\left(1-\frac{4z^2}{(2n-1)^2}\right)$. Here is what I have after that, \begin{equation*}\begin{split} \cos(\pi z) & = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\left(1+\frac{2z}{(2n-1)}\right)\\ & = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\prod_{n=1}^{\infty}\left(1+\frac{2z}{(2n-1)}\right)\\ & = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\prod_{-\infty}^{n=-1}\left(1-\frac{2z}{(2n+1)}\right)\\ & = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\prod_{-\infty}^{n=0}\left(1-\frac{2z}{(2n-1)}\right)\\ & = \prod_{n=-\infty}^{\infty}\left(1-\frac{2z}{(2n-1)}\right) \end{split}\end{equation*} I'm not sure where the exponential comes from.

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    $\begingroup$ It turns out that the exponential factors are absolutely necessary for convergence purposes. $\endgroup$ – Tony Apr 11 '16 at 17:58
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    $\begingroup$ you cannot split $\prod_{n=1}^{\infty} \left(1-\frac{2z}{(2n-1)}\right)\left(1+\frac{2z}{(2n-1)}\right)$ into $\prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)\prod_{m=1}^{\infty}\left(1+\frac{2z}{(2m-1)}\right)$ without checking that it converges, and it doesn't, $\endgroup$ – reuns Apr 16 '16 at 6:55
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    $\begingroup$ this is why you need the $e^{\frac{2z}{(2n-1)}}$ terms : $\prod_{n=1}^{\infty} \left(1-\frac{2z}{(2n-1)}\right)\left(1+\frac{2z}{(2n-1)}\right) $ $ = \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)e^{\frac{2z}{(2n-1)}}\left(1+\frac{2z}{(2n-1)}\right)e^{-\frac{2z}{(2n-1)}} =$ $ \prod_{n=1}^{\infty}\left(1-\frac{2z}{(2n-1)}\right)e^{\frac{2z}{(2n-1)}}\prod_{m=1}^{\infty} \left(1+\frac{2z}{(2m-1)}\right)e^{-\frac{2z}{(2m-1)}} $ $\endgroup$ – reuns Apr 16 '16 at 6:57
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We have $$ \begin{align} \csc\left(z\right)-\frac{1}{z}= &\sum_{k\in\mathbb{Z},\, k\neq0}\left(-1\right)^{k}\left(\frac{1}{k\pi}+\frac{1}{z-k\pi}\right) \\ = & \sum_{k\in\mathbb{Z},\, k\neq0}\left(\frac{1}{2k\pi}+\frac{1}{z-2k\pi}\right)-\sum_{k\in\mathbb{Z}}\left(\frac{1}{\left(2k-1\right)\pi}+\frac{1}{z-\left(2k-1\right)\pi}\right) \end{align}$$ so if we integrate we have $$ \begin{align} \left.\log\left(\frac{\tan\left(z/2\right)}{z}\right)\right|_{0}^{2x}= & \sum_{k=-\infty}^{\infty}\left.\left(\log\left(z-2k\pi\right)+\frac{z}{2k\pi}\right)\right|_{0}^{2x} \\ - & \sum_{k=-\infty}^{\infty}\left.\left(\log\left(z-\left(2k-1\right)\pi\right)+\frac{z}{\left(2k-1\right)\pi}\right)\right|_{0}^{2x} \end{align} $$ then $$\frac{\sin\left(x\right)}{x\cos\left(x\right)}=\frac{\prod_{k=-\infty}^{\infty}\left(1-\frac{x}{\pi k}\right)\exp\left(\frac{x}{k\pi}\right)}{\prod_{k=-\infty}^{\infty}\left(1-\frac{2x}{\pi\left(2k-1\right)}\right)\exp\left(\frac{2x}{\pi\left(2k-1\right)}\right)}. $$ With the same method, from $$\cot\left(z\right)-\frac{1}{z}=\sum_{k\in\mathbb{Z},\, k\neq0}\left(\frac{1}{k\pi}+\frac{1}{z-k\pi}\right) $$ it is possible to prove that $$\frac{\sin\left(x\right)}{x}=\prod_{k=-\infty}^{\infty}\left(1-\frac{x}{\pi k}\right)\exp\left(\frac{x}{\pi k}\right) $$ (and we recall that the term $k=0 $ is omitted) hence $$\cos\left(x\right)=\prod_{k=-\infty}^{\infty}\left(1-\frac{2x}{\pi\left(2k-1\right)}\right)\exp\left(\frac{2x}{\pi\left(2k-1\right)}\right). $$

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  • $\begingroup$ and why would you assume that we know such a pole decomposition of $\csc(z)$ ?... $\endgroup$ – reuns Apr 16 '16 at 6:52
  • $\begingroup$ @user1952009 I can't predict what you know or not. The question is to prove an identity and this is a way. $\endgroup$ – Marco Cantarini Apr 16 '16 at 7:00
  • $\begingroup$ this is not the question, and if you wanted to give a complete proof, you didn't since your formula for $\csc(z)$ is "magic" (at least explain why starting from $\csc(z)$ or $\cot(z)$ would be easier...) $\endgroup$ – reuns Apr 16 '16 at 7:03
  • $\begingroup$ Title: "Show $\cos(\pi z)$=...". In the question: "I'm not sure where the exponential comes from." So I think my answer shows how the exponential comes from. We always need to start from something to start a proof. I can find a reference for the two identities (with the proofs) but I wouldn't call it "magic". $\endgroup$ – Marco Cantarini Apr 16 '16 at 7:10
  • $\begingroup$ I'm sorry but the exponential only comes to make the product convergent... the true product is without the exponential, by grouping the terms by two. when you change the order of summation, you have to add the exponentials to make the logarithm of the product absolutely convergent. you didn't show that, (read my comment above to see what was the answer to his question) and I'm ensure you know all the possible ways to prove the product formula for $\cos(z)$ and the series formula for $\cot(z)$ (with the residue theorem or the en.wikipedia.org/wiki/Weierstrass_factorization_theorem ) $\endgroup$ – reuns Apr 16 '16 at 7:12

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