# Verify the following equation: $\sin (x+iy)=\sin x \cosh y+i \cos x \sinh y$

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?

• It is when you took ${smth\over2}\cdot{smth\over2}$ and isolated just $1\over2$. – Ivan Neretin Sep 15 '16 at 5:07
• You should have a factor of $\frac 14$ not $\frac 12$ in front of your first square bracket – David Quinn Sep 15 '16 at 5:07

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}