求详解:y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期
1个回答

∵2x+π/3+π/6-2x=π/2

∴y=tan(2x+π/3)-tan(π/6-2x)

=tan(2x+π/3)-cot[π/2-(π/6-2x)]

=tan(2x+π/3)-cot(2x+π/3)

=sin(2x+π/3)/cos(2x+π/3)-cos(2x+π/3)/sin(2x+π/3)

=[sin²(2x+π/3)-cos²(2x+π/3)]/sin(2x+π/3)cos(2x+π/3)

=-2cos2(2x+π/3)/sin2(2x+π/3)

=-2cos(4x+2π/3)/sin(4x+2π/3)

=-2cot(4x+2π/3)

∴y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期T=π/4