Q. 14

In a Δ ABC prove that

sin3 A cos (B – C) + sin3 B cos (C – A) + sin3 C cos (A – B) = 3 sin A sin B sin C

Answer :


The key point to solve the problem:


The idea of sine Formula:




Idea of projection Formula:


• c = a cos B + b cos A


• b = c cos A + a cos C


• a = c cos B + b cos C


As we have to prove:-


sin3 A cos (B – C) + sin3 B cos (C – A) + sin3 C cos (A – B) = 3 sin A sin B sin C


as there is no resemblance of above expression with any formula so first we need to simplify the expression


LHS = sin3 A cos (B – C) + sin3 B cos (C – A) + sin3 C cos (A – B)


LHS = sin2 A sin A cos (B – C) + sin2 B sin B cos (C – A) + sin2 C sin C cos (A - B)


LHS = sin2 A sin{π – (B+C)}cos (B – C) + sin2B sin{π – (A+C)}cos (C – A) + sin2C sin {π – (A + B)} cos (A - B)


LHS = sin2A sin (B+C) cos(B-C) + sin2B sin(A + C)cos(C – A) + sin2C sin(B + C) cos (A - B)


Using the relation sin ( X + Y )cos(X – Y) = sin 2X + sin 2Y , we have –


LHS = sin2A (sin 2B + sin 2C) + sin2B (sin 2A + sin 2C) + sin2C (sin 2B + sin 2A)


Using sin 2X = 2sin X cos X , we have –


LHS = sin2A (2sinB cosB + 2sinC cosC) + sin2B (2sinA cosA + 2sinC cosC) + sin2C (2sinBcosB + 2sinA cosA)


Using sine formula we have –



sin A = ka , sin B = kb and sin C = kc …eqn 1


Putting the values in LHS:-


LHS =


LHS =


Using projection formula


• c = a cos B + b cos A


• b = c cos A + a cos C


• a = c cos B + b cos C


We have


LHS =


= {using eqn 1}


= = RHS ….Hence proved


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