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Question

If A+B+C=π, then prove the following
sin2A+sin2B+sin2C=4sinA.sinB.sinC

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Solution

Given
A+B+C=π
Taking LHS
sin2A+sin2B+sin2C
=2sinAcosA+[2sin(B+C)cos(BC)] [sinC+sinD=2sin(C+D2)cos(CD2)]
=2sinAcosA+2sin(B+C)cos(BC)
=2sinAcosA+2sinAcos(BC) [A+B+C=π]
=2sinA[cos(B+C)+cos(BC)] [A+B+C=π;cosA=cos[π(B+C)]=cos(B+C)]
=2sinA(2sinBsinC) [cos(AB)cos(A+B)=2sinAsinB]
=4sinAsinBsinC
=RHS

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