Question

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.

Answer 1

(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