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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?

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    $\begingroup$ It is when you took ${smth\over2}\cdot{smth\over2}$ and isolated just $1\over2$. $\endgroup$ – Ivan Neretin Sep 15 '16 at 5:07
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    $\begingroup$ You should have a factor of $\frac 14$ not $\frac 12$ in front of your first square bracket $\endgroup$ – David Quinn Sep 15 '16 at 5:07
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Brute force: \begin{align*} z &= x+yi \\ \sin z &= \frac{e^{iz}-e^{-iz}}{2i} \\ &= \frac{e^{ix-y}-e^{-ix+y}}{2i} \\ &= \frac{e^{-y}(\cos x+i\sin x)-e^{y}(\cos x-i\sin x)}{2i} \\ &= \frac{(e^{y}+e^{-y})\sin x}{2}+ \frac{i(e^{y}-e^{-y})\cos x}{2} \\ \sin (x+yi) &= \sin x \cosh y+i\cos x \sinh y \end{align*}

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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 $$ \begin{align} \sin(x+iy) & = \sin(x)\cos(iy)+\cos(x)\sin(iy)\\ &= \sin(x)\cosh(y)+\cos(x)\Big(i\sinh(y)\Big)\\ &= \sin(x)\cosh(y)+i\cos(x)\sinh(y) \end{align} $$

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