函数y=arccos(sinx),x∈[-π/6,2π/3]的域值
当-π/6≦x≦2π/3时,-1/2≦sinx≦1;
因为y=arccos(sinx),故sinx=cosy,即有-1/2≦cosy≦1,故0≦y≦2π/3.这就是值域.