Q. 14

# In a Δ ABC 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

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:-

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

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

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

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

LHS = sin^{2} A sin{π – (B+C)}cos (B – C) + sin^{2}B sin{π – (A+C)}cos (C – A) + sin^{2}C sin {π – (A + B)} cos (A - B)

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

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

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

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

LHS = sin^{2}A (2sinB cosB + 2sinC cosC) + sin^{2}B (2sinA cosA + 2sinC cosC) + sin^{2}C (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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