题目

设函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,x属于[0,π/2],求f(x)的值域

提问时间：2020-10-10

答案

f(x)=2cosxsin(x+π/3) -√3sin²x+sinxcosx
=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3sin²x+sinxcosx
=2cosx[(1/2)sinx +(√3/2)cosx]-√3sin²x+sinxcosx
=sinxcosx +√3cos²x-√3sin²x+sinxcosx
=2sinxcosx+√3(cos²x-sin²x)
=sin(2x)+√3cos(2x)
=2[(1/2)sin(2x)+(√3/2)cos(2x)]
=2[sin(2x)cos(π/3)+cos(2x)sin(π/3)]
=2sin(2x+π/3)
x∈[0,π/2]
π/3≤2x+π/3≤4π/3
-√3/2≤sin(2x+π/3)≤1
-√3≤2sin(2x+π/3)≤2
-√3≤f(x)≤2
函数的值域为[-√3,2].