Vì `A;B;C` là $3$ góc của tam giác
`=>A+B+C=180°`
`=>C=180°-(A+B)`
Ta có: `sin[180°-(A+B)]=sin(A+B)`
`=>sinC=sin(A+B)`
`\qquad cos[180°-(A+B)]=-cos(A+B)`
`=>cosC=-cos(A+B)`
$\\$
`\qquad sin2A+sin2B+sin2C`
`=2sin\ {2A+2B}/2 \ cos\ {2A-2B}/2 +sin2C`
`=2sin (A+B).cos(A-B)+2sinCcosC`
`=2sinC.cos(A-B)+2sinC.cosC`
`=2sinC.[cos(A-B)+cosC]`
`=2sinC.[cos(A-B)-cos(A+B)]`
`=2sinC.[cosAcosB+sinAsinB-(cosAcosB-sinAsinB)]`
`=2sinC.2sinAsinB`
`=4sinAsinBsinC`
Vậy: `sin2A+sin2B+sin2C=4sinAsinBsinC`
