在三角ABC中,a,b,c分别是角A,角B,角C的对边.求证：acos^2 B/2+bcos^2 A/2=1/2(a+b+c)RT
网友回答
2[acos^2 B/2+bcos^2 A/2]
=2[a(cosB+1)/2+b(cosA+1)/2]
=acosB+bcosA+a+b
=a*(a^2+c^2-b^2)/(2ac)+b(b^2+c^2-a^2)/(2bc)+(a+b)
=(a^2+c^2-b^2+b^2+c^2-a^2)/(2c)+(a+b)
=2c^2/(2c)+(a+b)
=a+b+c
即：2[acos^2 B/2+bcos^2 A/2]=a+b+c
所以：acos^2 B/2+bcos^2 A/2=1/2（a+b+c）