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Question

If A+B+C=π, prove that sin2A+sin2B+sin2C=4sinAsinBsinC

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Solution

If A+B+C=π,C=π(A+B)
sinC=sin(A+B) and cosC=cos(A+B)

sin2A+sin2B+sin2C=2sin(A+B)cos(AB)+2sinCcosC

=2sinCcos(AB)+2sinCcosC
=2sinCcos(AB)2sinCcos(A+B)=2sinC(cos(AB)cos(A+B))
=2sinC(2sin(A)sin(B))

=4sinAsinBsinC

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