Question

已知函数f(x)=根3sinxcosx_cos^2x+1/2。求最小正周期及值域

Answer 1

f(x)=(√3)sinxcosx－cos²x＋(1/2)
=(√3/2)sin2x－(1/2)[cos2x＋1]＋(1/2)
=(√3/2)sin2x－(1/2)cos2x
=sin(2x－π/6)
则f(x)最小正周期是2π/2=π，值域是：[－1，1]

Answer 2

f(x)=√3sinxcosx－cos²x+1/2
=√3*1/2sin2x-[(cos2x＋1)/2]+1/2
=√3/2sin2x-1/2cos2x+1/2-1/2
=√3/2sin2x-1/2cos2x
=cos(π/6)sin2x-sin(π/6)cos2x
=sin(2x-π/6)
所以最小正周期是2π/w=2π/2=π，值域是：[－1，1]