# Simplifying this equation (trigonometric)

I was reading my notes and I came across these 2 lines, Im wondering how did it go from $\sin$ to $\sinh$?

$-5\sin i\pi$

$= -5i\sinh \pi$

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$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$
$$\sinh z=\frac{e^{z}-e^{-z}}{2}$$
In the first line, put $z=i\pi$ and note that $\dfrac{1}{i}=-i$.
You can start with the basic definitions $\sinh(x) = (e^x-e^{-x})/2$ and $e^{ix} = \sin(x)+i\cos(x)$, from the second one you can derive $\sin(x) = (e^{ix}-e^{-ix})/2$. Substitute $i\pi$ for $x$.