# How do you apply the domain, range, and quadrants to evaluate inverse trigonometric functions?

##### 1 Answer
Mar 19, 2015

Throughout the following answer, I will assume that you are asking about trigonometry restricted to real numbers.

Using Domain of $a r c \sin x$

Find $a r c \sin \left(3\right)$.

$3$ is not in the domain of $a r c \sin$, ($3$ is not in the range of $\sin$) so $a r c \sin \left(3\right)$ does not exist.

Using Range of $a r c \sin x$

Find $a r c \sin \left(\frac{1}{2}\right)$.

Although there are infinitely many $t$ with $\sin t = \frac{1}{2}$, the range of $a r c \sin$ restricts the value to those $t$ with $\frac{- \pi}{2} \le t \le \frac{\pi}{2}$, So the value we want is $\frac{\pi}{6}$.

Using the quadrant
This is the same as using the range, but it involves thinking about the problem more geometrically.

Find $a r c \sin \left(\frac{1}{2}\right)$.

$a r c \sin \left(\frac{1}{2}\right)$ is a number (an angle) in quadrant I or IV. It is the $t$ with smallest absolute value (the shortest path from the initial side).

Again, $a r c \sin \left(\frac{1}{2}\right) = \frac{\pi}{6}$.