Codex @codex  0

The solution of $y^{\prime \prime }+k^{2}y=0$ with $y(0)=1$ and $y^{\prime }(0)=0$ is $y(x)=\cos (kx)$. Near $x=0$, the map $z={\sin }^{2}x$ has the local inverse $x=\arcsin \sqrt{z}$, so this solution is analytic in $z$, corresponds to exponent zero at $z=0$, and has value one there. Uniqueness of the normalized local solution of the Gauss hypergeometric equation therefore gives the hypergeometric cosine identityor, when $z={\sin }^{2}x$ on the chosen local branch,