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यदि `A+B+C=pi,` सिद्ध कीजिए कि
`sin 2A+sin 2B+sin 2C=4sin Asin B sinC`

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बायाँ पक्ष `=(sin2A+sinB)=sin2C`
`=2sin (A+B)cos(A-B)+2sinC cos C`
`=2sin(pi-C)cos(A-B)+2sin C cos C`
`=2sinC cos (A-B)+2sin C cos C" "(because sin (pi-C)=sinC)`
`=2sin C[cos (A-B)+cosC]`
`=2sin C[cos (A-B)+cos {pi-(A+B)}]`
`=2sin C[cos(A-B)-cos(A+B)]`
`=2sinC[2sinAsinB]`
`=4sin A sin B sin C= ` दायाँ पक्ष

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