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

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

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Answer 1 · 2017-06-09T08:05:12

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

Explanation:

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

$- 1 \leq x^{2} + y^{2} - 2 \leq 1$ $-1 <= x^2+y^2-2 <= 1$

i.e. $x^{2} + y^{2} \geq 1$ $x^2+y^2 >=1$ and $x^{2} + y^{2} \leq 3$ $x^2+y^2 <=3$

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

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

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

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

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

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Explanation:

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