Question

Prove that:
sin 3A + sin 2A − sin A = 4 sin A cos \[\frac{A}{2}\] \[\frac{3A}{2}\]

 

Answer 1

Consider LHS:
\[ = \sin 3A + \sin 2A - \sin A\]
\[ = 2\sin \left( \frac{3A + 2A}{2} \right) cos \left( \frac{3A - 2A}{2} \right) - \sin A \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\sin \left( \frac{5}{2}A \right) \cos \left( \frac{A}{2} \right) - \sin A\]
\[= 2\sin \left( \frac{5}{2}A \right) cos \left( \frac{A}{2} \right) - 2\sin \frac{A}{2} \cos \frac{A}{2}\]
\[ = 2\cos \left( \frac{A}{2} \right) \left\{ \sin \frac{5}{2}A - \sin \frac{A}{2} \right\}\]
\[ = 2\cos \left( \frac{A}{2} \right) \times 2\sin \left( \frac{\frac{5}{2}A - \frac{A}{2}}{2} \right) \cos \left( \frac{\frac{5}{2}A + \frac{A}{2}}{2} \right)\]
\[ = 4\cos \left( \frac{A}{2} \right) \sin A \cos \left( \frac{3}{2}A \right)\]
\[ = 4\sin A \cos$\left( \frac{A}{2} \right)$\cos \left( \frac{3}{2}A \right)\]
 = RHS
Hence, LHS = RHS