# Inverse Trigonometric Functions

> The functions sin⁻¹ x, cos⁻¹ x and tan⁻¹ x. Their domains, ranges, graphs, exact values and key properties.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/inverse-trig-functions/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The ramp inspector
A building inspector checks wheelchair ramps. She measures that a ramp rises 1 m over a sloping length of 12 m, so $\sin\theta = \dfrac{1}{12}$. But the rules are written as angles: a ramp must be less than about $4.8^\circ$.

She needs to go backwards, from a sine value to the angle. Her calculator's $\sin^{-1}$ button does that. In this chapter you learn exactly what that button does, and why it always gives one particular answer.

**In short:** The inverse trig functions find an angle from a ratio, with a restricted range so each gives one answer. $\sin^{-1} x$ and $\cos^{-1} x$ take $x$ from $-1$ to $1$, with ranges $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ and $[0, \pi]$, and $\tan^{-1} x$ takes any $x$.

## Starting point: what you already know

- **Inverse functions** (Year 11): $f^{-1}$ undoes $f$, exists only for one-to-one functions, and its graph is the reflection of $f$ in $y = x$. Domain and range swap.
- Exact values on the unit circle, for example $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, $\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$, $\tan\dfrac{\pi}{3} = \sqrt{3}$.
- The graphs of $\sin x$, $\cos x$ and $\tan x$, in radians.

## The problem

$\sin x = \dfrac{1}{2}$ has infinitely many solutions: $\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{13\pi}{6}, \dots$ So $\sin x$ is not one-to-one, and it has no inverse on all real numbers.

The fix is to **restrict the domain** to one piece where the function goes through every possible value exactly once.

| Function | Restricted domain | So the inverse has domain | and range |
|---|---|---|---|
| $\sin x$ | $-\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}$ | $-1 \le x \le 1$ | $-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}$ |
| $\cos x$ | $0 \le x \le \pi$ | $-1 \le x \le 1$ | $0 \le y \le \pi$ |
| $\tan x$ | $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ | all real $x$ | $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ |

## What the symbols mean

$\sin^{-1} x$ is the angle between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ whose sine is $x$. It is also written $\arcsin x$.

$\cos^{-1} x$ is the angle between 0 and $\pi$ whose cosine is $x$. $\tan^{-1} x$ is the angle strictly between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ whose tangent is $x$.

### Worked example: Exact values
- $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$
- $\cos^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3}$, which is in the range 0 to $\pi$
- $\tan^{-1}(-1) = -\dfrac{\pi}{4}$
- $\sin^{-1} 2$ does not exist, because no sine is bigger than 1

### Watch out: Not one over sine
$\sin^{-1} x$ is not $\dfrac{1}{\sin x}$. That is $\operatorname{cosec} x$.

### Calculator tip: The mode decides the answer
In radian mode, `SHIFT sin 0.5` gives 0.5236, which is $\frac{\pi}{6}$. In degree mode it gives 30. The calculator always answers in the principal range: $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ for $\sin^{-1}$ and $\tan^{-1}$, and $0$ to $\pi$ for $\cos^{-1}$.

## The graphs

Reflect each restricted piece in $y = x$.

%%FIG:arcsincos%%

%%FIG:arctan%%

## Key properties

### Key idea: Odd and even behaviour
$$\sin^{-1}(-x) = -\sin^{-1} x \qquad \tan^{-1}(-x) = -\tan^{-1} x$$
$$\cos^{-1}(-x) = \pi - \cos^{-1} x$$
$$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$$

**Do and undo.** For $-1 \le x \le 1$, $\sin\left(\sin^{-1} x\right) = x$ always. But $\sin^{-1}(\sin x) = x$ only when $x$ is in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

### Worked example: A trap
$\sin^{-1}\left(\sin\dfrac{5\pi}{6}\right) = \sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$, not $\dfrac{5\pi}{6}$, because the answer must be in the range of $\sin^{-1}$.

**Mind map**

```text
Inverse trig functions
  Why restrict
    sin, cos, tan not one-to-one
  sin⁻¹ x
    Domain -1 to 1
    Range -π/2 to π/2
  cos⁻¹ x
    Domain -1 to 1
    Range 0 to π
  tan⁻¹ x
    Any x
    Range between -π/2 and π/2
  Properties
    sin⁻¹(-x) = -sin⁻¹ x
    cos⁻¹(-x) = π - cos⁻¹ x
    sin⁻¹ x + cos⁻¹ x = π/2
```

### Year 11: Why those pieces?
Each restricted piece is chosen to include the acute angles $0$ to $\dfrac{\pi}{2}$ (so triangle answers come out naturally) and to cover every output exactly once. For cosine, the piece $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$ would not work, because $\cos x$ is even and repeats each value twice there.

