Chứng minh với mọi tam giác ABC ta có: cos2A + cos2B + cos2C

Xét vế trái:

cos2A + cos2B + cos2C

= (cos2A + cos2B) + 2cos²C – 1

= 2cos(A + B).cos(A − B) + 2cos²C – 1

= −2cosC.cos(A − B) + 2cos²C – 1

= −2cosC[cos(A − B) − cosC] − 1

= −2cosC[cos(A − B) + cos(A + B)] − 1

= −4cosC.cosA.cosB − 1

Vậy cos2A + cos2B + cos2C = –1 – 4cosA.cosB.cosC