求y=4/3x+√(9-x^2)的值域∵9-x^2≥0∴-3≤x≤3令x=3sina,a∈[-π/2,π/2]∴y=3cosa+4sina
问题描述：
求y=4/3x+√(9-x^2)的值域
∵9-x^2≥0
∴-3≤x≤3
令x=3sina,a∈[-π/2,π/2]
∴y=3cosa+4sina
答
y=f(a)=3cosa+4sina,a∈[-π/2,π/2]
设Sinm=3/5,Cosm=4/5,则tanm=3/4
y=f(a)=5Sin(a+m)
f(a)最小值为f(-π/2)=-4
f(a)最大值为5
则值域为[-4,5]