证明:
左边=2sin(П+θ)cosθ-1/1-2sin^2θ
=(-2sinθcosθ-1)/cos2θ
=-(2sinθcosθ+sin^2 θ+cos^2 θ)/(cos^2 θ-sin^2 θ)
=-(sinθ+cosθ)^2/(cosθ-sinθ)(cosθ+sinθ)
=-(sinθ+cosθ)/(cosθ-sinθ)
=-[(sinθ/cosθ)+1]/[1-(sinθ/cosθ)]
=-(tanθ+1)/(1-tanθ)
=(tanθ+1)/(tanθ-1)
右边=tan(9П+θ)-1/tan(П+θ)+1
=(tanθ-1)/(tanθ+1)
证明 2sin(П+θ)cosθ-1/1-2sin^2θ =tan(9П+θ)-1/tan(П+θ)+1
证明 2sin(П+θ)cosθ-1/1-2sin^2θ =tan(9П+θ)-1/tan(П+θ)+1
数学人气:505 ℃时间:2019-10-23 11:47:41
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