How does one show that Im$(2e^y\sin(x+iy))=(e^{2y}-1)\cos(x)$?

Question

I'm thinking of rewriting $(x+iy)$ into Euler's form $re^{i\theta}$, but I can't see how it will help simplify the expression further.

Answer 1

We could try the addition formula. I assume $x,y$ are real. $$\begin{align} \sin(x+iy) &= \sin(x)\cos(iy)+\cos(x)\sin(iy) =\sin(x)\cosh(y)+i\cos(x)\sinh(y) \\ \text{Im}\big(\sin(x+iy)\big) &= \cos(x)\sinh(y) = \cos(x)\frac{e^y-e^{-y}}{2} \\ \text{Im}\big(2e^y\sin(x+iy)\big) &= \cos(x)e^y(e^y-e^{-y})= \cos(x)(e^{2y}-1) \end{align}$$