Question

求函数y=sinx/(2-cosx)的值域

Answer 1

y=sinx/(2-cosx)
y'=cosx/(2-cosx)-(sinx)^2/(2-cosx)^2
=(2cosx-1)/(2-cosx)^2
令y'=0 得　cosx=1/2
当co sx=1/2 sinx=√3/2 时　y有最大值　　ymax=√3/2/(2-1/2)=√3/3
当co sx=1/2 sinx=-√3/2 时　y有最小值　　ymax=-√3/2/(2-1/2)=-√3/3
所以　函数y=sinx/(2-cosx)的值域为　［-√3/3，√3/3］

Answer 2

y=sinx/(2-cosx)
(2-cosx)y=sinx
2y-ycosx=sinx
2y=sinx+ycosx
2y=√(y^2+1)sin(x+ψ)
sin(x+ψ)=2y/√(y^2+1)∈[-1,1]
-1≤2y/√(y^2+1)≤1
4y^2/(y^2+1)≤1
4y^2≤y^2+1
-√3/3≤y≤√3/3