# Prove Sin A + Sin B + Sin C - Sin(A+B+C) = 4 Sin((A+B)/2) Sin((B+C)/2) Sin ((C+A)/2) without assuming that the sum of angles is A+B+C=180 degrees

We answer the given question we use the formulas:

2 sin A sin B = - cos (A + B) + cos (A - B) and 2 cos A sin B = sin (A + B) - sin (A - B)

Now 4*sin ((A+B)/2)*sin ((B+C)/2)*sin ((C+A)/2)

=> 2*[-cos ((A+B)/2+ (B+C)/2) + cos ((A+B)/2-(B+C)/2)]*sin ((C+A)/2)

=> 2*[-cos (A/2+B+C/2) + cos (A/2-C/2)]*sin ((C+A)/2)

=> 2*[-cos (A/2+B+C/2)*sin ((C+A)/2)] + 2*cos (A/2-C/2)*sin ((C+A)/2)

=> - [sin (A/2+B+C/2 - (C+A)/2) - sin (A/2+B+C/2 - (C+A)/2))] + [sin (A/2-C/2 + (C+A)/2) - sin (A/2-C/2 - (C+A)/2)]

=> [-sin (A/2+B+C/2 +C/2+A/2) + sin (A/2+B+C/2 -C/2-A/2)] + [sin (A/2-C/2 +C/2+A/2) - sin (A/2-C/2 -C/2-A/2)]

=> - [sin (A+B+C) + sin (B] + [sin (A) - sin (-C)]

=> -sin (A+B+C) + sin B + sin A + sin C

=> sin A + sin B + sin C - sin (A+B+C)

Therefore sin A + sin B + sin C - sin (A+B+C) = 4*sin ((A+B)/2)*sin ((B+C)/2)*sin ((C+A)/2)

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