algebra precalculus - Help with showing how $sinalphacosbeta$ $=$ $frac{1}{2}(sin (alpha + beta) + sin(alpha-beta))$ using Eulers formula.

I need help with understanding how one can rewrite:

$\sin\alpha\cos\beta$

to be equal to: $\frac{1}{2}(\sin (\alpha + \beta) + \sin(\alpha-\beta))$ using Eulers formula.

I know that it probably is quite simple but I cannot get my head around this... Frustrating!

Eulers formulas:

$\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$

$\sin \theta = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})$

Thank you kindly for your help!

Answer

$$\begin{align*}
\sin\alpha\cos\beta&=\frac1{4i}(e^{i\alpha}-e^{-i\alpha})(e^{i\beta}+e^{-i\beta})\\
&=\frac1{4i}\Big(e^{i\alpha}e^{i\beta}+e^{i\alpha}e^{-i\beta}-e^{-i\alpha}e^{i\beta}-e^{-i\alpha}e^{-i\beta}\Big)\\
&=\frac1{4i}\Big(e^{i\alpha}e^{i\beta}-e^{-i\alpha}e^{-i\beta}+e^{i\alpha}e^{-i\beta}-e^{-i\alpha}e^{i\beta}\Big)\\
&=\frac1{4i}\Big(e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)}+e^{i(\alpha-\beta)}-e^{-i(\alpha-\beta)}\Big)\\
&=\frac12\Big(\sin(\alpha+\beta)+\sin(\alpha-\beta)\Big)
\end{align*}$$

I actually got this by working from each end towards the middle, however.