Proving a result for $\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)$ $$\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)=\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)}$$
According to WA. I attempted using
$$\prod_{k=0}^{\infty}\Bigl(1-\frac{x^2}{\pi^2k^2}\Bigr)=\frac{\sin x}{x}$$
But I couldn’t reindex the product appropriately to use it the way I wanted to (factoring out a 4 and then continuing from there). I’d like to have at least a direction to go in or an idea on how to do the product.

Use (proved by induction)
$$
\prod_{k=0}^{n-1} (a+4k) = \frac{4^n\Gamma(n+\frac{a}{4})}{\Gamma(\frac{a}{4})}
$$
together with
$$
1-\frac{4}{(4\,k+a)^2}={\frac { \left( a+4\,k+2 \right) 
 \left( a+4\,k-2 \right) }{ \left( 4\,k+a \right) ^{2}}}
$$
to get
$$
\prod _{k=0}^{n-1}{\frac { \left( a+4\,k+2 \right)  \left( a+4\,k-2
 \right) }{ \left( 4\,k+a \right) ^{2}}}
={\frac { \left( a-2
 \right)  \left( a+4\,n-2 \right)   \Gamma \left(\frac{a-2}{4}+n
 \right) ^{2} \;\Gamma \left( \frac{a-2}{4} \right) ^{2}
}{16 \; \Gamma \left( \frac{a}{4} +n\right) ^{2} \; \Gamma
 \left( \frac{a+2}{4} \right)  ^{2}}}
$$
Than take limit as $n \to \infty$.  My result is
$$
\frac{\displaystyle\Gamma\left(\frac{a}{4}\right)^2}{\displaystyle \Gamma\left(\frac{a-2}{4}\right)\Gamma\left(\frac{a+2}{4}\right)}
$$
which agrees with Polygon.
