三角形ABC,求sinA+sinB+sinC的值域.

题目

三角形ABC,求sinA+sinB+sinC的值域.

答案

sinA+sinB+sinC=sinB(1+cosC)+sinC(1+cosB)
=4cos(B/2)cos(C/2)[sin(B/2)cos(C/2)+cos(B/2)sin(C/2)]
=4cos(A/2)cos(B/2)cos(C/2)
比较不等式xyz≤(x^3+y^3+z^3)/3
A=B=C时,cos(A/2)=cos(B/2)=cos(C/2)
sinA+sinB+sinC最大=3√3/2
sinA,sinB,sinC>0
3√3/2 ≥sinA+sinB+sinC>0