Question:

Show that \(\frac{\sin(\alpha + \beta)\sin(\alpha – \beta)}{\sin\alpha – \sin\beta}=\sin\alpha + \sin\beta\).

Answer:

Use \(\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha\).

The given expression will become

\[ \frac{\sin(\alpha + \beta)\sin(\alpha – \beta)}{\sin\alpha – \sin\beta} = \frac{(\sin\alpha\cos\beta + \sin\beta\cos\alpha)(\sin\alpha\cos\beta – \sin\beta\cos\alpha)}{\sin\alpha – \sin\beta} \]

\[ = \frac{\sin^2\alpha\cos^2\beta – \sin^2\beta\cos^2\alpha}{\sin\alpha – \sin\beta} \]

Use \(\cos^2\theta=1-\sin^2\theta\) to eliminate \(\cos^2\alpha\) and \(\cos^2\beta\).

\[ \frac{\sin^2\alpha\cos^2\beta – \sin^2\beta\cos^2\alpha}{\sin\alpha – \sin\beta} \]

\[ = \frac{\sin^2\alpha(1-\sin^2\beta) – \sin^2\beta(1-\sin^2\alpha)}{\sin\alpha – \sin\beta} \]

\[ =  \frac{\sin^2\alpha-\sin^2\alpha\sin^2\beta – \sin^2\beta + \sin^2\beta\sin^2\alpha}{\sin\alpha – \sin\beta} \]

\[ = \frac{\sin^2\alpha – \sin^2\beta}{\sin\alpha – \sin\beta} \]

\[ = \frac{(\sin\alpha + \sin\beta)(\sin\alpha – \sin\beta)}{\sin\alpha – \sin\beta} \]

\[ = \sin\alpha + \sin\beta \]

It has been proven! Moreover, we can also say, \( \sin(\alpha + \beta)\sin(\alpha – \beta) = \sin^2\alpha – \sin^2\beta \)