If A + B + C = π, prove that

sin (B + C – A) + sin (C + A – B) – sin (A + B – C) = 4cos A cos B sin C


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


Using,



= 2sinC cos(B-A) – sin(A+B-C)


since A + B + C = π



= 2sinCcos(B-A) – sin(π – C – C)


= 2sinCcos(B-A) – sin2C


Since , sin2A = 2sinAcosA,


= 2sinCcos(B-A) – 2sinCcosC


= 2sinC{cos(B-A) – cosC}


Using ,





= 4cosAcosBsinC


= R.H.S


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