已知tanθ=2,则sinθ/(sin3θ-cos3θ)=

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已知tanθ=2,则sinθ/(sin3θ-cos3θ)=
已知tanθ=2,则sinθ/(sin3θ-cos3θ)=

已知tanθ=2,则sinθ/(sin3θ-cos3θ)=
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

10/21

sinθ/(sin3θ-cos3θ)
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
θ在第一象限
tanθ=2
sinθ=2√5/5
cos=√5/5
sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=1/[(1-cotθ)(1+2/5)]
=1/(1/2+7/5)
=10/(5+14)

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sinθ/(sin3θ-cos3θ)
=sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
θ在第一象限
tanθ=2
sinθ=2√5/5
cos=√5/5
sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=1/[(1-cotθ)(1+2/5)]
=1/(1/2+7/5)
=10/(5+14)
=10/19
θ在第二象限
tanθ=2
sinθ=-2√5/5
cos=-√5/5
sinθ/[(sinθ-cosθ)(1+sinθcosθ)]
=1/[(1-cotθ)(1+2/5)]
=1/(1/2+7/5)
=10/(5+14)
=10/19

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