在三角形ABC中,acos^2*C/2+ccos^2*A/2=3b/2,已求得a.b.c成等差数列,求角B的取值范围.

(1)
acos^C/2+ccos^A/2=3/2b
LZ的意思应该是a(cos(C/2))^2+c(cos(A/2))^2=3/2b吧~
a(cos(C/2))^2+c(cos(A/2))^2
=a*(1+cosC)/2+c(1+cosA)/2
=a/2+c/2+(acosC+ccosA)/2
=a/2+c/2+b/2 (做高BH,则acosC+ccosA=AH+CH=AC=b)
=3/2*b
a+c=2b
故a,b,c成等差数列
cosB=(a^2+c^2-b^2)/(2*a*c)
=(a^2+c^2-((a+c)/2)^2)/(2*a*c)
=(3*a^2+3*c^2-2*a*c)/(8*a*c)
=(3*a^2+3*c^2-6*a*c)/(8*a*c)+1/2
=3*(a-c)^2/(8*a*c)+1/2
>=1/2
所以 0