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Question
Prove that:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]
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Solution
\[ \frac{13\pi}{3} = 780^\circ, \frac{8\pi}{3} = 480^\circ, \frac{2\pi}{3} = 120^\circ, \frac{5\pi}{6} = 150^\circ\]
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.
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