Help with showing how $\sin\alpha\cos\beta$ $=$ $\frac{1}{2}(\sin (\alpha + \beta) + \sin(\alpha-\beta))$ 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!
 A: Hint: You have the right formulas, so now try to FOIL their product and remember $$e^{i\alpha}\cdot e^{i\beta}=e^{i(\alpha+\beta)}\quad\text{and}\quad e^{i\alpha}\cdot e^{-i\beta}=e^{i(\alpha-\beta)}.$$
A: $$\begin{align*}
\sin\alpha\cos\beta&=\frac1{4i}(e^{i\alpha}-e^{-i\alpha})(e^{i\beta}+e^{-i\beta})\\
&=\frac1{4i}\Big(e^{i\alpha}e^{i\beta}+e^{i\alpha}e^{-i\beta}-e^{-i\alpha}e^{i\beta}-e^{-i\alpha}e^{-i\beta}\Big)\\
&=\frac1{4i}\Big(e^{i\alpha}e^{i\beta}-e^{-i\alpha}e^{-i\beta}+e^{i\alpha}e^{-i\beta}-e^{-i\alpha}e^{i\beta}\Big)\\
&=\frac1{4i}\Big(e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)}+e^{i(\alpha-\beta)}-e^{-i(\alpha-\beta)}\Big)\\
&=\frac12\Big(\sin(\alpha+\beta)+\sin(\alpha-\beta)\Big)
\end{align*}$$
I actually got this by working from each end towards the middle, however.
A: An alternative proof is first showing the addition theorems: 
$$
\begin{align*}\cos(x+y) + i \sin(x+y)&= e^{i(x+y)}\\
&=e^{ix} \cdot e^{iy}\\
&= (\cos(x)+ i \sin(x)) \cdot (\cos(y)+ i \sin(y))\\ & = \cos(x)\cos(y)-\sin(x)\sin(y) + i( \cos(x)\sin(y)+\cos(y) \sin(x))
\end{align*}$$
Comparing real parts and imaginary parts give you both addition theorems. Now we take the right hand side 
$$\begin{align}
\sin(x+y)+\sin(x-y)&=\cos(x)\sin(y)+\cos(y)\sin(x) + 
 \cos(x)\sin(-y)+\cos(-y)\sin(x)\\ &= 
\cos(x)\sin(y)+\cos(y)\sin(x) - \cos(x)\sin(y) +\cos(y)\sin(x)\\&=2 \sin(x)\cos(y)\end{align}$$
A: $(\sin\alpha)(\cos\beta) = \frac{1}{4i}(e^{i\alpha}-e^{-i\alpha})(e^{i\beta}+e^{-i\beta})$
by expanding the right hand side we get
$\frac{1}{4i}(e^{i\alpha+i\beta}-e^{-i\alpha-i\beta}+e^{i\alpha-i\beta}-e^{-i\alpha+i\beta}) = \frac{1}{4i}\big((e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)})+(e^{i(\alpha-\beta)} - e^{-i(\alpha-\beta)})\big)$
I think I'll stop here. =)
