Question

Prove that:

\[\sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6} = \frac{1}{2}\]

 

Answer 1

 LHS = \[\sin\frac{8\pi}{3}\cos\frac{23\pi}{6} + \cos\frac{13\pi}{3}\sin\frac{35\pi}{6}\]
\[ = \sin \left( \frac{8}{3} \times 180^\circ \right) \cos \left( \frac{23}{6} \times 180^\circ \right) + \cos\left( \frac{13}{3} \times 180^\circ \right)\sin\left( \frac{35}{6} \times 180^\circ \right)\]
\[ = \sin \left( 480^\circ \right) \cos \left( 690^\circ \right) + \cos \left( 780^\circ \right) \sin \left( 1050^\circ \right)\]
\[ = \sin \left( 90^\circ \times 5 + 30^\circ \right) \cos \left( 90^\circ \times 7 + 60^\circ \right) + \cos \left( 90^\circ \times 8 + 60^\circ \right)\sin \left( 90^\circ \times 11 + 60^\circ \right)\]
\[ = \cos \left( 30^\circ \right) \sin \left( 60^\circ \right) + \cos \left( 60^\circ \right)\left[ - \cos \left( 60^\circ \right) \right]\]
\[ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \left( - \frac{1}{2} \right)\]
\[ = \frac{3}{4} - \frac{1}{4}\]
\[ = \frac{2}{4}\]
\[ = \frac{1}{2}\]
= RHS
Hence proved .