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