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Prove that: sin \(\cfrac{13\pi}3\) sin \(\cfrac{2\pi}3\) + cos \(\cfrac{4\pi}3\) sin \(\cfrac{13\pi}6\) = \(\cfrac12\) |
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Answer» LHS = sin \(\cfrac{13\pi}3\) sin \(\cfrac{2\pi}3\) + cos \(\cfrac{4\pi}3\) sin \(\cfrac{13\pi}6\) = sin 780° sin 120° + cos 240° sin 390° = sin (90° × 8 + 60°) sin (90° × 1 + 30°) + cos (90° × 2 + 60°) sin (90° × 4 + 30°) We know that when n is odd, sin → cos. = sin 60° cos 30° + [-cos 60°] sin 30° = sin 60° cos 30° - sin 30° cos 60° We know that sin A cos B – cos A sin B = sin (A – B) = sin (60° - 30°) = sin 30° = 1/2 = RHS Hence proved. |
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