证明:1、 tan(2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ+π)sin(5π+θ)=tanθ
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证明:1、 tan(2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ+π)sin(5π+θ)=tanθ
2、 2sin(π+θ)×cosθ-1/1-2(sinθ平方)=tan(9π+θ)-1/tan(π+θ)+1
2、 2sin(π+θ)×cosθ-1/1-2(sinθ平方)=tan(9π+θ)-1/tan(π+θ)+1
![证明:1、 tan(2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ+π)sin(5π+θ)=tanθ](/uploads/image/z/5682704-32-4.jpg?t=%E8%AF%81%E6%98%8E%EF%BC%9A1%E3%80%81+tan%EF%BC%882%CF%80-%CE%B8%EF%BC%89sin%EF%BC%88-2%CF%80-%CE%B8%EF%BC%89cos%EF%BC%886%CF%80-%CE%B8%EF%BC%89%EF%BC%8Fcos%EF%BC%88%CE%B8%2B%CF%80%EF%BC%89sin%EF%BC%885%CF%80%2B%CE%B8%EF%BC%89%3Dtan%CE%B8)
1、
tan(2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ+π)sin(5π+θ)
=(-tanθ)(-sinθ)cosθ)/(-cosθ)(-sinθ)
=tanθsinθcosθ/sinθcosθ
=tanθ
命题得证
2、
2sin(π+θ)×cosθ-1/1-2(sinθ平方
=(-2sinθcosθ-1)/[1-2(1-cos2θ)/2]
=-(sin²θ+cos²θ+2sinθcosθ)/(cos2θ)
=-+(sinθ+cosθ)²/(cos²θ-sin²θ)
=-(sinθ+cosθ)²/(cosθ+siθ)(cosθ-sinθ)
=(sinθ+cosθ)/(sinθ-cosθ)
上下除以cosθ
=(tanθ+1)/(tanθ-1)
右边=(tanθ-1)/(tanθ+1)
题目写错了吧
tan(2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ+π)sin(5π+θ)
=(-tanθ)(-sinθ)cosθ)/(-cosθ)(-sinθ)
=tanθsinθcosθ/sinθcosθ
=tanθ
命题得证
2、
2sin(π+θ)×cosθ-1/1-2(sinθ平方
=(-2sinθcosθ-1)/[1-2(1-cos2θ)/2]
=-(sin²θ+cos²θ+2sinθcosθ)/(cos2θ)
=-+(sinθ+cosθ)²/(cos²θ-sin²θ)
=-(sinθ+cosθ)²/(cosθ+siθ)(cosθ-sinθ)
=(sinθ+cosθ)/(sinθ-cosθ)
上下除以cosθ
=(tanθ+1)/(tanθ-1)
右边=(tanθ-1)/(tanθ+1)
题目写错了吧
证明:1、 tan(2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ+π)sin(5π+θ)=tanθ
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