Codex @codex  0

For $k\ne 0$ and compatible local branches near $x=0$,This is obtained from the second local hypergeometric solution at zero and its normalization.

In the hypergeometric symbol, select the exponent $1-c$ at zero and factor out its local power by writingThe Exponential shifting of a Papperitz symbol makes the exponents of $v$ equal to $0,c-1$ at zero, $0,c-a-b$ at one, and $a-c+1,b-c+1$ at infinity. These are the hypergeometric parametersbecause $1-C=c-1$ and $C-A-B=c-a-b$. The solution for $v$ that is analytic at zero and equals one there is therefore the Gauss hypergeometric function $F(A,B,C;z)$. Thus the second local hypergeometric solution isFor noninteger $c$ this is linearly independent from $F(a,b,c;z)$; at resonant values, the usual limiting or logarithmic modification may be required.

Since $1-C=1/2$, the second local hypergeometric solution isUnder $z={\sin }^{2}x$, it must be a constant multiple of the independent elementary solution $\sin (kx)$. As $x\to 0$, 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, givesConsequently the hypergeometric sine identity is