Question

Show that :

\[\sin 50^\circ \cos 85^\circ = \frac{1 - \sqrt{2} \sin 35^\circ}{2\sqrt{2}}\]

Answer 1

\[\text{ LHS }= 2 \sin50^\circ \cos 85^\circ\]
\[ = \frac{\sin \left( 50^\circ + 85^\circ \right) + \sin \left( 50^\circ - 85^\circ \right)}{2} \left[ \because \sin A \cos B = \frac{1}{2}\left\{ \sin (A + B) + \sin (A - B) \right\} \right]\]
\[ = \frac{\sin 135^\circ + \sin \left( - 35^\circ \right)}{2}\]
\[ = \frac{\sin 135^\circ - \sin 35^\circ}{2}\]
\[ = \frac{\cos 45^\circ - \sin 35^\circ}{2} \left[ \because \sin \left( 90^\circ + 45^\circ \right) = \cos 45^\circ \right]\]
\[ = \frac{1}{2}\left( \frac{1}{\sqrt{2}} - \sin 35^\circ \right)\]
\[ = \frac{1}{2}\left[ \frac{1 - \sqrt{2}\sin 35^\circ}{\sqrt{2}} \right]\]
\[ = \frac{1 - \sqrt{2}\sin 35^\circ}{2\sqrt{2}}\]
\[\text{ RHS }= \frac{1 - \sqrt{2}\sin 35^\circ}{2\sqrt{2}}\]
Hence, LHS = RHS