ta có sin2A+sin2B+sin2C =2sin(A+B)cos(A-B) + 2 sinCcosC =2sinCcos(A-B)+2sinCcosC =2sinC ( cos(A-B) + cosC) =2sinC ( cos(A-B) - cos(A+B)) =2sinC.2sinAsinB =4sinAsinBsinC
ta có
$sin2A+sin2B+sin2C
$ $=
$$2sin(A+B)cos(A-B) + 2 sinCcosC
$ $=2sinCcos(A-B)+2sinCcosC
$ $=2sinC ( cos(A-B) + cosC)
$ $=2sinC ( cos(A-B) - cos(A+B))
$ $=2sinC.2sinAsinB
$ $=4sinAsinBsinC
$