TITLE: Verify the following equation: $\sin (x+iy)=\sin x \cosh y+i \cos x \sinh y$
QUESTION [6 upvotes]: When I attempted this, I got an answer that was very close but not correct. I triple checked my work but I cannot find a mistake. Could someone help me out?
My attempt:
$$\sin x \cosh y+i \cos x \sinh y = (\frac{-ie^{ix}+ie^{-ix}}{2})(\frac{e^y+e^{-y}}{2})+(\frac{ie^{ix}+ie^{-ix}}{2})(\frac{e^y-e^{-y}}{2})$$
$$=\frac{1}{2}\left [(-ie^{ix+y}-ie^{ix-y}+ie^{-ix+y}+ie^{-ix-y}) + (ie^{ix+y}-ie^{ix-y}+ie^{-ix+y}-ie^{-ix-y}) \right ]$$
$$=\frac{1}{2} \left [ -2ie^{ix-y} + 2ie^{ix+y} \right ]$$
$$=ie^{y-ix}-ie^{-y+ix}$$
However, the correct final answer (from Wolfram) should be:
$$sin(x+iy)=\frac{1}{2}\left [ ie^{y-ix} - ie^{-y+ix} \right ]$$
Where am I going wrong?

REPLY [3 votes]: first I prove that $\cos(ix) = \cosh(x)$, consider that $e^x=\exp(x)$.
we khow that
$$\cos(x) = \frac{\exp(ix)+\exp(-ix)}{2}$$
so
$$
\begin{align}
\cos(ix) & = \frac{\exp\Big(i(ix)\Big)+\exp\Big(-i(ix)\Big)}{2}\\
& = \frac{\exp\Big(i^2x\Big)+\exp\Big(-i^2x\Big)}{2}\\
& = \frac{\exp(-x)+\exp(x)}{2}\\
& = \cosh(x)
\end{align}
$$
also
$$\sin(iy)=i\sinh(y)$$
and we know
$$\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)$$
now, if $a=x$ and $b=iy$ then
$$
\begin{align}
\sin(x+iy) & = \sin(x)\cos(iy)+\cos(x)\sin(iy)\\
&= \sin(x)\cosh(y)+\cos(x)\Big(i\sinh(y)\Big)\\
&= \sin(x)\cosh(y)+i\cos(x)\sinh(y)
\end{align}
$$