Trigonometric functions and their inverses: domain, range and parity – Trigonometric and Inverse Trigonometric Functions (Senior 5 Mathematics) | Ganzaa

Trigonometric and Inverse Trigonometric Functions

Lessons

Senior 5 · Mathematics

Trigonometric and Inverse Trigonometric Functions

5 lessons

Lesson 1 of 5

Trigonometric functions and their inverses: domain, range and parity

25 min

Undoing the sine

A ramp rises 1 m over a sloping length of 4 m, so the sine of its angle is 0.25. Which angle is it? The calculator key $\sin^{-1}$ answers $14.48^\circ$ . But $\sin 165.52^\circ = 0.25$ as well, and so does $\sin(14.48^\circ + 360^\circ)$ . A function must give one output for each input, so the inverse of the sine can only exist if we agree which angle to give. This lesson makes that agreement for the sine, cosine and tangent, and collects the domains, ranges and symmetries (parity) of all these functions. The rest of the unit studies their limits and derivatives.

Definition

Even function: $f (- x) = f (x)$ for every $x$ in the domain (which must be symmetric about 0); its graph is symmetric about the $y$ -axis.

Odd function: $f (- x) = - f (x)$ for every $x$ in the domain; its graph is symmetric about the origin.

Example: $\cos(-x) = \cos x$ , so cosine is even; $\sin(-x) = -\sin x$ , so sine is odd. Many functions, such as $\sin x + \cos x$ , are neither.

The six functions: domain, range and parity

Recall $\sec x = \frac{1}{\cos x}$ , $\csc x = \frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$ . A quotient is undefined where its denominator is 0:

Function	Domain	Range	Period	Parity
$\sin x$	$\mathbb{R}$	$[- 1, 1]$	$2\pi$	odd
$\cos x$	$\mathbb{R}$	$[- 1, 1]$	$2\pi$	even
$\tan x$	$x \ne \frac{\pi}{2} + k\pi$	$\mathbb{R}$	$π \pi$	odd
$\cot x$	$x \ne k\pi$	$\mathbb{R}$	$π \pi$	odd
$\sec x$	$x \ne \frac{\pi}{2} + k\pi$	$(-\infty, -1] \cup [1, +\infty)$	$2\pi$	even
$\csc x$	$x \ne k\pi$	$(-\infty, -1] \cup [1, +\infty)$	$2\pi$	odd

Parity combines like signs: the product or quotient of two odd functions is even (e.g. $\frac{\sin x}{x}$ ), and the product of an odd and an even function is odd (e.g. $x^2\tan x$ ).

Definition

Inverse sine $\arcsin$ (or $\sin^{-1}$ ): for $x \in [-1, 1]$ , $\arcsin x$ is the unique angle $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin y = x$ .

Inverse cosine $\arccos$ : for $x \in [-1, 1]$ , $\arccos x$ is the unique $y \in [0, \pi]$ with $\cos y = x$ .

Inverse tangent $\arctan$ : for every real $x$ , $\arctan x$ is the unique $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan y = x$ . (Similarly $\operatorname{arccot} x \in (0, \pi)$ .)

These chosen values are the principal values of Unit 1, Lesson 3.

Example: $\arcsin\frac{1}{2} = \frac{\pi}{6}$ , $\arccos\frac{1}{2} = \frac{\pi}{3}$ , $\arctan 1 = \frac{\pi}{4}$ .

Why these intervals?

A function has an inverse only if it is one-to-one. The sine repeats, but on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ it increases from $- 1$ to 1 and takes every value once (Figure 1). The cosine does the same on $[0, \pi]$ , decreasing, and the tangent on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , increasing. Each inverse graph is the reflection of the chosen piece in the line $y = x$ .

Figure 1. On

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

the sine is increasing, so it has an inverse. The graph of

\arcsin

is the reflection of that piece in

y = x

: domain

[- 1, 1]

, range

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Figure 2.

\arccos

decreases from

π \pi

to 0 on

[- 1, 1]

;

\arctan

is defined for every real

x

, increases, and stays strictly between

-\frac{\pi}{2}

and

\frac{\pi}{2}

Important

Properties of the inverse functions

$\sin(\arcsin x) = x$ for $x \in [-1, 1]$ , but $\arcsin(\sin x) = x$ only for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . The same holds for the others, each with its own interval.
Parity: $\arcsin(-x) = -\arcsin x$ and $\arctan(-x) = -\arctan x$ (odd); $\arccos(-x) = \pi - \arccos x$ , so $\arccos$ is neither even nor odd.
$\arcsin x + \arccos x = \frac{\pi}{2}$ for $x \in [-1, 1]$ .

Why the last one holds: let $\alpha = \arcsin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Then $\frac{\pi}{2} - \alpha \in [0, \pi]$ and $\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha = x$ , so $\arccos x = \frac{\pi}{2} - \alpha$ .

Worked example 1: exact values

Find (a) $\arcsin\left(-\frac{1}{2}\right)$ (b) $\arccos\left(-\frac{\sqrt{2}}{2}\right)$ (c) $\arctan(-\sqrt{3})$ (d) $\arcsin\left(\sin\frac{5\pi}{6}\right)$ .

Solution

The angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with sine $-\frac{1}{2}$ is $-\frac{\pi}{6}$ .
The angle in $[0, \pi]$ with cosine $-\frac{\sqrt{2}}{2}$ is $\frac{3\pi}{4}$ (second quadrant), not $-\frac{\pi}{4}$ .
$-\frac{\pi}{3}$ , since $\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}$ and $-\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .
$\sin\frac{5\pi}{6} = \frac{1}{2}$ , and $\arcsin\frac{1}{2} = \frac{\pi}{6}$ . The answer is $\frac{\pi}{6}$ , not $\frac{5\pi}{6}$ , because $\frac{5\pi}{6}$ is outside the range of $\arcsin$ .

