= Solution
Since $1-C=1/2$, the <second local hypergeometric solution> is
$$
z^{1/2}F(A-C+1,B-C+1;2-C;z)
=z^{1/2}F\left(\frac{1+k}{2},\frac{1-k}{2};\frac32;z\right).
$$
Under $z=\sin^2x$, it must be a constant multiple of the independent elementary solution $\sin(kx)$. As $x\to0$, both $z^{1/2}$ and $x$ are asymptotic to one another, while $sin(kx)\sim kx$. Matching the normalized hypergeometric factor, whose value at zero is one, gives
$$
z^{1/2}F\left(\frac{1+k}{2},\frac{1-k}{2};\frac32;z\right)
=\frac{\sin(kx)}{k}.
$$
Consequently the <hypergeometric sine identity> is
$$
\boxed{
F\left(\frac{1+k}{2},\frac{1-k}{2};\frac32;z\right)
=\frac{\sin\!\left(k\arcsin\sqrt z\right)}{k\sqrt z}
=\frac{\sin(kx)}{k\sin x}.
}
$$
Back to article page
