已知0
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 05:44:21
![已知0](/uploads/image/z/11468530-10-0.jpg?t=%E5%B7%B2%E7%9F%A50)
已知0
已知0
已知0
f(θ)=[2sin^2(θ)+1]/sin2θ
=[3sin^2(θ)+cos^2(θ)]/sin2θ
=3sin^2(θ)/sin2θ+cos^2(θ)/2sin(2θ)
=3sin^2(θ)/2sinθcosθ+cos^2(θ)/2sinθcosθ
=3sinθ/2cosθ+cosθ/2sinθ
>=√3
所以最小值为√3
f(θ)
=[2sin^2(θ)+1]/sin2θ
=[3sin^2(θ)+cos^2(θ)]/sin2θ
=3sin^2(θ)/2cos^2(θ)+cos(θ)/sin(θ)
>=√3√(sin(θ))