Worked example 2: a ratio of an inverse ratio

Find (a) $\cos\left(\arcsin\frac{3}{5}\right)$ (b) $\tan\left(\arccos\left(-\frac{5}{13}\right)\right)$ .

Solution

Let $\theta = \arcsin\frac{3}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , where the cosine is never negative: $\cos\theta = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ .
Let $\theta = \arccos\left(-\frac{5}{13}\right) \in [0, \pi]$ , where the sine is never negative: $\sin\theta = \frac{12}{13}$ , so $\tan\theta = \frac{12}{13} \div \left(-\frac{5}{13}\right) = -\frac{12}{5}$ .

Worked example 3: domains and parity

(a) Find the domain of $f(x) = \arcsin(2x - 1)$ and of $g(x) = \arccos(x^2 - 3)$ . (b) Decide whether $h(x) = x^2\tan x$ and $k(x) = \frac{\sin x}{x}$ are even or odd.

Solution

$-1 \le 2x - 1 \le 1$ gives $0 \le x \le 1$ . For $g$ : $-1 \le x^2 - 3 \le 1$ , so $2 \le x^2 \le 4$ : $x \in [-2, -\sqrt{2}] \cup [\sqrt{2}, 2]$ .
$h(-x) = (-x)^2\tan(-x) = -x^2\tan x = -h(x)$ : odd. $k(-x) = \frac{\sin(-x)}{-x} = \frac{-\sin x}{-x} = k(x)$ : even (domain $x \ne 0$ ).

Warning

$\sin^{-1} x$ is not $\frac{1}{\sin x}$ . The notation $\sin^{-1}$ means the inverse function $\arcsin$ ; the reciprocal is $(\sin x)^{-1} = \csc x$ . And $\arcsin(\sin x)$ "cancels" only inside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ : $\arcsin(\sin \pi) = \arcsin 0 = 0$ , not $π \pi$ .

Tip

Draw the triangle. To simplify an expression such as $\cos(\arcsin x)$ , draw a right-angled triangle with an angle $\theta$ whose sine is $\frac{x}{1}$ : the third side is $\sqrt{1 - x^2}$ , so $\cos(\arcsin x) = \sqrt{1 - x^2}$ . Then check the sign from the range of the inverse function.

Khan Academy: Inverse trig functions: arcsin. Watch how Sal finds arcsin of a value on the unit circle, and why the answers are restricted to the first and fourth quadrants so that arcsin is a function.

✦

Practice

Exact answers. The last two are harder.

Give the domain, range and parity of $\sec x$ and of $\csc x$ .
Even, odd or neither? (a) $x^3\cos x$ (b) $\sin^2 x + \cos x$ (c) $\sin x + \cos x$ (d) $\frac{\tan x}{x}$ .
Find (a) $\arcsin\frac{\sqrt{3}}{2}$ (b) $\arccos\left(-\frac{1}{2}\right)$ (c) $\arctan 1$ (d) $\arccos 0$ .
Find (a) $\sin\left(\arccos\left(-\frac{4}{5}\right)\right)$ (b) $\arccos\left(\cos\left(-\frac{\pi}{3}\right)\right)$ (c) $\arctan\left(\tan\frac{3\pi}{4}\right)$ .
Find the domain and range of $f(x) = \arcsin\frac{x - 1}{2}$ .
Show that $\arccos(-x) = \pi - \arccos x$ , and explain why $\arccos$ is neither even nor odd.
Show that $\sin(2\arctan x) = \frac{2x}{1 + x^2}$ .
Solve $\arctan x + \arctan 2x = \frac{\pi}{4}$ .

Show answersHide answers

$\sec$ : domain $x \ne \frac{\pi}{2} + k\pi$ , range $(-\infty, -1] \cup [1, +\infty)$ , even. $\csc$ : domain $x \ne k\pi$ , same range, odd.
(a) odd (b) even (c) neither (d) even.
(a) $\frac{\pi}{3}$ (b) $\frac{2\pi}{3}$ (c) $\frac{\pi}{4}$ (d) $\frac{\pi}{2}$ .
(a) $\frac{3}{5}$ (the angle is in $[0, \pi]$ , where sine is positive) (b) $\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}$ , so $\frac{\pi}{3}$ (c) $\tan\frac{3\pi}{4} = -1$ , so $-\frac{\pi}{4}$ .
$-1 \le \frac{x - 1}{2} \le 1$ : domain $[- 1, 3]$ ; range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
Let $\alpha = \arccos x \in [0, \pi]$ . Then $\pi - \alpha \in [0, \pi]$ and $\cos(\pi - \alpha) = -\cos\alpha = -x$ , so $\arccos(-x) = \pi - \alpha$ . For example $\arccos(-1) = \pi$ , while $\arccos 1 = 0$ : the values are neither equal nor opposite.
Let $\theta = \arctan x$ . Then $\sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$ (a t-formula from Unit 1) $= \frac{2x}{1 + x^2}$ .
Take tangents: $\frac{x + 2x}{1 - 2x^2} = 1$ , so $2 x^{2} + 3 x - 1 = 0$ and $x = \frac{-3 \pm \sqrt{17}}{4}$ . The negative root makes both arctangents negative, so their sum cannot be $\frac{\pi}{4}$ . Answer: $x = \frac{-3 + \sqrt{17}}{4} \approx 0.281$ .

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S5 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Inverse trig functions: arcsin, Khan Academy, embedded from YouTube.

1Undoing the sine

2The six functions: domain, range and parity

3Why these intervals?

Undoing the sine

The six functions: domain, range and parity

Why these intervals?