Question

Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]

Answer 1

\[ \frac{13\pi}{3} = 780^\circ, \frac{2\pi}{3} = 120^\circ, \frac{4\pi}{3} = 240^\circ, \frac{13\pi}{6} = 390^\circ\]
LHS = \[\sin\left( 780^\circ \right) \sin\left( 120^\circ \right) + \cos\left( 240^\circ \right) \sin\left( 390^\circ \right)\]
\[ = \sin\left( 90^\circ \times 8 + 60^\circ \right) \sin\left( 90^\circ \times 1 + 30^\circ \right) + \cos\left( 90^\circ \times 2 + 60^\circ \right) \sin\left( 90^\circ \times 4 + 30^\circ \right)\]
\[ = \sin 60^\circ \cos 30^\circ + \left[ - \cos 60^\circ \right] \sin 30^\circ\]
\[ = \sin 60^\circ \cos 30^\circ - \cos 60^\circ\sin 30^\circ\]
\[ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2}\]
\[ = \frac{3}{4} - \frac{1}{4}\]
\[ = \frac{1}{2}\]
 = RHS
Hence proved.