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