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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?

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$$a=cis({\alpha})=e^{i\alpha}$$ $$b=cis({\beta})=e^{i\beta}$$ $$RHS=\frac{b^2-a^2}{2ab}i=\frac{bi}{2a}-\frac{ai}{2b}=\frac12i(e^{i(\beta-\alpha)}-e^{i(\alpha-\beta)})=\frac12i(e^{i(\beta-\alpha)}-e^{-i(\beta-\alpha)})=\frac12i(2i\sin(\beta-\alpha))=\sin(\alpha-\beta)=LHS$$

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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?

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Recall that $$\def\cis{\operatorname{cis}} \sin\gamma=\frac{\cis\gamma-\cis(-\gamma)}{2i} $$ Then $$ \sin(\alpha-\beta)=\frac{\cis(\alpha-\beta)-\cis(\beta-\alpha)}{2i}= \frac{1}{2i}\left(\frac{\cis\alpha}{\cis\beta}-\frac{\cis\beta}{\cis\alpha}\right) $$

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