What is the range of ? [1 point]

College Board AP 2023 (style)-style practice: Section I, Part A (multiple choice, no calculator). What is (-(1)/(2))? (A) (π)/(3) (B) (2π)/(3) (C) (4π)/(3) (D) -(π)/(3)

The correct answer is (B), (2π)/(3). Arccosine returns an angle in [0, π] whose cosine is the input. We need cosθ = -(1)/(2) with θ in [0, π]. The reference angle is (π)/(3), and cosine is negative in Quadrant II, so θ = π - (π)/(3) = (2π)/(3). Choice (C) has the right cosine but lies outside the arccosine range.

United StatesPrecalculusSyllabus dot point

How are the inverse trigonometric functions defined on restricted domains, and what are their ranges?

Topic 3.9 Inverse Trigonometric Functions: define arcsine, arccosine and arctangent on restricted domains, and evaluate and interpret their outputs.

A focused answer to AP Precalculus Topic 3.9, covering why trig functions must be domain-restricted to have inverses, the ranges of arcsine, arccosine and arctangent, and how to evaluate and interpret inverse trig values.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

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

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Quick answer

The trigonometric functions repeat, so they fail the horizontal line test and cannot be inverted over all reals. To define an inverse, each is restricted to an interval where it is one-to-one. Arcsine inverts $\sin x$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , so $\arcsin$ returns an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ from an input in $[-1, 1]$ . Arccosine inverts $\cos x$ on $[0, \pi]$ , so $\arccos$ returns an angle in $[0, \pi]$ from an input in $[-1, 1]$ . Arctangent inverts $\tan x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so $\arctan$ returns an angle in that open interval from any real input. The output of an inverse trig function is always an angle, and it always lies in the function's restricted range, even when the original equation has many solutions.

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What this topic is asking
Why restriction is needed
The three inverse functions and their ranges
Interpreting the output
Try this

What this topic is asking

The College Board (Topic 3.9) wants you to define and use the inverse trigonometric functions arcsine, arccosine and arctangent. Because sine, cosine and tangent repeat, they are not one-to-one and have no inverse over their full domains. Restricting each to an interval where it is one-to-one creates an invertible piece; the inverse then returns an angle in that restricted range.

Why restriction is needed

This is the inverse-function logic of Topic 2.8 applied to periodic functions: the restriction is what makes the inverse well defined.

The three inverse functions and their ranges

Key fact

Inverse	Domain (input)	Range (output angle)
$\arcsin x$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\arccos x$	$[-1, 1]$	$[0, \pi]$
$\arctan x$	all reals	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

The range is the restricted interval of the original function; the inverse never returns an angle outside it.

The arccosine range $[0, \pi]$ spans Quadrants I and II; the arcsine and arctangent ranges span Quadrants IV and I. This is why a negative input to arccosine gives a Quadrant II angle, while a negative input to arcsine gives a negative (Quadrant IV) angle.

Evaluating an inverse trig value

Evaluate $\arcsin\left(-\frac{1}{2}\right)$ and explain why the answer is not $\frac{7\pi}{6}$ .

step 1 State what arcsine returns

$\arcsin$ returns the unique angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is the input.

step 2 Find the reference angle

We need $\sin\theta = -\frac{1}{2}$ . The reference angle is $\frac{\pi}{6}$ , since $\sin\frac{\pi}{6} = \frac{1}{2}$ .

step 3 Place it in the arcsine range

The output must lie in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , and the sine is negative, so the angle is negative: $\theta = -\frac{\pi}{6}$ .

step 4 Reject the out-of-range solution

The angle $\frac{7\pi}{6}$ also has sine $-\frac{1}{2}$ , but it lies in $[\pi, 2\pi]$ , outside the arcsine range. So $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ , the only valid output.

Interpreting the output

A point worth stating once is the difference between solving a trig equation and evaluating an inverse trig function. Evaluating $\arcsin x$ returns exactly one angle, the one in the restricted range. Solving $\sin\theta = x$ over all reals returns infinitely many angles, found by adding the period and using symmetry (the subject of Topic 3.10). The inverse function is the gatekeeper that picks the single principal value; the full solution set is built from it. Confusing the two leads to giving the principal value when all solutions are wanted, or vice versa, so always check which the question asks for.

Try this

Q1. What is the range of $\arccos$ ? [1 point]

Cue. $[0, \pi]$ , the interval on which cosine is restricted to be one-to-one.

Q2. Evaluate $\arctan(0)$ . [1 point]

Cue. The angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with tangent $0$ is $0$ , so $\arctan(0) = 0$ .

Exam-style practice questions

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

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). What is

\arccos\left(-\frac{1}{2}\right)

? (A)

\frac{\pi}{3}

(B)

\frac{2\pi}{3}

(C)

\frac{4\pi}{3}

(D)

-\frac{\pi}{3}

Show worked answer →

The correct answer is (B), $\frac{2\pi}{3}$ .

Arccosine returns an angle in $[0, \pi]$ whose cosine is the input. We need $\cos\theta = -\frac{1}{2}$ with $\theta$ in $[0, \pi]$ . The reference angle is $\frac{\pi}{3}$ , and cosine is negative in Quadrant II, so $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . Choice (C) has the right cosine but lies outside the arccosine range.

AP 2024 (style)3 marksSection II (free response, no calculator). (a) Explain why

y = \sin x

must be restricted to

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

to define

\arcsin

. (b) Evaluate

\arcsin\left(\frac{\sqrt{2}}{2}\right)

and

\arctan(1)

Show worked answer →

A 3-point question on inverse trig definitions and values.

(a) Restriction (1 point): $\sin x$ is not one-to-one over all reals (it repeats), so it fails the horizontal line test and has no inverse. Restricting to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ makes it one-to-one (passing through every output in $[-1, 1]$ exactly once), so an inverse exists there.
(b) Values (2 points): $\arcsin\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ (the angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $\frac{\sqrt{2}}{2}$ ); $\arctan(1) = \frac{\pi}{4}$ (the angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ whose tangent is $1$ ).

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)