证明 2sin(П+θ)cosθ-1/1-2sin^2θ =tan(9П+θ)-1/tan(П+θ)+1
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/14 23:35:00
证明 2sin(П+θ)cosθ-1/1-2sin^2θ =tan(9П+θ)-1/tan(П+θ)+1
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)
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1-2cos^2θ/tanθ-cotθ=sinθcosθ
证明下列恒等式(sinθ+cosθ)/(1-tan^2θ)+sin^2θ/(sinθ-cosθ)=sinθ+cosθ
证明下列恒等式tan^2 θ *(1-sinθ)/(1+cosθ)=(1-cosθ)/(1+sinθ)
已知tanθ=根号2,求(1)(cosθ+sinθ)/(cosθ-sinθ);(2)sin²θ-sinθcos
高一三角函数证明题 已知sin^2 (α)/sin^2( β)+cos^2(θ)=1 求证tan^2 (α)=sin^2
已知sinθ+cosθ=((√3)+1)/2,求(sinθ)/(1-(1/tanθ))+(cosθ)/(1-tanθ)的
sin^2θ/sinθ-cosθ + cosθ/1-tanθ = sin^2θ/sinθ-cosθ + cosθ/1-(
tanθ=2,sinθ平方+2sinθcosθ+1的值
证明:1、 tan(2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ+π)sin(5π+θ)=tanθ
证明tanα/2=1-cosɑ/sinɑ
证明tan a/2=sin a/(1+cos a)