← Trigonometric Functions (ME-T1, T2, T3)

NSWMaths Extension 1Syllabus dot point

What are the inverse trigonometric functions, and what are their domains, ranges and graphs?

Define and sketch the inverse trigonometric functions $\arcsin$ , $\arccos$ and $\arctan$ , including their domains and ranges

A focused answer to the HSC Maths Extension 1 dot point on inverse trigonometric functions. Restricted domains for $\sin$ , $\cos$ and $\tan$ to define $\arcsin$ , $\arccos$ and $\arctan$ , their graphs, exact values, and identities, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to know how the inverse trigonometric functions are defined as inverses of suitably restricted trig functions, their principal-value ranges, their graphs, and to evaluate or simplify expressions involving them.

The answer

Why we need to restrict

$\sin x$ , $\cos x$ and $\tan x$ are periodic and not one-to-one over $\mathbb{R}$ . To define inverses, we choose principal branches on which each is one-to-one.

IMATH_15 (or $\sin^{-1}$ )

Restrict $\sin x$ to $x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ . On this interval $\sin$ is strictly increasing from $-1$ to $1$ .

\arcsin: [-1, 1] \to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].

Graph: passes through $(-1, -\frac{\pi}{2})$ , $(0, 0)$ , $(1, \frac{\pi}{2})$ . Odd function, strictly increasing, with vertical tangents at the endpoints.

IMATH_25 (or $\cos^{-1}$ )

Restrict $\cos x$ to $x \in [0, \pi]$ . On this interval $\cos$ is strictly decreasing from $1$ to $-1$ .

\arccos: [-1, 1] \to [0, \pi].

Graph: passes through $(-1, \pi)$ , $(0, \frac{\pi}{2})$ , $(1, 0)$ . Strictly decreasing, with vertical tangents at the endpoints.

IMATH_35 (or $\tan^{-1}$ )

Restrict $\tan x$ to $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ . On this interval $\tan$ is strictly increasing from $-\infty$ to $+\infty$ .

\arctan: \mathbb{R} \to \left( -\frac{\pi}{2}, \frac{\pi}{2} \right).

Graph: passes through $(0, 0)$ . Odd function, strictly increasing, with horizontal asymptotes at $y = \pm \frac{\pi}{2}$ .

Identities

The complementary identity links $\arcsin$ and $\arccos$ :

\arcsin x + \arccos x = \frac{\pi}{2}, \qquad x \in [-1, 1].

For symmetric arguments:

\arcsin(-x) = -\arcsin x, \qquad \arctan(-x) = -\arctan x,

\arccos(-x) = \pi - \arccos x.

For positive $x$ ,

\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}.

For negative $x$ , the right-hand side is $-\frac{\pi}{2}$ .

Composing trig with inverse trig

For $x$ in the appropriate range,

\sin(\arcsin x) = x, \qquad \cos(\arccos x) = x, \qquad \tan(\arctan x) = x.

The reverse compositions only return $x$ if $x$ is already in the principal range:

\arcsin(\sin x) = x \text{ only when } x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].

For $x$ outside, you must use periodicity and reflection to reduce first.

Worked example

Exact value

Find $\arcsin\!\left( \frac{1}{2} \right)$ .

$\sin \frac{\pi}{6} = \frac{1}{2}$ and $\frac{\pi}{6} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ , so $\arcsin\!\left( \frac{1}{2} \right) = \frac{\pi}{6}$ .

Use the complementary identity

Find $\arccos\!\left( \frac{3}{5} \right) + \arcsin\!\left( \frac{3}{5} \right)$ .

By the complementary identity, the sum is $\frac{\pi}{2}$ regardless of the value of $\frac{3}{5}$ .

Composition with reduction

Find $\arcsin\!\left( \sin \frac{5 \pi}{6} \right)$ .

$\frac{5 \pi}{6}$ is not in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ . Use $\sin \frac{5 \pi}{6} = \sin\!\left( \pi - \frac{5 \pi}{6} \right) = \sin \frac{\pi}{6} = \frac{1}{2}$ .

