Question

Prove that:
sin 105° + cos 105° = cos 45°

Answer 1

Consider LHS: 
\[\sin 105^\circ + \cos 105^\circ\]
\[ = \sin 105^\circ + \cos \left( 90^\circ + 15^\circ \right)\]
\[ = \sin 105^\circ - \sin 15^\circ\]
\[ = 2\sin \left( \frac{105^\circ  - 15^\circ}{2} \right) \cos \left( \frac{105^\circ + 15^\circ}{2} \right) \left\{ \because \sin A + \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right\}\]
\[ = 2\sin 45^\circ\cos 60^\circ\]
\[ = 2\sin \left( 90^\circ - 45^\circ \right) \cos 60^\circ\]
\[ = 2 \times \frac{1}{2}\cos\left( 45^\circ \right)\]
\[ = \cos 45^\circ\]
Hence, LHS = RHS.