What is wrong sign for of a negative?

of a negative number lies in the second quadrant ((π)/(2), π), never negative.

State the domain and range of y = x. [2 marks]

Find the exact value of \!(-(sqrt(2))/(2)). [2 marks]

Evaluate sin\!((5)/(13)). [3 marks]

VCAA VCAA 2022 Exam 1-style practice: Find the exact value of (a) \!(-sqrt(3)2), (b) \!(-1sqrt(2)), and (c) cos\!( 2).

**(a)** Need the angle in [-(π)/(2), (π)/(2)] with sine -(sqrt(3))/(2). The reference angle is (π)/(3) and sine is negative, so \!(-(sqrt(3))/(2)) = -(π)/(3). **(b)** Need the angle in [0, π] with cosine -(1)/(sqrt(2)). Cosine is negative in the second quadrant, so the angle is π - (π)/(4) = (3π)/(4). **(c)** Let θ = 2, so tanθ = 2 with θ (0, (π)/(2)). Draw a right triangle with opposite 2, adjacent 1, hypotenuse sqrt(5). Then cosθ = (1)/(sqrt(5)) = (sqrt(5))/(5). Markers reward staying within each range, the correct quadrant for of a negative, and the triangle method for the composite.

§-Syllabus dot point

VICSpecialist MathematicsSyllabus dot point

How are the inverse circular functions defined by restricting the domains of sine, cosine and tangent, and what are the resulting domains, ranges and graphs?

The inverse circular functions $\arcsin$ , $\arccos$ and $\arctan$ , the domain restrictions needed to define them, their domains, ranges and graphs, and the evaluation of exact values and composite expressions

A focused answer to the VCE Specialist Mathematics Unit 3 key-knowledge point on inverse circular functions. Domain restrictions for sine, cosine and tangent, the domains, ranges and graphs of arcsin, arccos and arctan, and exact-value and composite evaluations.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Sine, cosine and tangent are not one-to-one, so each is restricted before inverting. With $\sin$ restricted to $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , $\cos$ to $[0, \pi]$ , and $\tan$ to $(-\frac{\pi}{2}, \frac{\pi}{2})$ , the inverses are: $\arcsin$ with domain $[-1, 1]$ and range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ ; $\arccos$ with domain $[-1, 1]$ and range $[0, \pi]$ ; $\arctan$ with domain $\mathbb{R}$ and range $(-\frac{\pi}{2}, \frac{\pi}{2})$ . The graph of an inverse is the reflection of the restricted function in the line $y = x$ . The range of an inverse is the only allowable set of output angles, so every answer must lie in that interval.

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What this dot point is asking
Why we restrict the domain
Domains, ranges and graphs
Examples in context
Try this

What this dot point is asking

VCAA wants you to define the inverse circular functions $\arcsin$ , $\arccos$ and $\arctan$ (also written $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ), to state and justify the domain restrictions that make sine, cosine and tangent invertible, to know the exact domains and ranges of the inverses, to sketch their graphs, and to evaluate exact values and composite expressions. These functions then appear as antiderivatives and as the results of integration later in the course.

Why we restrict the domain

A function has an inverse only if it is one-to-one. Sine, cosine and tangent are periodic, so each horizontal line meets the full graph infinitely often. To create an invertible function we keep a single interval on which the function takes every output value exactly once.

$\sin$ is restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , where it increases from $-1$ to $1$ .
$\cos$ is restricted to $[0, \pi]$ , where it decreases from $1$ to $-1$ .
$\tan$ is restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , where it increases over all of $\mathbb{R}$ .

These chosen intervals become the ranges of the inverse functions, which is why answers must lie inside them.

Domains, ranges and graphs

\arcsin: [-1, 1] \to \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], \qquad \arccos: [-1, 1] \to [0, \pi], \qquad \arctan: \mathbb{R} \to \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right).

Each inverse graph is the reflection of the corresponding restricted function in the line $y = x$ .

$y = \arcsin x$ increases from $\left(-1, -\frac{\pi}{2}\right)$ through the origin to $\left(1, \frac{\pi}{2}\right)$ .
$y = \arccos x$ decreases from $(-1, \pi)$ through $\left(0, \frac{\pi}{2}\right)$ to $(1, 0)$ .
$y = \arctan x$ increases through the origin with horizontal asymptotes $y = \pm\frac{\pi}{2}$ .

A useful identity is $\arcsin x + \arccos x = \frac{\pi}{2}$ for all $x \in [-1, 1]$ .

Worked example

Exact values and a composite

Evaluate (a) $\arccos\!\left(-\frac{1}{2}\right)$ , (b) $\arctan(\sqrt{3})$ , and (c) $\cos\!\left(\arcsin\frac{3}{5}\right)$ .

