Inverse trigonometric functions (HL only)

IB Mathematics: Analysis and Approaches HL· 2.10 Inverse Trigonometric Functions· 15 min read

1. 1. Definition and Domain Restrictions★★☆☆☆HL only⏱ 4 min

For a function to have an inverse, it must be one-to-one (pass the horizontal line test). Standard trigonometric functions are periodic and not one-to-one over their full natural domains, so we restrict their domains to make them invertible.

📘 Definition

Inverse trigonometric function

$arcsin x$ , $arccos x$ , $arctan x$ (also written $sin^{- 1} x$ , $cos^{- 1} x$ , $tan^{- 1} x$ )

The inverse of a trigonometric function with a restricted domain that maps trig output values back to a unique input angle (called the principal value) in a restricted range.

Example:

$sin (\frac{π}{6}) = \frac{1}{2}$ so $arcsin (\frac{1}{2}) = \frac{π}{6}$

Function	Original restricted domain	Inverse domain	Inverse range (principal values)
$arcsin x$	$[- \frac{π}{2}, \frac{π}{2}]$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$
$arccos x$	$[0, π]$	$[- 1, 1]$	$[0, π]$
$arctan x$	$(- \frac{π}{2}, \frac{π}{2})$	$R$	$(- \frac{π}{2}, \frac{π}{2})$

📐 Worked Example

Find the exact value of $arccos (- \frac{1}{2})$

1
Recall that the range of $arccos x$ is $[0, π]$ , so we need an angle in this interval with cosine equal to $- \frac{1}{2}$ .
2
We know that $cos (\frac{2 π}{3}) = cos (π - \frac{π}{3}) = - cos (\frac{π}{3}) = - \frac{1}{2}$
3
Check that the angle is in the required range:
4
$\frac{2\pi}{3} \in [0, \pi], so \arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

2. 2. Graphs of Inverse Trigonometric Functions★★☆☆☆HL only⏱ 3 min

The graph of any inverse function is the reflection of the original function's graph over the line $y = x$ . Each inverse trigonometric graph has key features you should remember for sketching questions:

$arcsin x$ : Increasing, crosses the origin, endpoints at $(- 1, - \frac{π}{2})$ and $(1, \frac{π}{2})$
$arccos x$ : Decreasing, endpoints at $(- 1, π)$ and $(1, 0)$
$arctan x$ : Increasing over all real $x$ , horizontal asymptotes at $y = - \frac{π}{2}$ and $y = \frac{π}{2}$

📐 Worked Example

For the transformed function $y = 2 arcsin (x - 1)$ , state the domain and range

1
Start with the parent function $y = arcsin x$ , which has domain $[- 1, 1]$ and range $[- \frac{π}{2}, \frac{π}{2}]$
2
The $(x - 1)$ term shifts the graph 1 unit right. Adjust the domain:
3
$-1 \leq x - 1 \leq 1 \implies 0 \leq x \leq 2$
4
The leading coefficient 2 stretches the graph vertically by a factor of 2. Adjust the range:
5
$2 \times \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] = [-\pi, \pi]$

3. 3. Simplifying Composite Expressions★★★☆☆HL only⏱ 5 min

One of the most common HL exam questions asks you to simplify composite expressions of the form $sin (arccos x)$ or $arcsin (sin θ)$ , where $θ$ may be outside the principal range. We use right triangle trigonometry and Pythagorean identities to simplify these.

📐 Worked Example

Simplify $sin (arccos (\frac{1}{3}))$ to exact form

1
Let $θ = arccos (\frac{1}{3})$ . By definition, $cos θ = \frac{1}{3}$ and $θ \in [0, π]$ , where $sin θ \geq 0$ .
2
Use the Pythagorean identity $sin^{2} θ + cos^{2} θ = 1$ :
3
$\sin^2 \theta = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}$
4
Since $sin θ$ is non-negative in $[0, π]$ , take the positive root:
5
$\sin \theta = \frac{2\sqrt{2}}{3}, so \sin\left(\arccos\left(\frac{1}{3}\right)\right) = \frac{2\sqrt{2}}{3}$

✓ Quick check

Test your understanding of principal values

What is the value of $arcsin (sin (\frac{5 π}{6}))$ ?
- $\frac{5 π}{6}$
- $\frac{π}{6}$
- $- \frac{π}{6}$
- $\frac{7 π}{6}$
Reveal answer
$\frac{\pi}{6}$ —
The range of $arcsin$ is $[- \frac{π}{2}, \frac{π}{2}]$ . $sin (\frac{5 π}{6}) = sin (\frac{π}{6}) = \frac{1}{2}$ , so $arcsin (\frac{1}{2}) = \frac{π}{6}$ .

