I would like to derive the identity $B(z,z)=2^{1-2z}B(z,\frac{1}{2})$ somehow.
The Beta function is defined as $B(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1}dt$ where $Re(p), Re(q)>0$
I used the substitution $t=\sin^2\theta$ to get $B(p,q)=2\int_0^{\frac{\pi}{2}}\sin^{2p-1}\theta\cos^{2q-1}d\theta$ but it does not really help me for $B(z,z)$, any ideas?
It follows from the duplication formula for the gamma function.
$$B(z,z) = \frac{\Gamma(z) \Gamma(z)}{\Gamma(2z)} = \Gamma(z) \Gamma(z) \frac{\sqrt{\pi}}{2^{2z-1} \Gamma(z) \Gamma (z+1/2)}$$
$$= 2^{1-2z} \frac{\Gamma (z) \Gamma(1/2)}{\Gamma(z+1/2)} = 2^{1-2z}B \left(z, \frac{1}{2} \right)$$
http://mathworld.wolfram.com/LegendreDuplicationFormula.html