= 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