Derive $\sin(\alpha - \beta)$ Here's me trying to do that. 

 After $\frac{BP}{AC} = \frac{BP}{BC}  \frac{BC}{AC} = \cos(\alpha - \beta)\tan(\beta)$, I didn't know what to do next. Theoretically, it should become $\cos(\alpha)\sin(\beta)$ and be something like $\frac{AQ}{AB}  \frac{BC}{AB}$, but I don't know how to get it there.
Thanks.
 A: One way to do this is to use Euler's formula,
$e^{i \theta}=\cos \theta+i\sin\theta.$
With this famous formula,
$$e^{i(\alpha-\beta)}=\cos({\alpha-\beta})+i\sin({\alpha-\beta}).$$
Since by law of exponents $e^{i(\alpha-\beta)}=e^{i\alpha}e^{-i\beta}$, we can also write that
$$
\begin{align*}
e^{i(\alpha-\beta)}&=e^{i\alpha}e^{-i\beta} \\
&=(\cos \alpha+i\sin \alpha)(\cos \beta - i \sin \beta) \\
&=\cos \alpha \cos \beta + \sin \alpha \sin \beta + i(\sin \alpha \cos \beta - \sin \beta \cos \alpha).
\end{align*}
$$
As a result,
$$\cos({\alpha-\beta})+i\sin({\alpha-\beta})=\cos \alpha \cos \beta + \sin \alpha \sin \beta + i(\sin \alpha \cos \beta - \sin \beta \cos \alpha).$$
When two complex numbers are equal, the real parts equal real parts, and the imaginary parts equal imaginary parts. Therefore we can conclude, by comparing imaginary parts of the last equation, that
$$\sin({\alpha-\beta})=\sin \alpha \cos \beta - \sin \beta \cos \alpha.$$
A: 
From the diagram
\begin{align}
AD&=AB\cos(BAD)=AB\cos(a)=\sin(b)\cos(a)\\
DE&=BC=OB\sin(a)=\cos(b)\sin(a)
\end{align}
thus
$$
\sin(a+b)=AE=DE+AD=\sin(a)\cos(b)+\cos(a)\sin(b).
$$
Replacing $b$ with $-b$ and using $\cos(-b)=\cos(b)$ while $\sin(-b)=-\sin(b)$, you get
$$
\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b).
$$
PS The argument is common, but I want to give credit where it is due, especially for the diagram.
