Deriving sine difference formula

Question

I’m stuck proofing the difference formula $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$.

Given this diagram:

$$\sin (\alpha - \beta) = CD/AC = PQ/AC = (BQ-BP)/AC=BQ/AC-BP/AC$$

Now we need to relate that back to sine and cosine of alpha and beta:

$$BQ/AC = ???$$

I’m stuck here because $\sin \alpha = BQ/AB$ and $\cos \beta = AC/AB$. Multiplying those fractions together won’t give me $BQ/AC$.

I thought I had the right idea, but I’m stuck. Hints are appreciated.

Answer 1

Let's instead prove the sum formula: $\sin(\alpha)=\sin(\alpha-\beta)\cos(\beta) + \sin(\beta)\cos(\alpha-\beta)$.

To do this, explain why the first summand on the right equals $CD/AB$, while the second summand on the right equals $BP/AB$ (hint: use similar triangles!). Combining these two gives you the sum formula.

To get the difference formula, draw the same diagram but with $\beta$ going clockwise.