Question

三角形中cos(A/2)^2 + cos(B/2)^2 + cos(C/2)^2 > 2

Answer 1

cos(A/2)^2 + cos(B/2)^2 + cos(C/2)^2

= (1/2)(1+cosA) + (1/2)(1+cosB)c + (1/2)(1+cosC)

= 3/2 + (1/2)(cosA + cosB + cosC)

= 3/2 + (1/2)(cosA + cosB - cos(A+B))

= 2 + cos[(A+B)/2)]{cos[(A-B)/2)] - cos[(A+B)/2)]

> 2