Proving that $|\sin(z)|^2 = \sin(x)^2+\sinh(y)^2$

Question

Goal: Prove $|\sin(z)|^2 = \sin(x)^2+\sinh(y)^2$

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I have \begin{align*} \sin(x+iy) &=\sin(x)\cos(iy)+\cos(x)\sin(iy)\\ &=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\\ |\sin(z)|^2 &=\sin(x)^2 \cosh(x)^2+\cos(x)^2\sinh(y)^2 \end{align*}

How do I proceed?

Answer 1

How to proceed:

Replace $\cos^2$ with $1-\sin^2$ and $\cosh^2$ with $1+\sinh^2$

(Also note that $\cos(iy)=\cosh(\color{red}y)$.)