2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/19 14:27:42
2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简2除以s
2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简
2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简
2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简
2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)
=2/sin2θ+cos(π+θ)/cos(π/2-θ)
=2/sin2θ-cosθ/cos(π/2-θ)
=2/sin2θ-cosθ/sinθ
=2/sin2θ-2cosθcosθ/2sinθcosθ
=2/sin2θ-2cos^2θ/sin2θ
=(2-2cos^2θ)/sin2θ
=2(1-cos^2θ)/sin2θ
=2sin^2θ/sin2θ
=2sin^2θ/2sinθcosθ
=sinθ/cosθ
=tanθ
(2/sin2θ)+[cos(π+θ)/cos(π/2-θ)]化简
原式=(1/sinθcosθ)+(-cosθ/sinθ)=(1-cos²θ)/sinθcosθ=sin²θ/sinθcosθ=sinθ/cosθ=tanθ
2除以sin2θ+cos(π+θ)除以cos(π除以2-θ)化简
设θ为第二象限角,则根号下1-sin2θ除以cosθ-sinθ的值是多少?
cosθ>0 ,sin2θ
sin2θ>0,cosθ
求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以(1+2sinθcosθ)
【(sinα+cosα-1) (sinα-cosα+1)】除以sin2α
根号下1减sin(π+2)cos(π+2)化简是多少啊A sin2-cos2 B cos2-sin2 C +-(sin2-cos2) D sin2+cos2
求证(2cosθ-sin2θ)/(2cosθ+sin2θ)=tan(π/4-θ/2)右边应该是tan(π/4-θ/2)^2
sin2除以sin1结果是2cos1怎么求的
证明2sinθcosθ=sin2θ.
为什么2cosθ*sinθ=sin2θ?
Θ∈(π/2,3π/4) sin2Θ∈a sinΘ+cosΘ
已知tan(π÷4+θ)=3,则sin2θ-2cos²θ=
若cosθ>0 且sin2θ
若cosθ>0,且sin2θ
sinθ除以(cosθ-1)化简
cosθ除以sinθ等于多少?我知道sinθ除以cosθ等于tanθ
若tan(6+π/4)=2 则cos²θ+1/2sin2θ=?A -6/5 B -4/5 C 4/5 D 6/5