题目

函数f(x)=√3sin^(wx/2)+sin(wx/2)cos(wx/2) (w>0)的周期为π,求w的值和函数f(x)的单调递增区间

提问时间：2020-11-23

答案

f(x)=√3sin²(wx/2)+sin(wx/2)cos(wx/2)=-(√3/2)*[1- 2sin²(wx/2) -1] +(1/2)*2sin(wx/2)cos(wx/2)=-(√3/2)*[cos(wx) -1]+(1/2)*sin(wx)=sin(wx)*cos(π/3) -cos(wx)*sin(π/3)+ (√3/2)=sin(wx - π/3)...