Deriving $\sin{ix}=i\sinh{x}$ Derive $\sin{ix}=i\sinh{x}$ from $(5)$. What is $\sin{i}$?
  $$\cos{x}=\frac{1}{2}\left(e^{ix}+e^{-ix}\right)\quad\text{and}\quad\sin{x}=\frac{1}{2i}\left(e^{ix}-e^{-ix}\right)\tag 5$$


We have
$$\begin{align}
\sin{ix}&amp;=\frac{1}{2i}\left(e^{i^2x}-e^{-i^2x}\right)\\
&amp;=\frac{1}{2i}\left(e^{-x}-e^x\right)\\
&amp;=-\frac{\sinh{x}}{i}
\end{align}$$
It does not look like what we are deriving. What went wrong?
Multiply
 the denominator and the numerator by $i.$
We know that $sinh(ix)=i sin(x)$.
                          Put $i=ix$
                   $$Sinh(i.ix)=isin(ix)$$
                  $$ Sinh(-x)=i sin(ix)$$
                   $$-sinhx=i sin(ix)$$
                  $$i\times isinhx=i sin(ix)$$
                    $$i sinhx=sin(ix)$$