# inverse trigonometric functions

The inverse trigonometric functions are , and . These functions perform the reverse operations to the original trigonometric functions , and respectively. Recall that a function is invertible if it is one-to-one. Click here to revise inverse functions. Hence, before we can sketch the graphs of the inverse trigonometric functions, we must choose a domain for them for which they are one-to-one. Note that the original trigonometric functions work on angles and so each of the inverse trigonometric functions will return an angle. We use radians for all angles in the following – see more on radians. Also note that we use instead of , for example, as this can be confused with , the reciprocal trigonometric function.Â

## Inverse Trigonometric Function: arcsin(x)

Since is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be . The range of is .  It follows that the domain of is and the range is . The graphs of and are reflections of one another in the line .

## Inverse Trigonometric Function: arccos(x)

Since is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be . The range of is .  It follows that the domain of is and the range is . The graphs of and are reflections of one another in the line .

## Inverse Trigonometric Function: arctan(x)

Since is periodic, there are infinitely many regions for which it is one-to-one. We choose the default domain to be . The range of is all of .  It follows that the domain of is and the range is . The graphs of and are reflections of one another in the line .

## Examples

Find the values (in radians) of the following without using a calculator:

Solution:

1. We can see from the graph or use the fact that on this interval to find .
2. and are inverses of one another and so the result is .
3. and . Hence

Sketch the graph of , stating its domain and range.

Solution:

Recall from the transformations page that replacing with reflects the original graph across the -axis. Multiplying by 2 stretches the graph by a factor of 2 in the -direction. It follows that the graph of is given by:

The domain of is and the range is .

Given that where is an acute angle, find an expression for in terms of .

Solution:

If then . Using (see more trigonometric identities), and so . It follows that . Note that is between and (for acute angles it is actually between 0 and 1) and so . It follows that in the expression for we are square rooting a nonnegative number, as we would hope. However, for acute , must be positive and so .