Taking the usual convention that $z=x+iy$, show that $|\sinh(y)|\leq|\sin(z)|\leq|\cosh(y)|$
What I tried so far was to expand for $\sin(z)$ in terms of complex exponentials and substituting $z=x+iy$ and got the following identity:
$\sin(z)=i\sinh(y)\cos(x)+\cosh(y)\sin(x)$
This means that the absolute value of $\sin(z)$ is given by
$|\sin(z)|^2=\sinh^2(y)\cos^2(x)+\cosh^2(y)\sin^2(x)$
Given that this identity relates all the terms covered by the inequality I suspect that I may be on the right track however I do not really know how to proceed from here (assuming that I can).
$\sin^2+\cos^2=1$, so let $θ = \cos^2 x \in [0,1]$ to see that $|\sin z|^2 = θ \sinh^2 y + (1-θ) \cosh^2 y $ is expressed as a convex combination of $\sinh^2 y$ and $\cosh^2 y$. Finally, on $\mathbb R$, $|\sinh| ≤ |\cosh|$, so the inequality follows from the fact that convex combinations are in between the two endpoints.