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.
 A: 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.
A: The key to this problem is to express all sides in terms of the sines and cosines of $\alpha$ and $\beta$. Doing a bit of angle-chasing, then it can be shown that $\angle PBC=\alpha-\beta$. Therefore, we can express $BP$ as
\begin{align*}
BP & =BC\cos(\alpha-\beta)\\ & =AB\sin\beta\cos(\alpha-\beta)
\end{align*}
Where the last line was obtained since $BC=AB\sin\beta$. Therefore, our expression becomes
$$\sin(\alpha-\beta)\cos\beta=\sin\alpha-\sin\beta\cos(\alpha-\beta)$$
Now, we need to find an expression for $\cos(\alpha-\beta)$ in terms of the sines and cosines of $\alpha$ and $\beta$. Referring back to the diagram, then it can be shown that
\begin{align*}
\cos(\alpha-\beta) & =\frac {AQ+QD}{AC}\\ & =\frac {AB}{AC}\cos\alpha+\frac {AB}{AC}\sin\beta\sin(\alpha-\beta)
\end{align*}
Thus, we get that
\begin{align*}
\sin(\alpha-\beta) & =\frac {\sin\alpha}{\cos\beta}-\tan\beta\cos(\alpha-\beta)\\\cos(\alpha-\beta) & =\frac {\cos\alpha}{\cos\beta}+\tan\beta\sin(\alpha-\beta)
\end{align*}
Substituting the equation for $\cos(\alpha-\beta)$ into $\sin(\alpha-\beta)$ and using the identity $1+\tan^2\theta=\sec^2\theta$, then you arrive at your result. A similar method can be used to prove the expansion for $\cos(\alpha-\beta)$.
A: Another perspective on this question: how we can also get the sine and cosine addition formulae from this construction.

Gelfand's method is very clever: splitting a ratio of unrelated lengths into multiplying two ratios, each of which is a trigonometric ratio. However, I do not find this very natural to attempt at first glance.
More natural would be to use the line segments $BQ$ and $AD$, and their parallel segments $CD$ and $PC$. Using this we can observe that:
$$AQ + QD = AD \implies AQ + PC = AD$$
$$BP + PQ = BQ \implies BP + CD = BQ$$
Now as $AD$ is parallel to $PC$, $\angle PCA = \alpha - \beta \implies \angle PBC = \alpha - \beta$, so $\Delta ACD \sim \angle \Delta BCP$. Thus:
$$AB \cos \alpha + BC \sin(\alpha - \beta) = AC \cos(\alpha - \beta) \tag{1}$$
$$BC \cos(\alpha - \beta) + AC \sin(\alpha - \beta) = AB \sin \alpha \tag{2}$$
and as $BC = AB \sin \beta, AC = AB \cos \beta$:
$$\cos \alpha + \sin \beta \sin(\alpha - \beta) = \cos \beta \cos(\alpha - \beta) \tag{3}$$
$$\sin \beta \cos(\alpha - \beta) + \cos \beta \sin(\alpha - \beta) = \sin \alpha \tag{4}$$
These formulae look awfully close to what we want to prove. So indeed, relabelling $\alpha - \beta = \alpha', \beta = \beta'$, we have $\alpha = \alpha' + \beta'$ and making $\cos \alpha$ the subject in equation $(3)$:
$$\cos(\alpha' + \beta') = \cos \beta' \cos \alpha' - \sin \beta' \sin \alpha'$$
$$\sin(\alpha' + \beta') = \sin \beta' \cos \alpha' + \cos \beta' \sin \alpha'$$
which yield the cosine and sine addition formulae.
A: 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.
