函数f(x)=sin²(wx+π/6)-cos²(wx+π/6)(w>0)的最小正周期为2π,求w的值;若tanx=4/3,
问题描述:
函数f(x)=sin²(wx+π/6)-cos²(wx+π/6)(w>0)的最小正周期为2π,求w的值;若tanx=4/3,
且x∈(π,3π/2),求f(x)的值
答
f(x)=sin²(wx+π/6)-cos²(wx+π/6)=-cos[2(wx+π/6)]=-cos(2wx+π/3)最小正周期为2π=2π/(2w)w=1/2f(x)=-cos(x+π/3)tanx=4/3且x∈(π,3π/2)sinx=-4/5,cosx=-3/5f(x)=-cos(x+π/3)=cosxcos(π...