求证:[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1
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求证:[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1
求证:[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1
求证:[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ=[tan(9π+θ)+1]/tanθ-1
左式=[2sin(θ+p/2)cos(θ+p/2)-1]/[1-2(sinθ)^2]=[sin(p+2θ)-1]/[1-2(sinθ)^2]
=-(sin2θ+1)/cos2θ=-(cosθ+sinθ)^2/[(cosθ-sinθ)(cosθ+sinθ)]
=-(cosθ+sinθ)/(cosθ-sinθ)=-(1+tanθ)/(1-tanθ)
右式=(tanθ+1)/(tanθ-1)=-(tanθ+1)/(1-tanθ)
所以:左式=右式
其中:p表示派
[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ
=(-2scosθinθ-1)/cos2θ
再利用万能公式,这样比较直观
得到=(-2tanθ-1-tan^2θ)/(1-tan^2θ)=(2tanθ+1+tan^2θ)/(tan^2θ-1)
[tan(9π+θ)+1]/tanθ-1
=(tanθ+1)/(tanθ-1 )...
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[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ
=(-2scosθinθ-1)/cos2θ
再利用万能公式,这样比较直观
得到=(-2tanθ-1-tan^2θ)/(1-tan^2θ)=(2tanθ+1+tan^2θ)/(tan^2θ-1)
[tan(9π+θ)+1]/tanθ-1
=(tanθ+1)/(tanθ-1 )=(2tanθ+1+tan^2θ)/(tan^2θ-1)
左边等于右边 得证
或者
[2sin(θ-3π/2)cos(θ+π/2)-1]/1-2sin^2 θ
=-(sin2θ+1)/cos2θ
=-(cosθ+sinθ)^2/[(cosθ-sinθ)(cosθ+sinθ)]
=-(cosθ+sinθ)/(cosθ-sinθ)
=(tanθ+1)/(tanθ-1)
同样可以
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