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Hello Sudarshan
sinA + sinB +sinC=2sin[(A+B)/2].cos[(A-B)/2]+2sin(C/2).cos(C/2)
A+B+C=0=>A+B=-C
Therefore
sinA + sinB +sinC=2sin(C/2).cos[(A+B)/2]-2Sin(C/2).cos[(A-B)/2]
=2sin(C/2).{cos[(A+B)/2]-cos[(A-B)/2}
=2sin(A/2).sin(B/2).sin(C/2)
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