Prove: $\sin (\alpha – \beta) = \sin \alpha \cos \beta – \sin \beta \cos \alpha$

solution-verificationtrigonometry

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: enter image description here

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?

Best Answer

This website has a lot of cool stuff about trigonometry.

https://trigonography.com/2015/09/28/angle-sum-and-difference-for-sine-and-cosine/ enter image description here

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