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来源：学生作业帮助网 编辑：作业帮 时间：2024/12/03 02:05:10
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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))