Question

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

Answer 1

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]。

Answer 2

f(x)=2cosx[(1/2)sinx+(√3/2)cosx]-√3sin²x+sinxcosx
=cosxsinx+√3cos²x-√3sin²x+sinxcosx
=2sinxcosx+√3(cos²x-sin²x)
=sin(2x)+√3cos(2x)
=2[(1/2)sin(2x)+(√3/2)cos(2x)]
=2sin(2x+π/3)
xE[0,π/2]，
2xE[0,π]
2x+π/3E[0,4π/3] 请画个y=sina的图。
sin(2x+π/3)E[sin(4π/3), 1]
2sin(2x+π/3)E[2*(-根号3/2),2]
即值域：[-根号3,2]

Answer 3

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] 即值域：[-根号3,2]