Question

17已知函数f（x）=2sinxcos(x+兀/6)-cos2x+m求函数fx最小正周期

Answer 1

f(x)=2sinxcosxcosπ/6-2sinxsinxsinπ/6-cos2x+m
=√3/2*sin2x-1/2+1/2*cos2x-cos2x+m
=√3/2*sin2x-1/2cos2x-1/2+m
=cosπ/6sin2x-sinπ/6cos2x-1/2+m
=sin(2x-π/6)-1/2+m
则最小正周期为2π/2=π

Answer 2

原式化简=2sinx(√3/2.cosx-sinx/2)-cos2x+m
=√3/2.sin2x-(sinx)^2-cos2x+m
=√3/2.sin2x-cos2x.1/2+m-1/2 (二倍角)
=sin(2x-兀/6)+m-1/2
最小正周期：兀