$\displaystyle \mbox{if}\ A+B+C=\pi\ \mbox{then prove,}\ sin2A+sin2B+sin2C=4sinAsinBsinC$
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sinn2a + sin2b = 2sin(a+b)cos(a-b) =2sinc{cos(a-b)} lhs becomes =2sinc{cos(a-b)} + sin2c =2sinc{cos(a-b) + cosc} =2sinc{cos(a-b) - cos(a+b)} =4sina.sinb.sinc=rhs
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