TITLE: Prove: $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$
QUESTION [7 upvotes]: This is an exercise in Gelfand's Trigonometry, It is not that difficult but I am doing something wrong that is preventing me from proving the identity.
We need to use the following diagram to prove it: 
My attempt: 
$$
\begin{eqnarray*} 
\sin (\alpha - \beta) = \frac{CD}{AC} \\
= \frac{PQ}{AC} \\
= \frac{BQ - BP}{AC} \\
= \frac{BQ}{AC} - \frac{BP}{AC} \\ 
\end{eqnarray*}
$$
Now in the following step we should use an intermediary to make this equal to the required identity, but for the first fraction I can't find anything rather than $AB$}
$$
= \frac{BQ}{AB} \cdot \frac{AB}{AC} \\
$$
My problem here is I don't see how $\frac{AB}{AC}$ would simplify to $\cos \beta$ to me this seems like $\sec \beta$ How could this be fixed?

REPLY [2 votes]: This website has a lot of cool stuff about trigonometry.
https://trigonography.com/2015/09/28/angle-sum-and-difference-for-sine-and-cosine/