In `DeltaABC`, prove that: `sin2A+sin2B+sin2C=4sinAsinBsinC`

1.	In `DeltaABC`, prove that: `sin2A+sin2B+sin2C=4sinAsinBsinC`
Answer» LHS=`sin2A+sin2B+sin2C` `=2sin(180^(@)-C)cos(A-B)+2sinC).cos{180^(@)-A(A+B)}(therefore A+B+C-180^(@))` `=2sinC.cos(A-B)-2sinC.cos(A+B)` `=2sinc[cos(A-B)-cos(A+B)]` `=2sinC.2sinAsinB` `=4sinAsinBsinC`=RHS Hence Proved.

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