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In DeltaABC, prove that: sin2A+sin2B+s...

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

Text Solution

Verified by Experts

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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