Q6 of 12 Page 595

If A + B + C = π, prove that


= sin2A + sin2B + sin2C


Using,



Sin2A = 2sinAcosA


= 2sinAcosA + 2sin(B+C)cos(B - C)
since A + B + C = π



= 2sinAcosA + 2sin(π - A)cos(B - C )
= 2sinAcosA + 2sinAcos(B - C)
= 2sinA{cosA + cos (B-C)}
( but cos A = cos { 180 - ( B + C ) } = - cos ( B + C )


And now using
= 2sinA{2sinBsinC}
= 4sinAsinBsinC



Now,


= sinA + sinB + sinC


Using,











Therefore,




= R.H.S


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