已知cosθ=-根号2/3,θ∈(π/2,π),求2/sin2θ－cosθ/sinθ
问题描述：
已知cosθ=-根号2/3,θ∈(π/2,π),求2/sin2θ－cosθ/sinθ
答
cosθ=-根号2/3,θ∈(π/2,π),
所以
sinθ=根号【1-cos²θ】=根号5/3
所以
2/sin2θ－cosθ/sinθ
=1/sinθcosθ-cosθ/sinθ
=[1-cos²θ]/sinθcosθ
=sinθ/cosθ
=-根号5/根号2
=-根号10/2