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