已知函数f(x)=sin^2wx+根号3sinwxsin(wx+π/2)的最小正周期为π,当x属于[-π/12,π/2]时,f(x)的值域
问题描述：
已知函数f(x)=sin^2wx+根号3sinwxsin(wx+π/2)的最小正周期为π,当x属于[-π/12,π/2]时,f(x)的值域
答
sin(wx+π/2)=sinwxcosπ/2+coswxsinπ/2=coswx
f(x)=sin^2wx+根号3sinwxsin(wx+π/2)
=sin²wx+√3sinwxcoswx
=(1-cos 2wx)/2+√3/2sin 2wx
=1/2-1/2cos 2wx+√3/2sin2wx
=1/2+sin(2wx-30)
T=2π/(2W)=π
w=1
f(x)=1/2+sin(2x-30)
x属于[-π/12,π/2]
-π/3≤(2x-30)≤5π/6
-√3/2≤sin(2wx-30)≤1
(1-√3)/2≤f(x)≤3/2