I am curious how to calculate the infinite product
$$ \prod_{k=1}^\infty \left( 1 + \frac{a}{k^2} \right). $$
WolframAlpha reports that it is equal to approximately
$$ \prod_{k=1}^\infty \left( 1 + \frac{a}{k^2} \right) \approx \frac{0.31831 \sinh (3.14159 \sqrt{a})}{\sqrt{a}} $$
which I suspect might be given by
$$ \frac{\sinh (\pi \sqrt{a})}{\pi \sqrt{a}}. $$
This in turn suggests that
$$ \sinh a = a \prod_{k=1}^\infty \left( 1 + \frac{a^2}{\pi^2 k^2} \right), $$
which is confirmed by WolframAlpha.
However, I don't know how this is derived.