<s>[INST] Hypergeometric Function Value Wolfram gives the following: 
$$_2F_1 \left(a,b;\frac{a+b+1}{2};\frac{1}{2}\right) = \frac{\sqrt{\pi} 
 \Gamma \left(\frac{a+b+1}{2} \right)} {\Gamma \left(\frac{a+1}{2} \right) \Gamma \left(\frac{b+1}{2} \right) }$$ and I am struggling to rederive it.
My ultimate aim is to allow the final parameter (the 1/2) to vary. I am fairly certain that the function is increasing up to 1/2, decreasing afterwards, but I am struggling to prove it. Any help in that direction would also be great.
Thanks! 
 [/INST] Here is a derivation of your formula using a transformation and the 
Gauss's Hypergeometric Theorem. The quadratic transformation
http://dlmf.nist.gov/15.8.E18
$${_2}F_1 \left(a,b;\frac{a+b+1}{2};z\right)={_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};4z(1-z) \right)$$
gives
$${_2}F_1 \left(a,b;\frac{a+b+1}{2};\frac{1}{2}\right)={_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right)$$
With the well known Gauss's Hypergeometric Theorem
$${_2}F_1\left(a,b;c;1\right)=\frac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{%
\Gamma\left(c-a\right)\Gamma\left(c-b\right)}.$$
you get 
$$
{_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right) =
\frac{\Gamma\left(\frac{a+b+1}{2}\right)\Gamma\left(\frac{a+b+1}{2}-\frac{a}{2}-\frac{b}{2}\right)}
{\Gamma\left(\frac{a+b+1}{2}-\frac{a}{2}\right)\Gamma\left(\frac{a+b+1}{2}-\frac{b}{2}\right)}$$
$$
{_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right) =
\frac{\Gamma\left(\frac{a+b+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}
{\Gamma\left(\frac{b+1}{2}\right)\Gamma\left(\frac{a+1}{2}\right)}$$
$$
{_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right) =
\frac{\sqrt{\pi}\;\Gamma\left(\frac{a+b+1}{2}\right)}
{\Gamma\left(\frac{b+1}{2}\right)\Gamma\left(\frac{a+1}{2}\right)}$$
Since $4z(1-z)$ has a maximum for $z=\frac{1}{2}$ the first formula should be
a good starting point for your conjecture, "that the function is increasing up to 1/2, decreasing afterwards".
</s>