y=tanx-1/tanx的最小正周期(求分析过程)
问题描述:
y=tanx-1/tanx的最小正周期(求分析过程)
答
y=sinx/cosx-cosx/sina
=-(cos²x-sin²x)/sinxcosx
=-2cos2x/sin2x
=-2cot2x
所以T=π/2
答
y=sinx/cosx-cosx/sinx
=(sin²x-cos²x)/sinx*cosx
=-2cos2x/sin2x
=-2cot2x
最小正周期是π/2
答
求y=1/tanx-tanx的最小正周期(用切化弦的方法做)请写出具体过程,谢谢y=1/tanx-tanx =cosx/sinx-sinx/cosx =(cos^2 x-sin^2 x)/(sinx
答
y=tanx-1/tanx
=[(sinx-cosx)/cosx]/(sinx/cosx)
=(sinx-cosx)/sinx
=1-cotx
最小正周期=π
答
y=-1/tanx+tanx
=-cosx/sinx+sinx/cosx
=-(cos^2 x-sin^2 x)/(sinxcosx)
=-2cos2x/sin2x
=-2cot2x
T=π/2