### Year 12: Transformed graphs
Sketch $y = 2\sin^{-1}(3x)$.

The domain needs $-1 \le 3x \le 1$, so $-\dfrac{1}{3} \le x \le \dfrac{1}{3}$. The range is doubled: $-\pi \le y \le \pi$. The shape is the $\sin^{-1}$ curve squashed horizontally and stretched vertically.

**Working back from the graph.** If $y = a\cos^{-1}(bx)$ has domain $-3 \leq x \leq 3$ and range $0 \leq y \leq 2\pi$, then $3b = 1$ gives $b = \frac{1}{3}$, and $a\pi = 2\pi$ gives $a = 2$.

**Ranges of combinations.** For $f(x) = \sin^{-1}x + \tan^{-1}x$, both parts increase, so $f$ increases on $[-1, 1]$ and its range runs from $f(-1) = -\frac{3\pi}{4}$ to $f(1) = \frac{3\pi}{4}$. When the parts pull in opposite directions, check $f'(x)$ to see whether $f$ still increases.

### Year 12: Using a triangle
Find $\cos\left(\sin^{-1}\dfrac{3}{5}\right)$ exactly.

Let $\theta = \sin^{-1}\dfrac{3}{5}$, so $\sin\theta = \dfrac{3}{5}$ with $\theta$ acute. Draw a right triangle with opposite 3 and hypotenuse 5. The adjacent side is 4, so the answer is $\dfrac{4}{5}$.

**Double angles.** Let $\theta = \tan^{-1}\frac{1}{3}$, so $\tan\theta = \frac{1}{3}$. Then $\cos 2\theta = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \dfrac{8/9}{10/9} = \dfrac{4}{5}$. And $\sin(2\sin^{-1}x) = 2\sin\theta\cos\theta = 2x\sqrt{1 - x^2}$ with $\theta = \sin^{-1}x$ (the root is positive because $\theta$ is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$).

**Sums of inverse tangents.** For $\tan^{-1}1 + \tan^{-1}2$: the tan of the sum is $\dfrac{1 + 2}{1 - 2} = -3$. Each angle is between $0$ and $\frac{\pi}{2}$, and $\tan^{-1}2 > \frac{\pi}{4}$, so the sum is between $\frac{\pi}{2}$ and $\pi$. The angle there with tangent $-3$ is $\pi - \tan^{-1}3$.

**Equations.** Use $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$. If $\sin^{-1}x = \cos^{-1}x$, each is $\frac{\pi}{4}$, so $x = \frac{1}{\sqrt{2}}$. Similarly $\tan^{-1}x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2}$ for $x > 0$ and $-\frac{\pi}{2}$ for $x < 0$.

### Year 12: Calculus with inverse trig
In the further calculus chapters you will meet
$$\frac{d}{dx}\sin^{-1} x = \frac{1}{\sqrt{1 - x^2}} \qquad \frac{d}{dx}\tan^{-1} x = \frac{1}{1 + x^2}$$
These make some hard-looking integrals easy.

## Quick summary

- $\sin^{-1} x$: domain $[-1, 1]$, range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
- $\cos^{-1} x$: domain $[-1, 1]$, range $[0, \pi]$.
- $\tan^{-1} x$: domain all real $x$, range strictly between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$.
- $\sin^{-1}(-x) = -\sin^{-1} x$, $\cos^{-1}(-x) = \pi - \cos^{-1} x$, $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$.
- $\sin^{-1}(\sin x) = x$ only inside the range of $\sin^{-1}$.

## Quiz

Try these questions from the chapter quiz.

1. Find $\sin^{-1}\dfrac{\sqrt{3}}{2}$.
   - A. $\dfrac{\pi}{6}$
   - B. $\dfrac{\pi}{3}$
   - C. $\dfrac{2\pi}{3}$
   - D. $\dfrac{\pi}{4}$
   - Answer: B. $\dfrac{\pi}{3}$. $\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ and it is in the range.

2. Find $\cos^{-1} 0$.
   - A. 0
   - B. $\dfrac{\pi}{2}$
   - C. $\pi$
   - D. $-\dfrac{\pi}{2}$
   - Answer: B. $\dfrac{\pi}{2}$. $\cos\dfrac{\pi}{2} = 0$.

3. Find $\tan^{-1} 1$.
   - A. $\dfrac{\pi}{4}$
   - B. $\dfrac{\pi}{2}$
   - C. 1
   - D. $\dfrac{3\pi}{4}$
   - Answer: A. $\dfrac{\pi}{4}$. $\tan\dfrac{\pi}{4} = 1$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/inverse-trig-functions/#quiz