求y=arcsin(sinx),(-π/3≤x≤2π/3)的值域
问题描述：
求y=arcsin(sinx),(-π/3≤x≤2π/3)的值域
答
-π/3≤x≤2π/3
则sin(-π/3)≤sinx≤sin(π/2)
即-√3/2≤sinx≤1
所以arcsin(-√3/2≤arcsin(sinx)≤arcsin1
-π/3≤arcsin(sinx)≤π/2