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.
Best Answer
The given diagram unnecessarily complicates what should be an otherwise simple proof. I shall use this one, which has a right angle at $B$ instead:
The angles are same as in the original diagram, i.e. $\angle BAC=\beta,\angle BAD=\alpha$ and $\angle CAD=\alpha-\beta$. We now take $$\sin(\alpha-\beta)=\frac{CD}{AC}=\frac{BQ-BP}{AC}$$
$$\implies\sin(\alpha-\beta)=\frac{BQ}{AB}\cdot\frac{AB}{AC}-\frac{BP}{BC}\cdot\frac{BC}{AC}$$
The only thing left to notice is that $\angle PBC=\alpha$ and thus:
$$\bbox[5px,border:2px solid #C0A000]{\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta}$$
The formula for $\cos(\alpha-\beta)$ follows in a similar fashion.