# How do you find the domain & range for y=Arcsin x?

##### 1 Answer
Nov 23, 2015

Domain: $\left[- 1 , + 1\right]$
Range: $\left[- \frac{\pi}{2} , + \frac{\pi}{2}\right]$

#### Explanation:

The argument for $\arcsin \left(x\right)$ (i.e. $x$) must be a value which can be returned by the $\sin$ function;
since the $\sin$ function has a range of $\left[- 1 , + 1\right]$
$\arcsin$ has a domain of $\left[- 1 , + 1\right]$

The range of the $\arcsin$ function is a matter of definition.
In order for $\arcsin$ to be a function its range is defined to be $\left[- \frac{\pi}{2} , + \frac{\pi}{2}\right]$; without this defined restriction $\arcsin$ would generate multiple values for a single argument and therefore would not be a function.

For example, noting that $\sin \left(0\right) = 0$ and $\sin \left(\pi\right) = 0$, it might seem reasonable that $\arcsin \left(0\right) = 0$ and $\arcsin \left(0\right) = \pi$ but this would be contrary to the definition of a function which requires every argument to map into only one value.