How can I deduce $\cos\pi z=\prod_{n=0}^{\infty}(1-4z^2/(2n+1)^2)$? 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 $\cos\pi z$?
 A: 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.
A: Well, you can perform a logarithmic differentiation and get a series that may be summed using the residue theorem.
Let $p(z)$ be the product in question; we intend to prove that $p(z)=\cos{\pi z}$.  
$$\log{p} = \sum_{n=0}^{\infty} \log{\left ( 1-\frac{4 z^2}{(2 n+1)^2}\right)}$$
$$\frac{d}{dz} \log{p} = -z \sum_{n=-\infty}^{\infty} \frac{1}{(n+(1/2))^2-z^2}$$
Note that we were able to use the symmetry of the sum to change the lower limit to $-\infty$.  This sum is in a form that may be evaluated using the residue theorem:
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{Res}_{s=s_k} [\pi \cot{\pi s} \, f(s)]$$
where the $s_k$ are the non-integral poles of $f$.  In this case, $f(s) = 1/((s+(1/2))^2-z^2)$, so that the poles of $f$ are at $s_{\pm} = -1/2 \pm z$.  The residues of these poles are
$$\frac{\pi \cot{(-\pi/2 + \pi z)}}{2 z} - \frac{\pi \cot{(-\pi/2 - \pi z)}}{2 z} = -\frac{\pi \tan{\pi z}}{z}$$
Therefore
$$\frac{d}{dz} \log{p} = -\pi \tan{\pi z} \implies \log{p} = \log{\cos{\pi z}} + C$$
where $C$ is a constant of integration, which using $p(0)=1$ implies that $C=0$.  Then
$$p(z) = \cos{\pi z}$$
as was to be shown.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\prod_{n = 0}^{m}
\bracks{1 - 4z^{2}/\pars{2n + 1}^{2}}
\right\vert_{\,m\ \in\ \mathbb{N}_{\,\geq 1}}} =
\prod_{n = 0}^{m}
{\pars{n + 1/2}^{2} - z^{2} \over \pars{n + 1/2}^{2}}
\\[5mm] = &\
\on{f}_{m}\pars{z}\on{f}_{m}\pars{-z}
\end{align}
where
\begin{align}
\on{f}_{m}\pars{z} & \equiv \prod_{n = 0}^{m}
{n + 1/2 - z \over n + 1/2} =
{\pars{1/2 - z}^{\overline{m + 1}} \over
\pars{1/2}^{\overline{m + 1}}}
\\[5mm] & =
{\pars{1/2 - z + m}!\,/\,\Gamma\pars{1/2 - z} \over
\pars{1/2 + m}!\,/\,\Gamma\pars{1/2}}
\\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\,
{\root{\pi} \over \Gamma\pars{1/2 - z}}\ \times
\\[2mm] &\
{\root{2\pi}\pars{1/2 - z + m}^{1 - z + m}\,\,\,
\expo{-1/2 + z - m} \over
\root{2\pi}\pars{1/2 + m}^{1 + m}\,\,
\expo{-1/2 - m}}
\\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\,
{\root{\pi} \over \Gamma\pars{1/2 - z}}\ \times
\\[2mm] &\
{m^{1 - z + m}\,\,\,\,\bracks{1 + \pars{1/2 - z}/m}^{\,m}
 \over
m^{m + 1}\,\,\,\bracks{1 + \pars{1/2}/m}^{\,m}}\expo{z}
\\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\,
{\root{\pi} \over \Gamma\pars{1/2 - z}}\,m^{-z}
\end{align}
Then,
\begin{align}
&\bbox[5px,#ffd]{\left.\prod_{n = 0}^{m}
\bracks{1 - 4z^{2}/\pars{2n + 1}^{2}}
\right\vert_{\,m\ \in\ \mathbb{N}_{\,\geq 1}}}
\\[5mm] = &\
\lim_{m \to \infty}\braces{\bracks{{\root{\pi} \over \Gamma\pars{1/2 - z}}\,m^{-z}}\bracks{{\root{\pi} \over \Gamma\pars{1/2 + z}}\,m^{z}}}
\\[5mm] = &\
{\pi \over \pi/\sin\pars{\pi\bracks{1/2 + z}}} =
\bbx{\cos\pars{\pi z}} \\ &
\end{align}
