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Prove That: Sin 47° + Cos 77° = Cos 17°

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Question

Prove that:
sin 47° + cos 77° = cos 17°

Sum
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Solution

Consider LHS:
\[\sin 47^\circ + \cos 77^\circ\]
\[ = \sin 47^\circ + \cos \left( 90^\circ - 13^\circ \right)\]
\[ = \sin 47^\circ + \sin 13^\circ\]
\[ = 2\sin \left( \frac{47^\circ + 13^\circ}{2} \right) \cos \left( \frac{47^\circ - 13^\circ}{2} \right) \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\sin 30^\circ \cos 17^\circ\]
\[ = 2 \times \frac{1}{2}\cos 17^\circ\]
\[ = \cos 17^\circ\]
 = RHS
Hence, LHS = RHS.

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Transformation Formulae
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Chapter 8: Transformation formulae - Exercise 8.2 [Page 18]

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R.D. Sharma Mathematics [English] Class 11
Chapter 8 Transformation formulae
Exercise 8.2 | Q 5.2 | Page 18

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