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.

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