If $cis(\alpha)=a$ and $cis(\beta)=b$, prove $\sin(\alpha-\beta)=\frac{b^2-a^2}{2ab}i$. If $cis(\alpha)=a$ and $cis(\beta)=b$, prove $$\sin(\alpha-\beta)=\frac{b^2-a^2}{2ab}i$$
I started with the right side and tried "expanding" $cis$ and arrived at the following formula:
$RHS=\frac{i[\cos(2\beta)-\cos(2\alpha)]-[\sin(2\beta)+\sin(2\alpha)]}{2\cos(\alpha+\beta)+i\sin(\alpha+\beta)}$
I don't think it helps as the $\sin(\alpha-\beta)$ that I want to arrive at doesn't appear anywhere in this form.
Any help?
 A: $$a=cis({\alpha})=e^{i\alpha}$$
$$b=cis({\beta})=e^{i\beta}$$
$$RHS=\frac{b^2-a^2}{2ab}i=\frac{bi}{2a}-\frac{ai}{2b}=\frac12i(e^{i(\beta-\alpha)}-e^{i(\alpha-\beta)})=\frac12i(e^{i(\beta-\alpha)}-e^{-i(\beta-\alpha)})=\frac12i(2i\sin(\beta-\alpha))=\sin(\alpha-\beta)=LHS$$
A: Recall from trig that
$$ \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta.$$
We know that
$$ \text{cis}(\alpha-\beta) = \cos(\alpha - \beta) + i\sin(\alpha-\beta).$$
Also, we can use the general fact that $\text{cis } \theta = e^{i\theta}$, along with basic rules of exponents, to say
$$\text{cis}(\alpha-\beta) = e^{i(\alpha-\beta)} = \frac{e^{i\alpha}}{e^{i\beta}} = \frac{\cos\alpha + i\sin\alpha}{\cos\beta + i\sin\beta}.$$
So we see then that
$$ \cos(\alpha-\beta) + i\sin(\alpha-\beta) = \frac{\cos\alpha + i\sin\alpha}{\cos\beta + i\sin\beta}. $$
Can you take it from here?
A: Recall that
$$\def\cis{\operatorname{cis}}
\sin\gamma=\frac{\cis\gamma-\cis(-\gamma)}{2i}
$$
Then
$$
\sin(\alpha-\beta)=\frac{\cis(\alpha-\beta)-\cis(\beta-\alpha)}{2i}=
\frac{1}{2i}\left(\frac{\cis\alpha}{\cis\beta}-\frac{\cis\beta}{\cis\alpha}\right)
$$
