How to find this infinite product

$$\begin{eqnarray*}\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) &=& \prod_{n\geq 0}\left(\frac{4(2n+1)^2-1}{4(2n+1)^2+1}\right)\\&=&\color{purple}{\prod_{n\geq 0}\left(1-\frac{1}{4(2n+1)^2}\right)}\color{blue}{\prod_{n\geq 0}\left(1+\frac{1}{4(2n+1)^2}\right)^{-1}}\end{eqnarray*}$$

hence by exploiting the Weierstrass products for the $\color{purple}{\cos}$ and $\color{blue}{\cosh}$ function we get:

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

$$ \begin{align} &\prod_{n=0}^\infty\left(1-\frac2{4(2n+1)^2+1}\right)\\ &=\prod_{n=0}^\infty\frac{(4n+1)(4n+3)}{(4n+2-i)(4n+2+i)}\tag{1}\\ &=\prod_{n=0}^\infty\frac{(n+\frac14)(n+\frac34)}{(n+\frac{2-i}4)(n+\frac{2+i}4)}\tag{2}\\ &=\lim_{n\to\infty}\left.\frac{\Gamma(n+\frac14)}{\Gamma(\frac14)}\frac{\Gamma(n+\frac34)}{\Gamma(\frac34)}\middle/\frac{\Gamma(n+\frac{2-i}4)}{\Gamma(\frac{2-i}4)}\frac{\Gamma(n+\frac{2+i}4)}{\Gamma(\frac{2+i}4)}\right.\tag{3}\\ &=\lim_{n\to\infty}\frac{\Gamma(\frac{2-i}4)\Gamma(\frac{2+i}4)}{\Gamma(\frac14)\Gamma(\frac34)}\cdot\frac{\Gamma(n+\frac14)\Gamma(n+\frac34)}{\Gamma(n+\frac{2-i}4)\Gamma(n+\frac{2+i}4)}\tag{4}\\ &=\frac{\pi\csc\left(\pi\frac{2-i}4\right)}{\pi\csc\left(\frac\pi4\right)}\tag{5}\\ &=\frac1{\sqrt2\cosh(\frac\pi4)}\tag{6} \end{align} $$ Explanation:
$(1)$: subtract and factor the numerator and denominator
$(2)$: factor out the $4$ so that it is easier to see the relation to $\Gamma$
$(3)$: $\prod\limits_{k=0}^{n-1}(k+x)=\frac{\Gamma(n+x)}{\Gamma(x)}$ applied four times
$(4)$: rearrange terms for easier cancelling
$(5)$: Euler's Reflection Formula and Gautschi's Inequality
$(6)$: $\csc\left(\frac\pi2-i\frac\pi4\right)=\sec\left(i\frac\pi4\right)=\mathrm{sech}(\frac\pi4)$