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)
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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)
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)
化简:sin(α+β)cos(r-β)-cos(β+α)sin(β-r).
化简:sin(α-β)sin(β-r)-cos(α-β)cos(r-β)
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
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)
sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β)
sin αcosβ+cos α sin β等于?
三角函数题:设α,β,γ∈R,则sinαcosβ+sinβcosγ+sinγcosα的最大值是——
设α,β∈R,A(cosα,sinα),B(cosβ,sinβ),则|AB|max=
存在α,β∈R,使cos(α+β)=cosα+sinβ,对吗
求证:sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
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求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
求证sin(2α+β)/sinα-2cos(α+β)=sinβ/sinα
证明:sin(2α+β)/sinα - 2cos(α+β)=sinβ/sinα
化简sin^2α+sin^2β+2sinαsinβcos(α+β)
β,γ∈R,则sinαcosβ+sinβcosγ+sinγcosα的最大值为设α,β,γ∈R,则sinαcosβ+sinβcosγ+sinγcosα的最大为______________
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