How to show that $\sin(iy)=i\sinh y$

Question

I know that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Then substituting $x=iy$: $$\sin(iy)=\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^{-y}-e^{y}}{2i}$$ Then, according to my lecture handout (this step is the one I don't get) it follows that: $$\sin(iy)=i\frac{e^{y}-e^{-y}}{2}=i\sinh y$$ Whats going on there?

Answer 1

You use that $-1/i = i$, to get the result.

Answer 2

The trick is to multiply by $\frac{i}{i}$: $$\sin(iy)=\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=\frac{i}{i}\frac{e^{-y}-e^{y}}{2i}=i\frac{e^{y}-e^{-y}}{2}=i\sinh ( y).$$