Question

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

要有详细的解题过程，谢谢

Answer 1

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<=cos(α-β)<=1
所以-1<=k^2/2-3/4<=1
-4<=2k^2-3<=4
-1<=2k^2<=7
-1/2<=k^2<=7/2
0<=k^2<=7/2
-√14/2<=k<=√14/2

所以-√14/2<=cosα+cosβ<=√14/2

Answer 2

1、sin(x-π/3)-cos(x+π/6)+√3cosx
解：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
解：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)