Deriving $\sin{ix}=i\sinh{x}$

Question

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 $$\begin{align} \sin{ix}&=\frac{1}{2i}\left(e^{i^2x}-e^{-i^2x}\right)\\ &=\frac{1}{2i}\left(e^{-x}-e^x\right)\\ &=-\frac{\sinh{x}}{i} \end{align}$$ It does not look like what we are deriving. What went wrong?

Answer 1

Re-arrangement of the representation of your result:

$$\boxed{\begin{align} \sin{ix} = -\frac{\sinh{x}}{i} = (-1)\cdot \frac{\sinh{x}}{i} = (i^2)\cdot \frac{\sinh{x}}{i} = i\sinh x \end{align}}$$

Answer 2

Multiply the denominator and the numerator by $i.$

Answer 3

You are on the right track bro.

$$-\frac{\sinh x}{i} = i^2\frac{\sinh x}{i} = i\sinh x$$