Ta co
sinA+sinB+sinC=4cosA2cosB2cosC2(tự chứng minh)
sin2A+sin2B+sin2C=4sinAsinBsinC
Vậy cosA2cosB2cosC2=sinAsinBsinC=8cosA2cosB2cosC2.sinA2sinB2sinC2
⇒8sinA2sinB2sinC2=1
Ta có 8sinA2sinB2sinC2=4sinA2[cosB−C2−cosB+C2]=4sinA2[cosB−C2−sinA2]≤4sinA2[1−sinA2]
(Vi cosB−C2≤1; sinA2≥0))
=1−[1−2sinA2]2≤1
Dấu = xảy ra khi chỉ khi sinA2=12; cosB−C2=1
⇔A=B=C=π3