已知tanθ=2,则sinθ/(sin3θ-cos3θ)=A.10/7B.7/10C.9/7D9/10还有一种不用立方差的方法:∵ tanθ=2∴ sinθ=2cosθ,代入sin²θ+cos²θ=15cos²θ=1∴ cos²θ=1/5 ∴ sinθ/(sin³θ-cos³θ)=2cosθ/(8cos³
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![已知tanθ=2,则sinθ/(sin3θ-cos3θ)=A.10/7B.7/10C.9/7D9/10还有一种不用立方差的方法:∵ tanθ=2∴ sinθ=2cosθ,代入sin²θ+cos²θ=15cos²θ=1∴ cos²θ=1/5 ∴ sinθ/(sin³θ-cos³θ)=2cosθ/(8cos³](/uploads/image/z/1990269-45-9.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B8%3D2%2C%E5%88%99sin%CE%B8%2F%28sin3%CE%B8-cos3%CE%B8%29%3DA.10%2F7B.7%2F10C.9%2F7D9%2F10%E8%BF%98%E6%9C%89%E4%B8%80%E7%A7%8D%E4%B8%8D%E7%94%A8%E7%AB%8B%E6%96%B9%E5%B7%AE%E7%9A%84%E6%96%B9%E6%B3%95%EF%BC%9A%E2%88%B5+tan%CE%B8%3D2%E2%88%B4+sin%CE%B8%3D2cos%CE%B8%2C%E4%BB%A3%E5%85%A5sin%26%23178%3B%CE%B8%2Bcos%26%23178%3B%CE%B8%3D15cos%26%23178%3B%CE%B8%3D1%E2%88%B4+cos%26%23178%3B%CE%B8%3D1%2F5+%E2%88%B4+sin%CE%B8%2F%28sin%26%23179%3B%CE%B8-cos%26%23179%3B%CE%B8%29%3D2cos%CE%B8%2F%EF%BC%888cos%26%23179%3B)
已知tanθ=2,则sinθ/(sin3θ-cos3θ)=A.10/7B.7/10C.9/7D9/10还有一种不用立方差的方法:∵ tanθ=2∴ sinθ=2cosθ,代入sin²θ+cos²θ=15cos²θ=1∴ cos²θ=1/5 ∴ sinθ/(sin³θ-cos³θ)=2cosθ/(8cos³
已知tanθ=2,则sinθ/(sin3θ-cos3θ)=
A.10/7
B.7/10
C.9/7
D9/10
还有一种不用立方差的方法:
∵ tanθ=2
∴ sinθ=2cosθ,代入sin²θ+cos²θ=15cos²θ=1
∴ cos²θ=1/5
∴ sinθ/(sin³θ-cos³θ)
=2cosθ/(8cos³θ-cos³θ)
=2cosθ/7cos³θ
=2/cos²θ
将cos²θ=1/5 即可得=10/7
已知tanθ=2,则sinθ/(sin3θ-cos3θ)=A.10/7B.7/10C.9/7D9/10还有一种不用立方差的方法:∵ tanθ=2∴ sinθ=2cosθ,代入sin²θ+cos²θ=15cos²θ=1∴ cos²θ=1/5 ∴ sinθ/(sin³θ-cos³θ)=2cosθ/(8cos³
∵ tanθ=2
∴ sinθ=2cosθ,
代入sin²θ+cos²θ=1
5cos²θ=1
∴ cos²θ=1/5
∴ sinθ/(sin³θ-cos³θ)
=sinθ/[(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=2cosθ/[(2cosθ-cosθ)(1+2cosθcosθ)]
=2cosθ/[cosθ(1+2cos²θ)]
=2/(1+2cos²θ)
=2/(1+2/5)
=2/(7/5)
=10/7
tanθ=sinθ/cosθ=2
sinθ=2cosθ
sinθ/(sin^3θ-cos^3θ)
=sinθ/[(sinθ-cosθ)(sin^2θ+sinθcosθ+cos^2θ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=2cosθ/[(2cosθ-cosθ...
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tanθ=sinθ/cosθ=2
sinθ=2cosθ
sinθ/(sin^3θ-cos^3θ)
=sinθ/[(sinθ-cosθ)(sin^2θ+sinθcosθ+cos^2θ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=2cosθ/[(2cosθ-cosθ)(1+2cosθcosθ)]
=2cosθ/[cosθ(1+2cosθcosθ)]
=2/(2+2cos2θ)
=1/(1+cos2θ)
=1/[1+(1-tan^2θ)/(1+tan^2θ)]
=1/[1+(1-2^2)/(1+2^2)]
=1/[1-3/5]
=1/[2/5]
=5/2不懂追问,希望采纳
我这个答案肯定真确的
相信我
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