View source for 2F1(1/2,1/2;3/2;z^2)=arcsin(z)/z

==Theorem==
The following formula holds:
$${}_2F_1 \left( \dfrac{1}{2}, \dfrac{1}{2}; \dfrac{3}{2}; z^2 \right) = \dfrac{\arcsin(z)}{z},$$
where ${}_2F_1$ denotes the [[hypergeometric 2F1]] and $\arcsin$ denotes the [[arcsin|inverse sine]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=2F1(1/2,1;3/2;-z^2)=arctan(z)/z|next=Sqrt(1-z^2)2F1(1,1;3/2;z^2)=arcsin(z)/z}}: $15.1.6$

[[Category:Theorem]]
[[Category:Unproven]]