# What is the domain and range of y = arcsin x?

Sep 2, 2016

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

Domain: $\left[- 1 , 1\right]$

#### Explanation:

The following is a fragment from my lecture about $y = \arcsin x$ presented on UNIZOR.COM. If you go to this very useful Web site, click Trigonometry - Inverse Trigonometric Functions - y=arcsin(x).

The original sine function defined for any real argument does not have an inverse function because it does not establish a one-to-one correspondence between its domain and a range.

To be able to define an inverse function, we have to reduce the original definition of a sine function to an interval where this correspondence does take place. Any interval where sine is monotonic and takes all values in its range would fit this purpose.

For a function $y = \sin \left(x\right)$ an interval of monotonic behavior is usually chosen as [−π/2,π/2], where the function is monotonously increasing from −1 to $1$.

This variant of a sine function, reduced to an interval where it is monotonous and fills an entire range, has an inverse function called $y = \arcsin \left(x\right)$.

It has range [−π/2,π/2] and domain from $- 1$ to $1$.