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?
4 Answers
Re-arrangement of the representation of your result:
$$\boxed{\begin{align} \sin{ix} = -\frac{\sinh{x}}{i} = (-1)\cdot \frac{\sinh{x}}{i} = (i^2)\cdot \frac{\sinh{x}}{i} = i\sinh x \end{align}}$$
Multiply the denominator and the numerator by $i.$
You are on the right track bro.
$$-\frac{\sinh x}{i} = i^2\frac{\sinh x}{i} = i\sinh x$$
We know that $sinh(ix)=i sin(x)$.
Put $i=ix$
$$Sinh(i.ix)=isin(ix)$$
$$ Sinh(-x)=i sin(ix)$$
$$-sinhx=i sin(ix)$$
$$i\times isinhx=i sin(ix)$$
$$i sinhx=sin(ix)$$
