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化简:(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+cos2αcos2β

来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/16 00:34:37
化简:(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+cos2αcos2β
化简:(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+cos2αcos2β
sinα=(1-cos2α)/2
(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+cos2αcos2β
=(1-cos2α)/2+(1-cos2β)/2-(1-cos2α)/2*(1-cos2α)/2+cos2αcos2β
=(1+cos2αcos2β)*0.75
化简:sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β 化简(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β 化简sinα^2sinβ^2+cos^2cosβ^2－1/2cos2αcos2β 化简：sin²αsin²β+cos²αcos²β-1/2cos2αcos2β 化简：sin²αsin²β+cos²αcos²β—1/2cos2αcos2β 化简 sin²αsin²β+cos²αcos²β-1/2（cos2αcos2β） 已知sinθ+cosθ=2sinα,sinθcosθ=(sinβ)^2,求证4(cos2α)^2=(cos2β)^2 2sinα=sinθ+cosθ,sin²β==sinθcosθ.求证cos2β=2cos2α=2cos 已知sinθ+cosθ=2sinα,sinθ·cosθ=sin²β,求证：2cos2α=cos2β. 三角数列题：sinθ sinα cosθ成等差数列,sinθ sinβ cosθ为等比数列,求证2COS2α=cos2β 化解sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β Sin²αSin²β+Cos²αCos²β-1/2Cos2αCos2β