If A + B + C = π, prove that
Taking L.C.M
Multiplying and divide the above equation by 2, we get
Since , sin2A = 2sinAcosA
NOW,
= sin2A + sin2B + sin2C
= 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
Putting the above value in the equation, we get
= 2
= R.H.S