<s>[INST] Gelfands Trigonometry $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ Trying Prove the identity $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ using the figure provided in Gelfands trigonometry.

What I have so far
$\sin(\alpha - \beta) = \frac{CD}{AC} = \frac{PQ}{AC} = \frac{BQ}{AC} - \frac{BP}{AC}$
$\sin(\alpha) = \frac{BQ}{AB} \implies AB\sin(\alpha) = BQ$
$\sin(\alpha - \beta) = \frac{AB\sin(\alpha)}{AC} - \frac{BP}{AC}$
$\frac{AB}{AC} = \frac{1}{\cos(\beta)}$ #corrected
$\sin(\alpha - \beta) = \frac{\sin(\alpha)}{\cos(\beta)} - \frac{BP}{AC}$ # corrected
Im stuck on what to do with $\frac{BP}{AC}$. I've seen the posts here about the derivation of $\sin(\alpha + \beta)$ from the same diagram and I understand that proof perfectly well, but I am stuck on this one.
 [/INST] Try using the whole angle $\alpha$ as in the proof for the sin and cos addition formulas:
$$\sin \alpha = \frac{BQ}{AB} = \frac{BP}{AB} + \frac{CD}{AB}$$
$$=\frac{BP}{BC} \frac{BC}{AB} + \frac{CD}{AC} \frac{AC}{AB}$$
$$=\cos(\alpha - \beta) \sin \beta + \sin(\alpha - \beta) \cos \beta$$
and also:
$$\cos \alpha = \frac{AQ}{AB} = \frac{AD}{AB} - \frac{PC}{AB}$$
$$= \frac{AD}{AC} \frac{AC}{AB} - \frac{PC}{BC} \frac{BC}{AB}$$
$$= \cos(\alpha - \beta) \cos \beta - \sin(\alpha - \beta) \sin \beta$$
Now let $\cos(\alpha - \beta) = x, \sin(\alpha - \beta) = y$, and we have some simultaneous equations:
$$x \cos \beta - y \sin \beta = \cos \alpha$$
$$x \sin \beta + y \cos \beta = \sin \alpha$$
This means that:
$\sin \beta(x \cos \beta - y \sin \beta) = \cos \alpha \sin \beta \tag{1}$
$\cos \beta(x \sin \beta + y \cos \beta) = \sin \alpha \cos \beta \tag{2}.$
Thus $(1) - (2)$ gives:
$$-y \sin^2 \beta - y \cos^2 \beta = \cos \alpha \sin \beta - \sin \alpha \cos \beta, -y = \cos \alpha \sin \beta - \sin \alpha \cos \beta.$$
Remembering what $y$ is, we have the identity for $\sin(\alpha - \beta)$.
Solving similarly for $\cos(\alpha - \beta)$ gives the desired identity.
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