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Inverse Trigonometric Functions (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 24 April 2026

Arcsin, arccos & arctan

What are inverse trigonometric functions?

An inverse trigonometric function reverses the role of inputs and outputs compared to the corresponding trigonometric function
For a trigonometric function
- the input is an angle
- and the output is a numerical value
For an inverse trigonometric function
- the input is a numerical value
- and the output is an angle
For example, if $\sin \frac{π}{6} = \frac{1}{2}$
- then the inverse of sine will take an input of $\frac{1}{2}$
- and return and output of $\frac{π}{6}$

What are the three inverse trigonometric functions called?

The three inverse trigonometric functions are:
- Arcsine, the inverse of the sine function
  - The notation for this is $\sin^{- 1} x$ or $\arcsin x$
- Arccosine, the inverse of the cosine function
  - The notation for this is $\cos^{- 1} x$ or $arc \cos x$
- Arctangent, the inverse of the tangent function
  - The notation for this is $\tan^{- 1} x$ or $\arctan x$

Examiner Tips and Tricks

Be careful with the notation here!

The notation $\sin^{- 1} x$ means "the angle whose sine is $x$ "
It does not mean $\frac{1}{\sin x}$

Why do trigonometric functions need restricted domains to have inverses?

Because trigonometric functions are periodic, they are not one-to-one over their full domains
- More than one input value produces the same output value
- For example, $\sin \frac{π}{6} = \frac{1}{2}$ and $\sin \frac{5 π}{6} = \frac{1}{2}$
A function must be one-to-one (each output comes from exactly one input) in order to have an inverse
To define inverse trigonometric functions
- the domains of sine, cosine, and tangent are restricted to intervals where
  - each function is one-to-one
  - and takes on all of its possible output values exactly once

What are the standard domain restrictions?

Sine is restricted to the domain $[- \frac{π}{2}, \frac{π}{2}]$
- On this interval, sine is increasing
  - and takes every value from $- 1$ to $1$ exactly once
- Therefore $\sin^{- 1} x$ returns an angle in $[- \frac{π}{2}, \frac{π}{2}]$ as an output
- The domain of $\sin^{- 1}$ is $[- 1, 1]$

Graph of y equals sin x from -π/2 to π/2, showing a curve from -1 to 1 through the origin, with x and y axes labelled. — **Graph of y=sinx with restricted domain**

Graph of y equals arcsin x, showing a red curve from (-1, -π/2) to (1, π/2) on the Cartesian plane, with x and y axes labelled. — **Graph of y=arcsinx**

Cosine is restricted to the domain $[0, π]$
- On this interval, cosine is decreasing
  - and takes every value from $- 1$ to $1$ exactly once
- Therefore $\cos^{- 1} x$ returns an angle in $[0, π]$ as an output
- The domain of $\cos^{- 1}$ is $[- 1, 1]$

Graph of y = cos(x) with the x-axis ranging from 0 to π and the y-axis from -1 to 1, highlighting the curve's wave-like shape. — **Graph of y=cosx with restricted domain**

Graph of y=arccos(x) showing a curve from (1,0) through (0,π/2) to (-1,π), with x and y axes labelled. — **Graph of y=arccosx**

Tangent is restricted to the domain $(- \frac{π}{2}, \frac{π}{2})$
- On this interval, tangent is increasing
  - and takes every real number value exactly once
- Therefore $\tan^{- 1} x$ returns an angle in $(- \frac{π}{2}, \frac{π}{2})$
- The domain of $\tan^{- 1}$ is all real numbers

Graph of y = tan(x) with vertical asymptotes at -π/2 and π/2, passing through the origin, with x and y axes labeled. — **Graph of y=tan x with domain restriction**

Graph of y = arctan(x), showing an S-shaped curve centred at the origin. Asymptotes at y = π/2 and y = -π/2 are indicated by dashed lines. — **Graph of y=arctanx**

These domain restrictions, along with the domain and range of the inverse functions, are summarised in the table below

Function	Restricted domain	Range of inverse	Domain of inverse
$\sin$	$[- \frac{π}{2}, \frac{π}{2}]$	$[- \frac{π}{2}, \frac{π}{2}]$	$[- 1, 1]$
$\cos$	$[0, π]$	$[0, π]$	$[- 1, 1]$
$\tan$	$(- \frac{π}{2}, \frac{π}{2})$	$(- \frac{π}{2}, \frac{π}{2})$	All real numbers

Examiner Tips and Tricks

Both forms of notation for the inverse functions appear on the exam

$\sin^{- 1} x, \cos^{- 1} x$ and $\tan^{- 1} x$
as well as $\arcsin x, \arccos x$ and $\arctan x$

Make sure you are comfortable with both forms of notation.

How can inverse trigonometric functions be evaluated?

To evaluate an expression like $\sin^{- 1} (\frac{\sqrt{3}}{2})$ , ask: "What angle in $[- \frac{π}{2}, \frac{π}{2}]$ has sine equal to $\frac{\sqrt{3}}{2}$ ?"
- From the unit circle, $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$
  - and $\frac{π}{3}$ is in $[- \frac{π}{2}, \frac{π}{2}]$
- So $\sin^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3}$
To evaluate $\cos^{- 1} (- \frac{1}{2})$ , ask: "What angle in $[0, π]$ has cosine equal to $- \frac{1}{2}$ ?"
- $\cos \frac{2 π}{3} = - \frac{1}{2}$
  - and $\frac{2 π}{3}$ is in $[0, π]$
- So $\cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$
To evaluate $\tan^{- 1} (1)$ , ask: "What angle in $(- \frac{π}{2}, \frac{π}{2})$ has tangent equal to $1$ ?"
- $\tan \frac{π}{4} = 1$
  - and $\frac{π}{4}$ is in $(- \frac{π}{2}, \frac{π}{2})$
- So $\tan^{- 1} (1) = \frac{π}{4}$
Be careful with negative inputs
- the restricted domain determines which angle is returned
- E.g. $\sin^{- 1} (- \frac{\sqrt{2}}{2}) = - \frac{π}{4}$
  - not $\frac{5 π}{4}$ , which is outside the restricted domain

Examiner Tips and Tricks

The most common mistake with inverse trigonometric functions is returning an angle outside the restricted domain of the original trig function.

