How to show $\sin(x+iy)=\sin(x) \cosh(y) + i\cos(x) \sinh(y)$

Question

How to show $$\sin(x+iy)=\sin(x) \cosh(y) + i\cos(x) \sinh(y)$$

I begin with $$\sin(x+iy) = \frac{e^{x+iy}-e^{-x-iy}}{2i} = \frac{e^xe^{iy}-e^{-x}e^{-iy}}{2i}$$

$$ = \frac{e^xe^{iy}-e^{-x}e^{iy}+-e^{-x}e^{iy}-e^{-x}e^{-iy}}{2i}$$

$$ = \frac{e^{iy}(e^x-e^{-x})-e^{-x}e^{iy}-e^{-x}e^{-iy}}{2i}$$

$$ = e^{iy}\sin(x) + \frac{-e^{-x}e^{iy}-e^{-x}e^{-iy}}{2i}$$

At this point I could have $\displaystyle\frac{-e^{-x}}{i}cos(iy)$ in the right term but I want to make $\cos(x)$ appear, so I don't know how to continue.

Answer 1

For all $x,y\in \mathbb R$ the following holds: $$\begin{align} \sin(x+iy)&=\dfrac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}\\ &=\dfrac{e^{-y+ix}-e^{y-ix}}{2i}\\ &=\dfrac{e^{-y}(\cos(x)+i\sin(x))-e^{y}(\cos(x)-i\sin(x))}{2i}\\ &=\dfrac{(e^{-y}-e^y)\cos(x)+i(e^{{-y}}+e^y)\sin(x)}{2i}\\ &=\dfrac{(e^{-y}-e^y)\cos(x)}{2i}+\dfrac{i(e^{{-y}}+e^y)\sin(x)}{2i}\\ &=i\sinh(y)\cos(x)+\cosh(y)\sin(x). \end{align}$$

Answer 2

$$\sin(x+iy)= \sin(x) \, \cos(iy) + \cos(x) \, \sin(iy) = \sin(x) \, \cosh(y) + i \, \cos(x) \, \sinh(y).$$