# Inverse trigonometric functions (HL only)

> IB Mathematics: Analysis and Approaches HL · Functions
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u2-inverse-trigonometric-functions/

This sub-topic covers the definition, domain, range, and graphs of inverse sine, cosine, and tangent functions for IB AA HL. You will learn to simplify composite expressions and solve equations involving inverse trigonometric functions, a common exam topic.

**Prerequisites:** [Function inverses, domain and range](https://www.owlsprep.com/study/ib-math-aa-hl-u1-function-inverses/); [Core trigonometric function properties](https://www.owlsprep.com/study/ib-math-aa-hl-u2-trigonometric-functions/)

## Learning objectives

- Define inverse trigonometric functions with correct restricted domains and ranges
- Graph and transform inverse sine, cosine, and tangent functions
- Simplify composite expressions of trigonometric and inverse trigonometric functions
- Solve algebraic equations involving inverse trigonometric functions

## 1. Definition and Domain Restrictions

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.

**Inverse trigonometric function** — 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.

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

*Example:* $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ so $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$

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

> **warning**
>
> Note that $\sin^{-1}x$ is **not** the same as $(\sin x)^{-1} = \frac{1}{\sin x} = \csc x$. Always use $\arcsin x$ to avoid ambiguity.

**Worked example:** Find the exact value of $\arccos\left(-\frac{1}{2}\right)$

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

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

**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 $[-\frac{\pi}{2}, \frac{\pi}{2}]$
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. $$2 \times \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] = [-\pi, \pi]$$

> **tip**
>
> For exam sketch questions, always label endpoints, intercepts, and asymptotes to earn full marks.

## 3. Simplifying Composite Expressions

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

**Worked example:** Simplify $\sin\left(\arccos\left(\frac{1}{3}\right)\right)$ to exact form

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

**Check your understanding**

Test your understanding of principal values

1. What is the value of $\arcsin\left(\sin\left(\frac{5\pi}{6}\right)\right)$?

   - $\frac{5\pi}{6}$
   - $\frac{\pi}{6}$
   - $-\frac{\pi}{6}$
   - $\frac{7\pi}{6}$

   *Why:* The range of $\arcsin$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. $\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, so $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$.

## 4. Solving Inverse Trigonometric Equations

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 $2\arctan x = \frac{\pi}{3}$ for $x \in \mathbb{R}$

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 $x \in \mathbb{R}$:
4. $$x = \tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$$
5. Check the solution in the original equation: $2\arctan\left(\frac{\sqrt{3}}{3}\right) = 2 \times \frac{\pi}{6} = \frac{\pi}{3}$, which is valid.

> **warning**
>
> Always check your solutions at the end. Algebraic manipulations like squaring both sides can introduce extraneous solutions that do not satisfy the original equation.

## Common pitfalls

- **Wrong:** Claiming $\arcsin(\sin \theta) = \theta$ for any value of $\theta$
  - Why it fails: This identity only holds when $\theta$ is in the principal range of $\arcsin$, $[-\frac{\pi}{2}, \frac{\pi}{2}]$
  - Correct: If $\theta$ is outside the principal range, find an angle inside the range with the same sine value to get the correct result
- **Wrong:** Confusing $\arcsin x$ with the reciprocal $\frac{1}{\sin x} = \csc x$
  - Why it fails: The notation $\sin^{-1}x$ refers to the inverse function, not the reciprocal
  - Correct: Use $\arcsin x$ for the inverse function to avoid ambiguity, and write $\csc x$ or $\frac{1}{\sin x}$ for the reciprocal
- **Wrong:** Taking the negative root when simplifying $\cos(\arcsin x)$
  - Why it fails: The range of $\arcsin x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, where cosine is always non-negative
  - Correct: Always check the sign of the trigonometric function in the principal range of the inverse function before selecting your root
- **Wrong:** Stating the domain of $\arctan x$ is $[-1, 1]$
  - Why it fails: This is the domain of $\arcsin x$ and $\arccos x$, not $\arctan x$
  - Correct: The domain of $\arctan x$ is all real numbers $\mathbb{R}$, with range $(-\frac{\pi}{2}, \frac{\pi}{2})$
- **Wrong:** Giving a solution $\theta$ outside the principal range when evaluating $\arcsin x$
  - Why it fails: Inverse trigonometric functions always output a unique principal value in their defined range
  - Correct: Restrict your result to the principal range of the inverse function when evaluating or solving inverse trig problems

## Cheatsheet

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

## 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.

- [Trigonometric identities](https://www.owlsprep.com/study/ib-math-aa-hl-u2-trigonometric-identities/)
- [Function properties: parity and periodicity](https://www.owlsprep.com/study/ib-math-aa-hl-u2-function-properties-parity-and-periodicity/)
- [Solving function equations and inequalities](https://www.owlsprep.com/study/ib-math-aa-hl-u2-solving-function-equations-and-inequalities/)

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