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1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ

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解决时间 2021-01-25 22:17
  • 提问者网友:暗中人
  • 2021-01-25 05:16
1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ
最佳答案
  • 五星知识达人网友:舊物识亽
  • 2021-01-25 06:27
(tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)=(-tanθ)(-sinθ)cosθ/(-cosθ)(-sinθ)=tanθsinθcosθ/cosθsinθ=tanθsin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α)=sin(8π-π/2+α)cos(π/2-α)/sin(4π+π/2-α)cos(3π/2+α)=sin(-π/2+α)cos(π/2-α)/sin(π/2-α)cos(3π/2+α)=[-sin(π/2-α)]cos(π/2-α)/sin(π/2-α)cos(3π/2+α)=[-cosα]sinα/cosαsinα=-1[sin(π/2+θ)-cos(π-θ)]/[sin(π/2-θ)-sin(π-θ)]=(cosθ+cosθ)/(cosθ-sinθ)=2cosθ/(cosθ-sinθ)(分子分母同时除以cosθ)=(2cosθ/cosθ)/(cosθ/cosθ-sinθ/cosθ)=2/(1-tanθ)======以下答案可供参考======供参考答案1:1.tan(2π-b)sin(-2π-b)cos(6π-b)/(cos(b-π)sin(5π+b))=tan(-b)sin(-b)cos(-b)/(cos(π-b)sin(4π+π+b))=sin(-b)/cos(-b).sin(-b)cos(-b)/(-cosb.(-sinb))=sinb.sinb/(cosbsinb)=tanb供参考答案2:tan(2π-θ)=-tanθ,sin(-2π-θ)=-sinθ,cos(6π-θ)=cosθ,cos(θ-π)=-cosθ,sin(5π+θ)=-sinθ.所以(tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)=-tanθ
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  • 1楼网友:你哪知我潦倒为你
  • 2021-01-25 07:19
这个答案应该是对的
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