1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)2、化简 sin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α) 3、求证 sin(π/2+θ)-cos(π-θ)/sin(π/2-θ)-sin(π-θ)=2

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/29 20:38:06
1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)2、化简 sin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α) 3、求证 sin(π/2+θ)-cos(π-θ)/sin(π/2-θ)-sin(π-θ)=2

1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)2、化简 sin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α) 3、求证 sin(π/2+θ)-cos(π-θ)/sin(π/2-θ)-sin(π-θ)=2
1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)
2、化简 sin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α)
3、求证 sin(π/2+θ)-cos(π-θ)/sin(π/2-θ)-sin(π-θ)=2/1-tanθ

1、化简 (tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)2、化简 sin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α) 3、求证 sin(π/2+θ)-cos(π-θ)/sin(π/2-θ)-sin(π-θ)=2
(tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)
=(-tanθ)(-sinθ)cosθ/(-cosθ)(-sinθ)
=tanθsinθcosθ/cosθsinθ
=tanθ
sin(15π/2+α)cos(α-π/2)/sin(9π/2-α)cos(3π/2+α)
=sin(8π-π/2+α)cos(π/2-α)/sin(4π+π/2-α)cos(3π/2+α)
=sin(-π/2+α)cos(π/2-α)/sin(π/2-α)cos(3π/2+α)
=[-sin(π/2-α)]cos(π/2-α)/sin(π/2-α)cos(3π/2+α)
=[-cosα]sinα/cosαsinα
=-1
[sin(π/2+θ)-cos(π-θ)]/[sin(π/2-θ)-sin(π-θ)]
=(cosθ+cosθ)/(cosθ-sinθ)
=2cosθ/(cosθ-sinθ)(分子分母同时除以cosθ)
=(2cosθ/cosθ)/(cosθ/cosθ-sinθ/cosθ)
=2/(1-tanθ)

1.tan(2π-b)sin(-2π-b)cos(6π-b)/(cos(b-π)sin(5π+b))
=tan(-b)sin(-b)cos(-b)/(cos(π-b)sin(4π+π+b))
=sin(-b)/cos(-b).sin(-b)cos(-b)/(-cosb.(-sinb))
=sinb.sinb/(cosbsinb)
=tanb

tan(2π-θ)=-tanθ,sin(-2π-θ)=-sinθ,cos(6π-θ)=cosθ,cos(θ-π)=-cosθ,sin(5π+θ)=-sinθ.
所以(tan2π-θ)sin(-2π-θ)cos(6π-θ)/cos(θ-π)sin(5π+θ)=-tanθ