4. 4. Solving Inverse Trigonometric Equations★★★★☆HL only⏱ 5 min

Exam questions often ask you to solve equations that include inverse trigonometric functions. The standard method is to rearrange to isolate the inverse trig term, then apply the corresponding trigonometric function to both sides of the equation.

📐 Worked Example

Solve $2 arctan x = \frac{π}{3}$ for $x \in R$

1
Isolate the inverse trigonometric term by dividing both sides by 2:
2
$\arctan x = \frac{\pi}{6}$
3
Apply tangent to both sides, using the identity $tan (arctan x) = x$ for all $x \in R$ :
4
$x = \tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$
5
Check the solution in the original equation: $2 arctan (\frac{3}{3}) = 2 \times \frac{π}{6} = \frac{π}{3}$ , which is valid.

5. Common Pitfalls

Wrong move:

Claiming $arcsin (sin θ) = θ$ for any value of $θ$

Why:

This identity only holds when $θ$ is in the principal range of $arcsin$ , $[- \frac{π}{2}, \frac{π}{2}]$

Correct move:

If $θ$ is outside the principal range, find an angle inside the range with the same sine value to get the correct result

Wrong move:

Confusing $arcsin x$ with the reciprocal $\frac{1}{s i n x} = csc x$

Why:

The notation $sin^{- 1} x$ refers to the inverse function, not the reciprocal

Correct move:

Use $arcsin x$ for the inverse function to avoid ambiguity, and write $csc x$ or $\frac{1}{s i n x}$ for the reciprocal

Wrong move:

Taking the negative root when simplifying $cos (arcsin x)$

Why:

The range of $arcsin x$ is $[- \frac{π}{2}, \frac{π}{2}]$ , where cosine is always non-negative

Correct move:

Always check the sign of the trigonometric function in the principal range of the inverse function before selecting your root

Wrong move:

Stating the domain of $arctan x$ is $[- 1, 1]$

Why:

This is the domain of $arcsin x$ and $arccos x$ , not $arctan x$

Correct move:

The domain of $arctan x$ is all real numbers $R$ , with range $(- \frac{π}{2}, \frac{π}{2})$

Wrong move:

Giving a solution $θ$ outside the principal range when evaluating $arcsin x$

Why:

Inverse trigonometric functions always output a unique principal value in their defined range

Correct move:

Restrict your result to the principal range of the inverse function when evaluating or solving inverse trig problems

6. Quick Reference Cheatsheet

Inverse Function	Domain	Range	Key Identity
$arcsin x$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$	$sin (arcsin x) = x$
$arccos x$	$[- 1, 1]$	$[0, π]$	$cos (arccos x) = x$
$arctan x$	$R$	$(- \frac{π}{2}, \frac{π}{2})$	$tan (arctan x) = x$
$arcsin (- x)$			$- arcsin x$
$arccos (- x)$			$π - arccos x$
$arctan (- x)$			$- arctan x$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 1
Simplify composite trigonometric expression
2021 · Paper 2
Solve inverse trigonometric equation
2019 · Paper 1
Sketch transformed arcsine graph

Going deeper

official syllabusIB AA HL Topic 2 FunctionsOfficial syllabus reference for inverse trigonometric functions

What's Next

Inverse trigonometric functions are a foundational HL topic that reappears later in calculus, when you learn to differentiate inverse functions and integrate functions resulting in inverse trigonometric outputs. They are also used extensively in vector geometry to find angles between lines and planes, and to find the argument of complex numbers in the complex plane. Mastery of domain and range restrictions for inverse trig functions is critical to avoiding errors in later topics, particularly when evaluating definite integrals or solving differential equations. Building on this foundation, you will next extend your knowledge of calculus and applications of trigonometry.