Question

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

Answer 1

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

Using Domain of `arc sin x`

Find `arc sin (3)`.

`3` is not in the domain of `arc sin`, (`3` is not in the range of `sin`) so `arc sin (3)` does not exist.

Using Range of `arc sin x`

Find `arc sin (1/2)`.

Although there are infinitely many `t` with `sin t = 1/2`, the range of `arc sin` restricts the value to those `t` with `(-pi)/2 <= t <= pi/2`, So the value we want is `pi/6`.

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

Find `arc sin (1/2)`.

`arc sin (1/2)` 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, `arc sin (1/2) = pi/6`.