sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/05/05 06:23:23
sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)
3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
2.( tan4分之5π+tan12分之5π)/(1-tan12分之5π)
3.[ sin(α+β)-2sinαcosβ]/2sinαsinβ+cos(α+β)
sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
=-[cos(α-β)cos(r-β)-sin(α-β)sin(β-r)]
=-[cos(α-β)cos(β-r)-sin(α-β)sin(β-r)]
=-[cos(α-β)+(β-r)]
=-cos(α-β+β-r)
=-cos(α-r)
( tan5π/4+tan5π/12)/(1-tan5π/12)
=[ tan(π+π/4)+tan5π/12]/(1-tan5π/12)
=[ tanπ/4+tan5π/12]/(1-tan5π/12)
=[ tanπ/4+tan5π/12]/(1-tanπ/4tan5π/12)
= tan(π/4+5π/12)
= tan(3π/4)
= tan(π-π/4)
=- tanπ/4
=-1
[ sin(α+β)-2sinαcosβ]/[2sinαsinβ+cos(α+β)]
= [ sinαcosβ+cosαsinβ-2sinαcosβ]/[2sinαsinβ+cosαcosβ-sinαsinβ]
= [ cosαsinβ-sinαcosβ]/[cosαcosβ+sinαsinβ]
=-[ sinαcosβ-cosαsinβ]/[cosαcosβ+sinαsinβ]
=- sin(α-β)/cos(α-β)
=-tan(α-β)
=-[cos(α-β)cos(r-β)-sin(α-β)sin(β-r)]
=-[cos(α-β)cos(β-r)-sin(α-β)sin(β-r)]
=-[cos(α-β)+(β-r)]
=-cos(α-β+β-r)
=-cos(α-r)
( tan5π/4+tan5π/12)/(1-tan5π/12)
=[ tan(π+π/4)+tan5π/12]/(1-tan5π/12)
=[ tanπ/4+tan5π/12]/(1-tan5π/12)
=[ tanπ/4+tan5π/12]/(1-tanπ/4tan5π/12)
= tan(π/4+5π/12)
= tan(3π/4)
= tan(π-π/4)
=- tanπ/4
=-1
[ sin(α+β)-2sinαcosβ]/[2sinαsinβ+cos(α+β)]
= [ sinαcosβ+cosαsinβ-2sinαcosβ]/[2sinαsinβ+cosαcosβ-sinαsinβ]
= [ cosαsinβ-sinαcosβ]/[cosαcosβ+sinαsinβ]
=-[ sinαcosβ-cosαsinβ]/[cosαcosβ+sinαsinβ]
=- sin(α-β)/cos(α-β)
=-tan(α-β)
化简:sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
sin(α+β).cos(r-β)-cos(β+α).sin(β-r)
化简:sin(α+β)cos(r-β)-cos(β+α)sin(β-r).
sin(α+β)cos(r-β)-cos(β+a)sin(β-r) 化简
存在α,β∈R,使cos(α+β)=cosα+sinβ,对吗
设α、β、γ∈R,且sinα+sinγ=sinβ,cosα+cosγ=cosβ,则α-β()
数学证明题,α = arc tan (2h / l ) R cos β = R cos α + h R sin β =
存在α,β属于R,使cos(α+β)=cosα+sinβ是假命题吗
已知sinα+sinβ+sinγ=0,cosα+cosβ+cosγ=0.求cos(β
sinα+sinβ=sinγ cosα+cosβ=cosγ 证明cos(α-γ)
已知a=(cosα,sinα),b=(cosβ,sinβ)(α,β∈R),若a=λb,则实数λ的值为