# 2F1(1/2,1/2;3/2;z^2)=arcsin(z)/z

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 inverse sine.