If $A + B + C = π$ then ${sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + {sin}^{3} C cos (A - B) = 3 sin A sin B sin C$

True

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False

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Solution

The correct option is A True

$A + B + C = π$

${sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + {sin}^{3} c . cos (a - b)$

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

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

$\frac{1}{2} {sin}^{2} A ({sin}^{2} B + {sin}^{2} C) + \frac{1}{2} {sin}^{2} B (sin 2 C + sin 2 A) + \frac{1}{2} {sin}^{2} C (sin 2 A + sin 2 B)$

${sin}^{2} A (sin B cos B + sin C cos C) + {sin}^{2} B (sin C cos C + sin A cos A) + {sin}^{2} C (sin A cos A + sin B cos B)$

${sin}^{2} A sin B cos B + {sin}^{2} A sin C cos C + {sin}^{2} B sin C cos C + {sin}^{2} B sin A cos A + {sin}^{2} C sin A cos A + {sin}^{2} C sin B cos B$

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

$sin A sin B sin C + sin A sin B sin C + sin A sin B sin C$

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

Suggest Corrections

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