Using the definitions prove that $$|\sinh y| \le |\cos z|\le \cosh y \ , \ |\sinh y|\le|\sin z|\le \cosh y$$ Conclude that the complex cosine and sine are not bounded in the whole complex plane.
So I used the identity that $|\cos z|^2 = \cos^2(x) + \sinh^2(y)\ge |\sinh y|^2=\sinh^2(y)$. However, I'm not sure how to show that $|\cos z|^2 \le \cosh^2 y$.
Similarly for $|\sin z|$ I used the identity $|\sin z|^2 = \sin^2(x) + \sinh^2(y)\ge |\sinh y|^2$ and because we have the absolute value this implies $|\sinh y| \le |\sin z|$. However, again I am unsure how to show that $|\sin z| \le \cosh y$.
Also how do these inequalities allow us to conclude that the complex cosine and sine are not bounded in the whole complex plane?