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Prove that :sin \(\cfrac{10\pi}3\) cos \(\cfrac{13\pi}6\) + cos \(\cfrac{8\pi}3\) sin \(\cfrac{5\pi}6\) = -1 |
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Answer» LHS = sin \(\cfrac{10\pi}3\) cos \(\cfrac{13\pi}6\) + cos \(\cfrac{8\pi}3\) sin \(\cfrac{5\pi}6\) = sin 600° cos 390° + cos 480° sin 150° = sin (90° × 6 + 60°) cos (90° × 4 + 30°) + cos (90° × 5 + 30°) sin (90° × 1 + 60°) We know that when n is odd, sin → cos and cos → sin. = [-sin 60°] cos 30° + [-sin 30°] cos 60° = -sin 60° cos 30° - sin 30° cos 60° = -[sin 60° cos 30° + cos 60° sin 30°] We know that sin A cos B + cos A sin B = sin (A + B) = -sin (60° + 30°) = -sin 90° = -1 = RHS Hence proved. |
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