化简：1、sin(x-π/3)-cos(x+π/6)+√3cosx=?2、已知,sinα+sinβ=√2/2,求cosα+cosβ的取值范围

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化简：1、sin(x-π/3)-cos(x+π/6)+√3cosx=?2、已知,sinα+sinβ=√2/2,求cosα+cosβ的取值范围化简：1、sin(x-π/3)-cos(x+π/6)+√3c

化简：1、sin(x-π/3)-cos(x+π/6)+√3cosx=?2、已知,sinα+sinβ=√2/2,求cosα+cosβ的取值范围
化简：1、sin(x-π/3)-cos(x+π/6)+√3cosx=?2、已知,sinα+sinβ=√2/2,求cosα+cosβ的取值范围

化简：1、sin(x-π/3)-cos(x+π/6)+√3cosx=?2、已知,sinα+sinβ=√2/2,求cosα+cosβ的取值范围
sin(x-π/3)-cos(x+π/6)+√3cosx
=sinxcosπ/3-cosxsinπ/3-cosxcosπ/6+sinxsinπ/6+√3cosx
=1/2*sinx-√3/2*cosx-√3/2*cosx+1/2*sinx+√3cosx
=sinx
sinα+sinβ=√2/2,求cosα+cosβ的取值范围
sinα+sinβ=√2/2(两边平方)
(sinα)^2+(sinβ)^2+2sinαsinβ=1/2.1
令cosα+cosβ=k(两边平方)
(cosα)^2+(cosβ)^2+2cosαcosβ=k^2.2
1式+2式得
所以2+2(cosαcosβ+sinαsinβ)=k^2+1/2
2(cosαcosβ+sinαsinβ)=k^2-3/2
2cos(α-β)=k^2-3/2
cos(α-β)=k^2/2-3/4
-1

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1、sin(x-π/3)-cos(x+π/6)+√3cosx
原式=sinxcos(π/3)-cosxsin(π/3)-[cosxcos(π/6)-sinxsin(π/6)+√3cosx
=1/2sinx-√3/2cosx-√3/2cosx+1/2sinx+√3cosx
=sinx
2、已知，sinα+sinβ=√2/2,求cosα+cosβ的取值范围
因为(sinα+sinβ)^2+(cosα+cosβ)^2
=(sinα)^2+(sinβ)^2+2sinαsinβ+(cosα)^2+(cosβ)^2+2cosαcosβ
=2+2cos(α-β)
又因为sinα+sinβ=√2/2，所以
0≤(cosα+cosβ)^2≤4-2√2
-√(4-2√2)≤cosα+cosβ≤√(4-2√2)

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