求证:(1/sin2θ)+(1/tan2θ)+(1/sinθ)=(1/(tanθ/2))
来源:学生作业帮 编辑:大师作文网作业帮 分类:综合作业 时间:2024/10/01 09:44:45
求证:(1/sin2θ)+(1/tan2θ)+(1/sinθ)=(1/(tanθ/2))
左边=1/sin2θ+cos2θ/sin2θ+1/sinθ
=1/sin2θ+(2cos²θ-1)/sin2θ+1/sinθ
=2cos²θ/(2sinθcosθ)+1/sinθ
=(cosθ+1)/sinθ
=[2cos²(θ/2)-1+1]/[2sin(θ/2)cos((θ/2)]
=2cos²(θ/2)/[2sin(θ/2)cos(θ/2)]
=cos(θ/2)/sin(θ/2)
=1/tan(θ/2)=右边
命题得证
=1/sin2θ+(2cos²θ-1)/sin2θ+1/sinθ
=2cos²θ/(2sinθcosθ)+1/sinθ
=(cosθ+1)/sinθ
=[2cos²(θ/2)-1+1]/[2sin(θ/2)cos((θ/2)]
=2cos²(θ/2)/[2sin(θ/2)cos(θ/2)]
=cos(θ/2)/sin(θ/2)
=1/tan(θ/2)=右边
命题得证
求证:(1-tanθ)/(1+tanθ)=(1-2sinθcosθ)/(cos2θ-sin2θ)
求证2(sin2α+1)/1+sinα+cosα=tanα+1
求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2
求证(sinθ+cosθ-1)(sinθ-cosθ+1)/sin2θ=tanθ/2
(sinθ+cosθ-1)(sinθ-cosθ+1)/sin2θ=tan2/θ
α,β在第二象限,且sinα=1/3,cosβ=-½,求sin2α,cos2β,tan(α-β),tan2β,
求证:2(sin2α+1)/1+sin2α+cos2α=tanα+1
若2sin(π/4+a)=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0.
若2sin(π/4+a)=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0
若2sin(π/4+θ)=sinθ+cosθ,2sin^2(β)=sin2θ,求证sin2α+1/2cos2β=0
必修四数学 求证(sin2θ+1)/(sin2θ+cos2θ+1)=1/2(tanθ+1)
三角恒等变换,急~求证(1-cosα+sinα )/(1+cosα+sinα)=tan(α/2) (3sin2α-4co