Question

Prove that:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]

Answer 1

\[ \frac{13\pi}{3} = 780^\circ, \frac{8\pi}{3} = 480^\circ, \frac{2\pi}{3} = 120^\circ, \frac{5\pi}{6} = 150^\circ\]

 LHS = \[\sin \left( 780^\circ \right) \sin \left( 480^\circ \right) + \cos \left( 120^\circ \right) \sin\left( 150^\circ \right)\]

\[ = \sin \left( 90^\circ \times 8 + 60^\circ \right) \sin \left( 90^\circ \times 5 + 30^\circ \right) + \cos \left( 90^\circ \times 1 + 30^\circ \right) \sin \left( 90^\circ \times 1 + 60^\circ \right)\]

\[ = \sin \left( 60^\circ \right) \cos \left( 30^\circ \right) + \left[ - \sin \left( 30^\circ \right) \right] \cos \left( 60^\circ \right)\]

\[ = \sin \left( 60^\circ \right) \cos \left( 30^\circ \right) - \sin \left( 30^\circ \right) \cos\left( 60^\circ \right) \]

\[ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2}\]

\[ = \frac{3}{4} - \frac{1}{4}\]

\[ = \frac{1}{2}\]

 = RHS

Hence proved.