Question

In ∆ABC, prove the following: 

\[\sin^3 A \cos \left( B - C \right) + \sin^3 B \cos \left( C - A \right) + \sin^3 C \cos \left( A - B \right) = 3 \sin A \sin B \sin C\]

Answer 1

\[\text{ Let } \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k . . . \left( 1 \right)\]

\[\text{ LHS } = \sin^3 A\cos\left( B - C \right) + \sin^3 B\cos\left( C - A \right) + \sin^3 C\cos\left( A - B \right)\]

\[ = \sin^2 A\left\{ \sin A\cos\left( B - C \right) \right\} + \sin^2 B\left\{ \sin B\cos\left( C - A \right) \right\} + \sin^2 A\left\{ \sinA\cos\left( A - B \right) \right\}\]

\[ = \frac{a^2}{k^2}\left\{ \sin A\cos\left( B - C \right) \right\} + \frac{b^2}{k^2}\left\{ \sin B\cos\left( C - A \right) \right\} + \frac{c^2}{k^2}\left\{ \sin A\cos\left( A - B \right) \right\} \left[ \text{ from } \left( 1 \right) \right]\]

\[ = \frac{1}{2 k^2}\left[ a^2 \left\{ 2\sin A\cos\left( B - C \right) \right\} + b^2 \left\{ 2\sin B\cos\left( C - A \right) \right\} + c^2 \left\{ 2\sinA\cos\left( A - B \right) \right\} \right]\]

\[ = \frac{1}{2 k^2}\left[ a^2 \left\{ 2\sin\left( \pi - \left( B + C \right) \right)\cos\left( B - C \right) \right\} + b^2 \left\{ 2\sin\left( \pi - \left( A + C \right) \right)\cos\left( C - A \right) \right\} + c^2 \left\{ 2\sin\left( \pi - \left( B + C \right) \right)\cos\left( A - B \right) \right\} \right]\]

\[ = \frac{1}{2 k^2}\left[ a^2 \left\{ 2\sin\left( B + C \right)\cos\left( B - C \right) \right\} + b^2 \left\{ 2\sin\left( C + A \right)\cos\left( C - A \right) \right\} + c^2 \left\{ 2\sin\left( A + B \right)\cos\left( A - B \right) \right\} \right]\]

\[ = \frac{1}{2 k^2}\left[ a^2 \left\{ \sin2B + \sin2C \right\} + b^2 \left\{ \sin2C + \sin2A \right\} + c^2 \left\{ \sin2A + \sin2B \right\} \right]\]

\[ = \frac{1}{2 k^2}\left[ 2 a^2 \left\{ \sin B\cos B + \sin C\cos C \right\} + 2 b^2 \left\{ \sin C\cos C + \sin A\cos A \right\} + 2 c^2 \left\{ \sin A\cos A + \sin B\cos B \right\} \right]\]

\[ = \frac{1}{2 k^3}\left[ 2 a^2 \left\{ k\sin B\cos B + k\sin C\cos C \right\} + 2 b^2 \left\{ k\sin C\cos C + k\sin A\cos A \right\} + 2 c^2 \left\{ k\sin A\cos A + k\sin B\cos B \right\} \right]\]

\[ = \frac{1}{k^3}\left[ a^2 \left\{ b\cos B + c\cos C \right\} + 2 b^2 \left\{ c\cos C + a\cos A \right\} + 2 c^2 \left\{ a\cos A + a\cos B \right\} \right]\]

\[ = \frac{1}{k^3}\left[ ab\left( a\cos B + b\cos A \right) + bc\left( b\cos C + c\cos B \right) + ac\left( a\cos C + c\cos A \right) \right]\]

\[ = \frac{1}{k^3}\left( abc + bca + acb \right)\]

\[ = 3abc \times \frac{1}{k^3}\]

\[ = 3\sin A\sin B\sin C \times \frac{1}{k^3} \times k^3 \]

\[ = 3\sin A\sin B\sin C\]

\[ = \text{ RHS }\]

\[\text{ Hence proved } .\]