$\arcsin \frac{1}{2} = \frac{\pi}{6}$ .

Find an exact value involving IMATH_12

Find $\tan\!\left( \arccos \frac{4}{5} \right)$ .

Let $\theta = \arccos \frac{4}{5}$ , so $\theta \in [0, \pi]$ and $\cos \theta = \frac{4}{5}$ .

Since $\theta \in [0, \pi]$ and $\cos \theta > 0$ , $\theta$ is in the first quadrant, so $\sin \theta > 0$ .

$\sin \theta = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$ .

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$ .

Inverse-trig sum

Show that $\arctan 1 + \arctan 2 + \arctan 3 = \pi$ .

Use $\arctan x + \arctan y = \arctan \frac{x + y}{1 - x y}$ when $x y < 1$ .

$\arctan 1 + \arctan 2 = \arctan \frac{3}{1 - 2} = \arctan(-3) + \pi$ (because $x y = 2 > 1$ , add $\pi$ to keep value in range).

Hmm, careful: $\arctan(-3) = -\arctan 3$ , so the chain gives $-\arctan 3 + \pi$ , plus $\arctan 3$ , equals $\pi$ . As required.

A cleaner argument: in the unit triangle picture, $\arctan 1 = \frac{\pi}{4}$ , and we can show $\arctan 2 + \arctan 3 = \frac{3 \pi}{4}$ by drawing the relevant right triangles. Sum is $\pi$ .

Common traps

Confusing $\sin^{-1}$ with $\frac{1}{\sin}$: IMATH_2 means the inverse function $\arcsin x$ , not $\csc x = \frac{1}{\sin x}$ .
Out-of-domain input: IMATH_5 is undefined because $2 \not\in [-1, 1]$ . Always check the argument is in range.
Forgetting the principal-value restriction: IMATH_7 , not $\pi$ , because the range of $\arcsin$ is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ .
Treating $\arccos$ as odd: IMATH_12 is neither even nor odd. The correct identity is $\arccos(-x) = \pi - \arccos x$ .
Sign errors in $\arctan$ sum identity: The identity $\arctan x + \arctan y = \arctan\!\left( \frac{x + y}{1 - x y} \right)$ requires $x y < 1$ . If $x y > 1$ and both positive, add $\pi$ to the right side; if both negative, subtract $\pi$ .

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2021 HSC Q42 marksFind the exact value of

\arccos\left( -\frac{1}{2} \right) + \arcsin\left( \frac{\sqrt{3}}{2} \right)

Show worked answer →

$\arccos\!\left( -\frac{1}{2} \right) = \frac{2 \pi}{3}$ because cosine of $\frac{2 \pi}{3}$ equals $-\frac{1}{2}$ and the principal range of $\arccos$ is $[0, \pi]$ .

$\arcsin\!\left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3}$ because sine of $\frac{\pi}{3}$ equals $\frac{\sqrt{3}}{2}$ and the principal range of $\arcsin$ is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ .

Sum: $\frac{2 \pi}{3} + \frac{\pi}{3} = \pi$ .

Markers reward stating the principal-value ranges, evaluating each inverse separately, and combining to a clean exact answer.

2023 HSC Q21 marksWhat is the domain of the function

f(x) = \arcsin(2 x - 1)

Show worked answer →

For $\arcsin u$ to be defined, $-1 \le u \le 1$ .

So $-1 \le 2 x - 1 \le 1$ .

Add $1$ throughout: $0 \le 2 x \le 2$ .

Divide by $2$ : $0 \le x \le 1$ .

Domain: $[0, 1]$ .

Markers reward the constraint on the inner argument, the algebra to isolate $x$ , and the correct interval.

What this dot point is asking

The answer

Why we need to restrict

IMATH_15 (or sin⁡−1\sin^{-1}sin−1)

IMATH_25 (or cos⁡−1\cos^{-1}cos−1)

IMATH_35 (or tan⁡−1\tan^{-1}tan−1)