Question

prove that sinA.sin(B-C)+sinB.sin(C-A)+sinC.sin(A-B)=0

Answer 1

SinA . Sin(B - C) + SinB . Sin(C - A) + SinC . Sin(A - B)=0

=>SinA . (SinBCosC - CosBSinC) + SinB . (SinCCosA - CosCSinA) + SinC . (SinACosB - CosASinB)

=>SinASinBCosC -SinACosBSinC + SinBSinCCosA - SinBCosCSinA + SinCSinACosB - SinCCosASinB

All get cancelled

=>0

Hence Proved

Answer 2

It should be:

sin(a) sin(b-c) + sin(b) sin(c-a) + sin(c) sin(a-b) = 0

Just use the addition formula for sine, i.e. sin(x+y) = sin(x)cos(y) + cos(x)sin(y) and the fact that sin(-x) = -sin(x) and cos(-x) = cos(x), and everything cancels out.

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