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
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(2wx+π/3);
T=2π/2w=2π;
w=1/2;
f(x)=-cos(x+π/3);
tanx=sinx/cosx=4/3;
sinx=(4/3)cosx;
sin²x+cos²x=1;
25cos²x/9=1;
cos²x=9/25;
∵x∈（π，3π/2）
∴cosx=-3/5;
sinx=-4/5;
f(x)=-(cosxcos(π/3)-sinxsin(π/3))=sinxsin(π/3)-cosxcos(π/3)=(-4/5)(√3/2)-(-3/5)(1/2)=-2√3/5+3/10;
答
w=0.5
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...