f（π/6）=-√3sin²(π/6)+sin(π/6)cos(π/6)
=-√3*(1/4)+1/2*(√3/2)
=0
f(x)=-√3/2(1-cos2x)+1/2sin2x
=1/2sin2x+√3/2cos2x-√3/2
=sin(2x+π/3)-√3/2
T=2π/2=π
当sin(2x+π/3)=1的时候
f(x)有最大值
此时f(x)的最大值=1-√3/2=（2-√3）/2