Evaluate the infinite product $\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$ $$\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$$
I've seen some similar questions asked. But this one is different from all these. Euler product does not apply. One cannot simply factorize $\left(1-\frac{2}{(2n+1)^2}\right)$ since the $\sqrt{2}$ on top will prevent terms from cancelling. Any help will be appreciated!
Note: we are expected to solve this in 2 mins.
 A: Maple says this is $$\sin(\pi (\sqrt{2}-1)/2)$$
and more generally
$$ \prod_{n=1}^\infty \left(1 - \frac{t^2}{(2n+1)^2} \right) = \frac{\sin(\pi (1+t)/2)}{1-t^2} $$
A: Hint. One may recall Euler's infinite product for the cosine function
$$\cos x =\prod_{n=0}^\infty \left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right),\qquad |x|<\frac \pi2.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\ds{%
\prod_{n = 1}^{\infty}\bracks{1 - {2 \over \pars{2n + 1}^{2}}}}} =
\prod_{n = 1}^{\infty}\bracks{1 - {1/2 \over \pars{n + 1/2}^{2}}}
\\[5mm] = &\
\lim_{N \to \infty}
{\bracks{\prod_{n = 1}^{N}\pars{n + 1/2 - \root{2}/2}}
\bracks{\prod_{n = 1}^{N}\pars{n + 1/2 + \root{2}/2}} \over
\bracks{\prod_{n = 1}^{N}\pars{n + 1/2}}^{2}}
\\[5mm] = &\
\lim_{N \to \infty}
{\pars{3/2 - \root{2}/2}^{\overline{N}}\pars{3/2 + \root{2}/2}^{\overline{N}} \over \bracks{\pars{3/2}^{\overline{N}}}^{2}}
\\[5mm] = &\
{\Gamma^{2}\pars{3/2} \over
\Gamma\pars{3/2 + \root{2}/2}\Gamma\pars{3/2 - \root{2}/2}}\,
\lim_{N \to \infty}
{\pars{N + 1/2 - \root{2}/2}!\pars{N + 1/2 + \root{2}/2}! \over
\bracks{\pars{N + 1/2}!}^{2}}
\\[5mm] = &\
{\pi/4 \over
\pars{-1/4}\Gamma\pars{1/2 + \root{2}/2}\Gamma\pars{1/2 - \root{2}/2}}\
\times
\\[2mm] &\
\lim_{N \to \infty}
{\bracks{\pars{N + 1/2 - \root{2}/2}^{N + 1 - \root{2}/2}}
\bracks{\pars{N + 1/2 + \root{2}/2}^{N + 1 + \root{2}/2}}
\expo{-2N - 1} \over
\pars{N + 1/2}^{2N + 2}\expo{-2N - 1}}
\\[5mm] = &\
-\sin\pars{\pi\bracks{{1 \over 2} + {\root{2} \over 2}}}
\times
\\[2mm] &\
\lim_{N \to \infty}
{\bracks{1 + \pars{1/2 - \root{2}/2}/N}^{N}
\bracks{1 + \pars{1/2 + \root{2}/2}/N}^{N}
 \over
\braces{\bracks{1 + \pars{1/2}/N}^{N}}^{2}}
\\[5mm] = &
-\cos\pars{\pi\,{\root{2} \over 2}}\,
{\exp\pars{1/2 - \root{2}/2}\exp\pars{1/2  \root{2}/2} \over \bracks{\exp\pars{1/2}}^{2}} = \bbx{-\,\cos\pars{{\root{2} \over 2}\,\pi}}
\approx 0.6057
\end{align}
A: Considering the partial products
$$A_p=\prod_{n=1}^{p} \left(1-\frac{2}{(2n+1)^2}\right)$$ a CAS produced 
$$A_p=-\cos \left(\frac{\pi }{\sqrt{2}}\right) \frac{\Gamma
   \left(p-\frac{1}{\sqrt{2}}+\frac{3}{2}\right) \Gamma
   \left(p+\frac{1}{\sqrt{2}}+\frac{3}{2}\right)}{\Gamma
   \left(p+\frac{3}{2}\right)^2}$$ and using Stirling approximetion for large $p$, this leads to $$\log\left(\frac{\Gamma
   \left(p-\frac{1}{\sqrt{2}}+\frac{3}{2}\right) \Gamma
   \left(p+\frac{1}{\sqrt{2}}+\frac{3}{2}\right)}{\Gamma
   \left(p+\frac{3}{2}\right)^2} \right)=\frac{1}{2p}+O\left(\frac{1}{p^2}\right)$$ then $$A_p =-\cos \left(\frac{\pi }{\sqrt{2}}\right)\left(1+\frac{1}{2p}+O\left(\frac{1}{p^2}\right)\right)$$
