Does there exist an analytic function such that $|f(z)|=|\sin z|$? Does there exist an analytic function $f(z)$ defined on $\Omega$ such that $|f(z)|=|\sin z|$ for all $z\in\Omega\subseteq\mathbb{C}$? Well I guess if there is a constant $c$ with $|c|=1$ and $f(z)=c\sin(z)$ for all $z\in\Omega$, am I right?

If $f$ is such a function, and assuming that $\Omega$ is a connected open set, then $g(z)=\frac{f(z)}{\sin(z)}$ has removable singularities at the zeros of $\sin$. Thus $g$ extends uniquely to define an analytic function (I'll still call it $g$) with $|g(z)|\equiv 1$ on $\Omega$.  Since $\Omega$ is connected, this implies that your "guess" not only gives correct examples (as mentioned by JLA), it gives all examples.
