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Prove That: Sin 38° + Sin 22° = Sin 82° - Mathematics

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Question

Prove that:
sin 38° + sin 22° = sin 82°

Sum
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Solution

Consider LHS:
\[\sin 38^\circ + \sin 22^\circ\]
\[ = 2\sin \left( \frac{38^\circ + 22^\circ}{2} \right) \cos \left( \frac{38^\circ - 22^\circ}{2} \right) \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\sin 30^\circ \cos 8^\circ\]
\[ = 2 \times \frac{1}{2}\cos(90^\circ - 8^\circ)\]
\[ = \sin 82^\circ\]
 = RHS
Hence, LHS = RHS .

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Transformation Formulae
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Chapter 8: Transformation formulae - Exercise 8.2 [Page 17]

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RD Sharma Mathematics [English] Class 11
Chapter 8 Transformation formulae
Exercise 8.2 | Q 2.1 | Page 17

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