Question

如果sin³θ-cos³θ>cosθ-sinθ，且θ∈（0,2π），那么角θ的取值范围是

Answer 1

解：
(sinθ)^3-(cosθ)^3
=(sinθ-cosθ)((sinθ)^2+sinθ*cosθ+(cosθ)^2)
=(sinθ-cosθ)(1+sinθ*cosθ)
(sinθ)^3-(cosθ)^3-(cosθ-sinθ)
=(sinθ-cosθ)(2+sinθ*cosθ)>0

因为2+sinθ*cosθ=2+sin2θ/2>0
所以sinθ-cosθ>0
所以sin(θ-π/4)=sinθ*√2/2-cosθ*√2/2>0
θ∈（0,2π）
θ-π/4∈(-π/4,7π/4)
sin(θ-π/4)>0
0<θ-π/4<π
π/4<θ<5π/4
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
祝你学习进步，更上一层楼！
不明白请及时追问，满意敬请采纳，O(∩_∩)O谢谢~~

Answer 2

sin³θ-cos³θ>cosθ-sinθ
sin³θ-cos³θ+sinθ-cosθ>0
∴(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ+1)>0
∵sin²θ+sinθcosθ+cos²θ+1>0
∴sinθ-cosθ>0
∴π/4<θ<5π/4

Answer 3