Question

Prove that:
 sin 50° − sin 70° + sin 10° = 0

Answer 1

Consider LHS: 
\[\sin 50^\circ - \sin 70^\circ + \sin 10^\circ\]
\[ = 2\sin \left( \frac{50^\circ - 70^\circ}{2} \right) \cos \left( \frac{50^\circ + 70^\circ}{2} \right) + \sin 10^\circ \left\{ \because \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) \right\}\]
\[ = 2\sin \left( - 10^\circ \right) \cos 60^\circ + \sin 10^\circ\]
\[ = 2 \times \frac{1}{2}\sin \left( - 10^\circ \right) + \sin 10^\circ\]
\[ = - \sin 10^\circ + \sin 10^\circ\]
\[ = 0\]
Hence, LHS = RHS.