Advertisement Remove all ads

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

Sum

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

Advertisement Remove all ads

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.

Concept: Transformation Formulae
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Exercise 8.2 | Q 5.2 | Page 18
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×