一共5小题,不要只给公式,全答出来会加赏10分.急,

问题描述:

一共5小题,不要只给公式,全答出来会加赏10分.急,
1,)sin164°sin224°+sin254°sin314°
2.)sin(α+β)cos(γ-β)-cos(β+α)sin(β-γ)
3.)sin(α-β)sin(β-γ)-cos(α-β)cos(γ-β)
4.) (tan4分之5π+tan12分之5π)除以(1-tan12分之5π)
5.) [sin(α+β)-2sinαcosβ]除以[2sinαsinβ=cosα+β]
1个回答 分类:数学 2014-11-29

问题解答:

我来补答
1,)sin164°sin224°+sin254°sin314°
=sin164°sin224°+sin(164°+90°)sin(224°+90°)
=sin164°sin224°+cos164°cos224°
=cos(224°-164°)=cos60°=1/2
2.)sin(α+β)cos(γ-β)-cos(β+α)sin(β-γ)
=sin(α+β)cos(γ-β)+cos(α+β)sin(γ-β)
=sin(α+β+γ-β)=sin(α+γ)
3.)sin(α-β)sin(β-γ)-cos(α-β)cos(γ-β)
=-sin(α-β)sin(γ-β)-cos(α-β)cos(γ-β)
=-[sin(α-β)sin(γ-β)+cos(α-β)cos(γ-β)]
=-[cos(α-β)cos(γ-β)+sin(α-β)sin(γ-β)]
=-cos(α-β-(γ-β))=-cos(α-γ)
4.) (tan5π/4+tan5π/12)/(1-tan5π/12)
=(tanπ/4+tan5π/12)/(1-tanπ/4tan5π/12)
=tan(π/4+π/12)=tan4π/12=tanπ/3=√3
5.) [sin(α+β)-2sinαcosβ]/[2sinαsinβ+cos(α+β)]
=[sinαcosβ+cosαsinβ-2sinαcosβ]/[2sinαsinβ+cosαcosβ-sinαsinβ]
=[cosαsinβ-sinαcosβ]/[sinαsinβ+cosαcosβ]
=-sin(α-β)/cos(α-β)=-tan(α-β)
 
 
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