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

Jun 9, 2017

Domain of $\left(x , y\right)$ is the ring between the circles ${x}^{2} + {y}^{2} = 1$ and ${x}^{2} + {y}^{2} = 3$ and range is from $- \frac{\pi}{2}$ to $\frac{\pi}{2}$.

#### Explanation:

As the range for $\sin x$ is $- 1 \le x \le 1$

$- 1 \le {x}^{2} + {y}^{2} - 2 \le 1$

i.e. ${x}^{2} + {y}^{2} \ge 1$ and ${x}^{2} + {y}^{2} \le 3$

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

in other words domain of $\left(x , y\right)$ 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 $\arcsin \left({x}^{2} + {y}^{2} - 2\right)$ is as that of any $\arcsin x$, that is between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$