Question

Prove that:

\[\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = - 1\]

Answer 1

\[ \frac{10\pi}{3} = 600^\circ, \frac{13\pi}{6} = 390^\circ, \frac{8\pi}{3} = 480^\circ, \frac{5\pi}{6} = 150^\circ\]

LHS = \[\sin 600^\circ\cos 390^\circ + \cos 480^\circ \sin 150^\circ\]

\[ = \sin \left( 90^\circ \times 6 + 60^\circ \right) \cos\left( 90^\circ \times 4 + 30^\circ \right) + \cos\left( 90^\circ \times 5 + 30^\circ \right) \sin\left( 90^\circ \times 1 + 60^\circ \right)\]

\[ = \left[ - \sin 60^\circ \right] \cos30^\circ + \left[ - \sin 30^\circ \right] \cos 60^\circ\]

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

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

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

\[ = - 1\]

 = RHS

Hence proved.