sin³θ-cos³θ>cosθ-sinθ,θ∈(0,2π),那么角θ的取值

sin³θ-cos³θ>cosθ-sinθ,θ∈(0,2π),那么角θ的取值
sin³θ-cos³θ>cosθ-sinθ,θ∈(0,2π),那么角θ的取值

sin³θ-cos³θ>cosθ-sinθ,θ∈(0,2π),那么角θ的取值
sin³θ-cos³θ=(sinθ-cosθ)(1+sinθcosθ)=(sinθ-cosθ)(1+1/2sin2θ)>cosθ-sinθ ,1+1/2sin2θ>0
sinθ-cosθ>cosθ-sinθ ,sinθ>cosθ
θ∈(π/4,5π/4)