How do you find the domain and range of $arcsin(x^2+y^2-2)$ ?

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Answer 1 · 2017-06-09T08:05:12

Domain of $(x,y)$ is the ring between the circles $x^2+y^2=1$ and $x^2+y^2=3$ and range is from $-pi/2$ to $pi/2$ .

As the range for $sinx$ is $-1 <= x <= 1$

$-1 <= x^2+y^2-2 <= 1$

i.e. $x^2+y^2 >=1$ and $x^2+y^2 <=3$

Hence $x$ and $y$ can take values between $1$ and $3$

in other words domain of $(x,y)$ is the ring between the circles $x^2+y^2=1$ and $x^2+y^2=3$

graph{(x^2+y^2-1)(x^2+y^2-3)=0 [-5, 5, -2.5, 2.5]}

and range for $arcsin(x^2+y^2-2)$ is as that of any $arcsinx$ , that is between $-pi/2$ and $pi/2$

How do you find the domain and range of arcsin(x^2+y^2-2)arcsin(x^2+y^2-2)?