In any triangle ABC,
${sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + {sin}^{3} C cos (A - B) = m sin A sin B sin C$ . Find $m$

Open in App

Solution

Given:
$A + B + C = π$

$∴ A = π - (B + C)$

LHS= ${sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + {sin}^{3} C cos (A - B)$

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

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

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

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

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

$= 3 sin A sin B sin C =$ RHS
Ans: 3

Suggest Corrections

Similar questions

{sin}^{3} A cos (B - C) + {sin}^{3} B cos (C - A) + s i n^{3} C cos (A - B) = 3 sin A sin B sin C

Δ A B C

{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

A + B + C = π

{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

A + B + C = π

{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

A + B + X = π

{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

Join BYJU'S Learning Program