Question

Show that :

\[\sin 25^\circ \cos 115^\circ = \frac{1}{2}\left( \sin 140^\circ - 1 \right)\]

Answer 1

\[LHS = 2\sin25^\circ \cos 115^\circ\]
\[ = \frac{\sin \left( 25^\circ + 115^\circ \right) + \sin \left( 25^\circ - 115^\circ \right)}{2} \left[ \because \sin A \cos B = \frac{1}{2}\left\{ \sin (A + B) + \sin (A - B) \right\} \right]\]
\[ = \frac{\sin 140^\circ + \sin \left( - 90^\circ \right)}{2}\]
\[ = \frac{\sin 140^\circ - \sin \left( 90^\circ \right)}{2}\]
\[ = \frac{\sin 140^\circ - 1}{2} \]
\[RHS = \frac{\sin 140^\circ - 1}{2}\]
Hence, LHS = RHS