求|sinx|+cos值域

若sinx>=0
此时2kπ<=x<=2kπ+π
则y=sinx+cosx=√2sin(x+π/4)
2kπ+π/4<=x+π/4<=2kπ+5π/4
所以最小是sin(2kπ+5π/4)=-√2/2
最大是sin(2kπ+π/2)=1
所以-1<=y<=√2

若sinx<=0
此时2kπ-π<=x<=2kπ
则y=-sinx+cosx=-√2sin(x-π/4)
2kπ-5π/4<=x-π/4<=2kπ-π/4
所以最小是sin(2kπ-π/2)=-1
最大是sin(2kπ-5π/4)=√2/2
所以也是-1<=y<=√2

所以值域[-1,√2]

-1=0-1<=|sinx|+cosx<=|sinx|+|cosx|<=根号2