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?
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2use $1/i=-i$ and the definition of $\sinh y = {e^y-e^{-y}\over 2}$ – danimal Jun 02 '15 at 10:35
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The trick is to multiply by $\frac{i}{i}$: $$\sin(iy)=\frac{e^{i(iy)}-e^{-i(iy)}}{2i}=\frac{e^{-y}-e^{y}}{2i}=\frac{i}{i}\frac{e^{-y}-e^{y}}{2i}=i\frac{e^{y}-e^{-y}}{2}=i\sinh ( y).$$
zoli
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You use that $-1/i = i$, to get the result.