For example, $\cos^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6}$ , not $- \frac{5 π}{6}$ or $\frac{7 π}{6}$
- because $\cos^{- 1}$ must return an angle in $[0, π]$

Always check that your answer falls within the correct interval for the inverse function you are using.

Note that using the $\sin^{- 1}, \cos^{- 1}$ and $\tan^{- 1}$ buttons on your calculator will automatically return a value in the correct interval.

Worked Example

Find the exact value of each of the following expressions.

(a) $\sin^{- 1} (- \frac{\sqrt{3}}{2})$

(b) $\cos^{- 1} (- \frac{\sqrt{2}}{2})$

(d) $\sin (\cos^{- 1} \frac{1}{2})$

Answer:

(a)

$\sin^{- 1} (- \frac{\sqrt{3}}{2})$ answers the question: what angle in $[- \frac{π}{2}, \frac{π}{2}]$ has sine equal to $- \frac{\sqrt{3}}{2}$ ?

From the unit circle, $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$
The negative value means the angle is in the lower half of the restricted domain
- And $\sin (- x) = - \sin x$ , so

$\sin^{- 1} (- \frac{\sqrt{3}}{2}) = - \frac{π}{3}$

(b)

$\cos^{- 1} (- \frac{\sqrt{2}}{2})$ answers the question: what angle in $[0, π]$ has cosine equal to $- \frac{\sqrt{2}}{2}$ ?

From the unit circle, $\cos \frac{π}{4} = \frac{\sqrt{2}}{2}$
The negative value means the angle is in the second quadrant (since cosine is negative in that quadrant)
By symmetry of the cosine function the value you want is

$π - \frac{π}{4} = \frac{3 π}{4}$ .

$\cos^{- 1} (- \frac{\sqrt{2}}{2}) = \frac{3 π}{4}$

(c)

$\tan^{- 1} (- 1)$ answers the question: what angle in $(- \frac{π}{2}, \frac{π}{2})$ has tangent equal to $- 1$ ?

From the unit circle, $\tan \frac{π}{4} = 1$
The negative value means the angle is in the lower half of the restricted domain
- And $\tan (- x) = - \tan x$ , so

$\tan^{- 1} (- 1) = - \frac{π}{4}$

(d)

Work from the inside out

First evaluate $\cos^{- 1} \frac{1}{2}$ :
I.e., what angle in $[0, π]$ has cosine equal to $\frac{1}{2}$ ?

$\cos^{- 1} \frac{1}{2} = \frac{π}{3}$

Then substitute into $\sin$

$\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$

$\sin (\cos^{- 1} \frac{1}{2}) = \frac{\sqrt{3}}{2}$

Worked Example

The function $f$ is given by $f (x) = \cos x + \sin x$ and has a period of $2 π$ . In order to define the inverse function of $f$ , which of the following specifies a restricted domain for $f$ and provides a rationale for why $f$ is invertible on that domain?

(A) $0 \leq x \leq π$ , because all possible values of $f (x)$ occur without repeating on this interval.

(B) $\frac{π}{4} \leq x \leq \frac{5 π}{4}$ , because all possible values of $f (x)$ occur without repeating on this interval.

(D) $- \frac{π}{4} \leq x \leq \frac{3 π}{4}$ , because the length of this interval is half the period.

Answer

A question like this will appear on the calculator section of the exam

Use your graphing calculator to graph the function $f$

Graph showing a sinusoidal curve, representing function cos x + sin x, with labelled key points along the horizontal axis from -π to 9π/4.

You are looking for an interval on which the function takes on all of its possible output values exactly once

Consider the options:

(A) $0 \leq x \leq π$
- Between 0 and $\frac{π}{2}$ the function returns the same output for more than one input (i.e., it fails the 'horizontal line test')
- And it doesn't include all possible output values of the function
  - So this can't provide a restricted domain for an inverse
(B) $\frac{π}{4} \leq x \leq \frac{5 π}{4}$
- Over this interval the function goes from a maximum to the subsequent minimum, including all the values in between
- And no output values occur more than once
  - So this must be the correct answer
(C) $0 \leq x \leq 2 π$
- This does have the same length as the period of $f$
- But it fails the 'horizontal line test' just like option (A)
  - So this can't provide a restricted domain for an inverse
(D) $- \frac{π}{4} \leq x \leq \frac{3 π}{4}$
- This fails the 'horizontal line test' just like option (A)
- And it doesn't include all possible output values of the function
  - So this can't provide a restricted domain for an inverse

So the correct answer is (B)

(B) $\frac{π}{4} \leq x \leq \frac{5 π}{4}$ , because all possible values of $f (x)$
occur without repeating on this interval

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Inverse Trigonometric Functions (College Board AP® Precalculus): Study Guide

Arcsin, arccos & arctan

What are inverse trigonometric functions?

What are the three inverse trigonometric functions called?

Examiner Tips and Tricks

Why do trigonometric functions need restricted domains to have inverses?

What are the standard domain restrictions?

Examiner Tips and Tricks

How can inverse trigonometric functions be evaluated?

Examiner Tips and Tricks

Worked Example

Worked Example

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Author: Roger B

Reviewer: Mark Curtis