A scary looking relationship between Euler constant $\gamma$ and $\pi$ I saw this scary looking relation between the Euler constant $\gamma$, the Gamma function and $\pi$ in Peter Luschny blog. Any proofs?
$$ 
\left({{\frac {\Gamma (\gamma )}{\Gamma (2\gamma )\Gamma (1/2-\gamma )}}+{\frac {2\Gamma (1-2\gamma )}{\Gamma (1-\gamma )\Gamma (1/2-\gamma )}}}\right)^{2}\left({\frac {\Gamma (\gamma )\Gamma (1/2-\gamma /2)}{\Gamma (\gamma /2)}}\right)^{4}=\pi 
$$
 A: One could say this is something of a joke: it's actually a heavily disguised trigonometric identity.
We have the reflection formula
$$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{\pi z}} , $$
or in the form that is more useful to us,
$$ \Gamma(1/2+z) \Gamma(1/2-z) = \frac{\pi}{\cos{\pi z}} , $$
and the duplication formula
$$ \Gamma(z) \Gamma(z+1/2) = 2^{1-2z} \sqrt{\pi} \, \Gamma(2z) . $$
You might already suspect what is about to happen: if we replace $\gamma$ by any $z$ the three fractions become
$$ \frac{\Gamma(z)}{\Gamma(2z)\Gamma(1/2-z)} = \frac{\Gamma(z)\Gamma(z+1/2)}{\Gamma(2z)} \frac{\cos{\pi z}}{\pi} = \frac{2^{1-2z}}{\sqrt{\pi}} \cos{\pi z} \\
\frac{2\Gamma(1-2z)}{\Gamma(1-z)\Gamma(1/2-z)} = \dotsb = \frac{2^{1-2z}}{\sqrt{\pi}} \\
\frac{\Gamma(z)\Gamma(1/2-z/2)}{\Gamma(z/2)} = \dotsb = \frac{2^{x-1}\sqrt{\pi}}{\cos{(\pi x/2)}} , $$
and therefore
$$ \left( \frac{\Gamma(z)}{\Gamma(2z)\Gamma(1/2-z)} +
\frac{2\Gamma(1-2z)}{\Gamma(1-z)\Gamma(1/2-z)} \right)
 \left( \frac{\Gamma(z)\Gamma(1/2-z/2)}{\Gamma(z/2)} \right)^2 = \frac{2^{1-2z}}{\sqrt{\pi}} ( \cos{\pi z} + 1 ) \left( \frac{2^{x-1}\sqrt{\pi}}{\cos{(\pi x/2)}} \right)^2 \\
= \sqrt{\pi} \frac{\cos{\pi z}+1}{2\cos^2{(\pi z/2)}} , $$
and the last fraction is $1$ using a double-angle formula.
