已知fx=(-2cos2x)/(1+tanx )+1求的最小正周期

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fx=(-2cos2x)/(1+tanx )+1
= -2（cos²x-sin²x)/[(cosx+sinx )/cosx] +1
= -2(cosx+sinx)（cosx-sinx)/[(cosx+sinx )/cosx] +1
= -2cosx（cosx-sinx) +1
=1- 2cos²x+2sinxcosx
= - cos2x+sin2x
= √2sin(2x-π/4)
最小正周期2π/2 = π


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