Question

Prove that:

\[\sin 65^\circ + \cos 65^\circ = \sqrt{2} \cos 20^\circ\]

Answer 1

Consider LHS:
\[\sin 65^\circ + \cos 65^\circ\]
\[ = \sin 65^\circ + \cos \left( 90^\circ - 25^\circ \right)\]
\[ = \sin 65^\circ + \sin 25^\circ\]
\[ = 2\sin \left( \frac{65^\circ + 25^\circ}{2} \right) \cos \left( \frac{65^\circ - 25^\circ}{2} \right) \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\sin 45^\circ \cos 20^\circ\]
\[ = 2 \times \frac{1}{\sqrt{2}} \cos 20^\circ\]
\[ = \sqrt{2}\cos 20^\circ\]
 = RHS
Hence, LHS = RHS.