Step 1: Evaluate $\arccos\!\left(-\frac{1}{2}\right)$

The range of $\arccos$ is $[0, \pi]$ , so we need the angle in that interval whose cosine equals $-\frac{1}{2}$ . Because cosine is negative, the angle lies in the second quadrant. The reference angle with cosine $\frac{1}{2}$ is $\frac{\pi}{3}$ , so the second-quadrant angle is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . Since $\frac{2\pi}{3} \in [0, \pi]$ , we have $\arccos\!\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ .

Step 2: Evaluate $\arctan(\sqrt{3})$

The range of $\arctan$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so we need the angle in that interval whose tangent equals $\sqrt{3}$ . Since $\tan\frac{\pi}{3} = \sqrt{3}$ and $\frac{\pi}{3}$ lies inside the range, no adjustment is needed: $\arctan(\sqrt{3}) = \frac{\pi}{3}$ .

Step 3: Evaluate $\cos\!\left(\arcsin\frac{3}{5}\right)$

Let $\theta = \arcsin\frac{3}{5}$ , so $\sin\theta = \frac{3}{5}$ with $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Within this range cosine is always non-negative, so applying the Pythagorean identity gives

\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.

The three answers are $\frac{2\pi}{3}$ , $\frac{\pi}{3}$ and $\frac{4}{5}$ .

Common exam traps

Answering outside the range: $\arcsin\frac{1}{2}$ is $\frac{\pi}{6}$ , not $\frac{5\pi}{6}$ , because the range of $\arcsin$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
Wrong sign for $\arccos$ of a negative: $\arccos$ of a negative number lies in the second quadrant $\left(\frac{\pi}{2}, \pi\right)$ , never negative.
Treating $\sin^{-1}$ as $\frac{1}{\sin}$: The notation $\sin^{-1} x$ means the inverse function, not the reciprocal $\csc x$ .
Dropping the domain: $\arcsin x$ and $\arccos x$ are only defined for $-1 \le x \le 1$ . Inputs outside this give no real value.

Examples in context

Example 1. $\arcsin(-1) = -\frac{\pi}{2}$ and $\arccos(-1) = \pi$ , the endpoints of their ranges.

Example 2. Using the identity, $\arccos\frac{3}{5} = \frac{\pi}{2} - \arcsin\frac{3}{5}$ .

Try this

Q1. State the domain and range of $y = \arctan x$ . [2 marks]

Cue. Domain $\mathbb{R}$ , range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

Q2. Find the exact value of $\arcsin\!\left(-\frac{\sqrt{2}}{2}\right)$ . [2 marks]

Cue. $-\frac{\pi}{4}$ , since it lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

Q3. Evaluate $\sin\!\left(\arccos\frac{5}{13}\right)$ . [3 marks]

Cue. With $\cos\theta = \frac{5}{13}$ and $\theta \in [0, \pi]$ , $\sin\theta = \frac{12}{13}$ .

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

VCAA 2022 Exam 14 marksFind the exact value of (a)

\arcsin\!\left(-\dfrac{\sqrt{3}}{2}\right)

, (b)

\arccos\!\left(-\dfrac{1}{\sqrt{2}}\right)

, and (c)

\cos\!\left(\arctan 2\right)

Show worked answer →

(a) Need the angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with sine $-\frac{\sqrt{3}}{2}$ . The reference angle is $\frac{\pi}{3}$ and sine is negative, so $\arcsin\!\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$ .

(b) Need the angle in $[0, \pi]$ with cosine $-\frac{1}{\sqrt{2}}$ . Cosine is negative in the second quadrant, so the angle is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$ .

(c) Let $\theta = \arctan 2$ , so $\tan\theta = 2$ with $\theta \in \left(0, \frac{\pi}{2}\right)$ . Draw a right triangle with opposite $2$ , adjacent $1$ , hypotenuse $\sqrt{5}$ . Then $\cos\theta = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$ .

Markers reward staying within each range, the correct quadrant for $\arccos$ of a negative, and the triangle method for the composite.

VCAA 2023 Exam 25 marksLet

f(x) = 2\arccos\!\left(\dfrac{x}{3}\right)

. (a) State the maximal domain and the range of

f

. (b) Find the exact value of

f(0)

and

f(3)

. (c) Determine the value of

x

for which

f(x) = \dfrac{\pi}{2}

Show worked answer →

(a) $\arccos$ requires $-1 \le \frac{x}{3} \le 1$ , so the maximal domain is $-3 \le x \le 3$ . The range of $\arccos$ is $[0, \pi]$ , so $2\arccos\!\left(\frac{x}{3}\right)$ has range $[0, 2\pi]$ .

(b) $f(0) = 2\arccos(0) = 2 \times \frac{\pi}{2} = \pi$ . $f(3) = 2\arccos(1) = 2 \times 0 = 0$ .

(c) Set $2\arccos\!\left(\frac{x}{3}\right) = \frac{\pi}{2}$ , so $\arccos\!\left(\frac{x}{3}\right) = \frac{\pi}{4}$ . Then $\frac{x}{3} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ , giving $x = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ .

Markers reward the domain from the $\arccos$ restriction, the doubled range, the exact endpoint values, and solving the equation.