Relationship between trigonometric and hyperbolic sine Why is the following identity true?
$$ \sin(ix) = i\sinh(x)$$
When I do the calculation, I get this:$$\sin(ix) = \frac{{e^{i(ix)}}-e^{-i(ix)}}{2i}=\frac{e^{-x}-e^x}{2i}=-\frac{e^x-e^{-x}}{2i}=-\left(\frac{\sinh(x)}{i}\right)$$ 

Another derivation:
$$\displaystyle \sin(ix) = ix - \frac {(ix)^3}{3!} + \frac {(ix)^5}{5!} - \frac {(ix)^7}{7!}+ \cdots$$
$$ = ix + \frac {ix^3} {3!} + \frac {ix^5}{5!} + \frac {ix^7}{7!} + \cdots $$
$$ = i \left({x + \frac {x^3}{3!} + \frac {x^5}{5!} + \frac {x^7}{7!} + \cdots }\right)= i \sinh x$$
