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人气:480 ℃ 时间:2019-11-04 09:59:31
解答
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...
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