Evaluate the infinite product $\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$

Question

$$\prod_{n=1}^{\infty} \left(1-\frac{2}{(2n+1)^2}\right)$$

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.

Answer 1

Hint. One may recall Euler's infinite product for the cosine function

$$\cos x =\prod_{n=0}^\infty \left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right),\qquad |x|<\frac \pi2.$$

Answer 2

Maple says this is $$\sin(\pi (\sqrt{2}-1)/2)$$ and more generally $$ \prod_{n=1}^\infty \left(1 - \frac{t^2}{(2n+1)^2} \right) = \frac{\sin(\pi (1+t)/2)}{1-t^2} $$

Answer 3

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 $$A_p =-\cos \left(\frac{\pi }{\sqrt{2}}\right)\left(1+\frac{1}{2p}+O\left(\frac{1}{p^2}\right)\right)$$