Prove that: ( sin2A + sin2B + sin2C ) / ( sinA + sinB + sinC ) = 8 sin(A/2) sin(B/2) sin(C/2) Share with your friends Share 14 Himanshu answered this -19 View Full Answer Faisal Alam answered this Actual question is: If A + B + C = π, prove that: ( sin2A + sin2B + sin2C ) / ( sinA + sinB + sinC ) = 8 sin(A/2) sin(B/2) sin(C/2) -5
