Question

已知函数f(x)=sin^2wx+根号3sinwxsin(wx+π/2)的最小正周期为π，

当x∈（-12分之π，2分之π）时，求函数f（x）的值域

Answer 1

f(x)=sin^2wx+√3sinwxsin(wx+π/2)
=sin^2wx+√3sinwxcoswx
=1/2(1-cos2wx)+√3/2sin2wx
=√3/2sin2wx-1/2cos2wx+1/2
=sin(2wx-π/6)+1/2
∵f(x)最小正周期T=π
∴2π/(2w)=π,w=1
f(x)=sin(2x-π/6)+1/2
∵x∈(-π/12,π/2)
∴2x∈(-π/6,π)
2x-π/6∈(-π/3,5π/6)
∴-√3/2<sin(2wx-π/6)≤1
∴（1-√3)/2<sin(2x-π/6)+1/2≤3/2
函数的 值域为( (1-√3)/2,3/2]