How to find this infinite product How to find this infinite product ?
$$\prod_{n=0}^\infty \left(1-\dfrac{2}{4(2n+1)^2+1}\right)$$
I try to use infinite product of $\cos{x}$ but it doesn't work.
Thank you.

$$\begin{eqnarray*}\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) &=& \prod_{n\geq 0}\left(\frac{4(2n+1)^2-1}{4(2n+1)^2+1}\right)\\&=&\color{purple}{\prod_{n\geq 0}\left(1-\frac{1}{4(2n+1)^2}\right)}\color{blue}{\prod_{n\geq 0}\left(1+\frac{1}{4(2n+1)^2}\right)^{-1}}\end{eqnarray*}$$
hence by exploiting the Weierstrass products for the $\color{purple}{\cos}$ and $\color{blue}{\cosh}$ function we get:

$$\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) = \frac{1}{\sqrt{2}\cosh\frac{\pi}{4}}=\color{red}{\frac{\sqrt{2}}{e^{\pi/4}+e^{-\pi/4}}}.$$

