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
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