In a $Δ A B C$ , prove that ${sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + {sin}^{3} C cos (A - B) = 3 sin A sin B sin C$

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Solution

$L H S = {sin}^{3} A cos (B - C) + {sin}^{3} B (C - A) + {sin}^{3} cos (A - B)$

$= {sin}^{2} A - sin (B + C) cos (B - C) + {sin}^{2} B sin (C + A) cos (C - A) + {sin}^{2} C sin (A + B) cos (A - B) [∵ A + B + C = π]$

$= \frac{1}{2} [{sin}^{2} A (sin 2 B + sin 2 C) + {sin}^{2} B (sin 2 C + sin 2 A) + {sin}^{2} C (sin 2 A + sin 2 B)] [∵ sin C + sin D = 2 sin \frac{C + D}{2} cos \frac{C + D}{2}]$

$= \frac{2}{2} [{sin}^{2} A sin B cos B + {sin}^{2} A sin C cos C + {sin}^{2} B sin C cos C + {sin}^{2} B sin A cos A + {sin}^{2} (sin A cos A + {sin}^{2} C sin B cos B)]$

$= sin A sin B (sin A cos B + cos A sin B) + sin B sin C (sin B cos C + cos B sin C) + sin C sin A (sin C cos A + cos C sin A)$

$= sin A sin B sin (A + B) + sin B sin C sin C sin B + C + sin (sin A sin (C + A))$

$= 3 sin A sin B sin C = R H S$

$[∵ A + B + C = π ∴ sin (π - θ) = sin θ]$

Hence, proved

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