Note that $\sin(\overline z)=\overline{\sin(z)}$. Since $\sin$ is analytic, if $g$ is analytic on some open set, then so is the real-valued function $\sin+g$. But a nonconstant real-valued function cannot be analytic (e.g., use the CR-equations or the open mapping theorem).
More generally, if $f$ is a nonconstant analytic function*, then $g(z)=f(\overline z)$ is not analytic on any open set. It is true that $\overline g$ is analytic, so if $g$ were analytic then the real-valued functions $\mathrm{Re}(g)=\frac{1}{2}(g+\overline g)$ and $\mathrm{Im}(g)=\frac{1}{2i}(g-\overline g)$ would be analytic, hence constant, which means that $g$ (and hence $f$) would be constant.
Furthermore, the only points at which the Cauchy-Riemann equations can be satisfied for $g$, being the complex conjugate of the analytic function $\overline g$, are at the zeros of the derivative of $\overline g$, which are isolated.
*(I am assuming that the domain of $f$ is symmetric about the real axis for the arguments used here.)