Inverse Trigonometric Functions — AP® Precalculus

Part 1 · Hook

Why this matters

A building code limits how steep a wheelchair ramp can be. The inspector measures the rise and the run, so she knows the tangent of the ramp's angle. To get the angle itself she has to run tangent backward. That backward function exists only if you agree, in advance, which angle to report.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\sin\frac{\pi}{6}$ ?

$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{2}}{2}$
$1$

Show the answer

On the unit circle the angle $\frac{\pi}{6}$ meets the point $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ , so its sine is $\frac{1}{2}$ .

Correct: $\frac{1}{2}$ :
$\frac{\sqrt{3}}{2}$ :
$\frac{\sqrt{2}}{2}$ :
$1$ :

2. $f$ is invertible and $f(2) = 5$ . What is $f^{-1}(5)$ ?

$2$
$5$
$\frac{1}{5}$
$\frac{1}{2}$

Show the answer

An inverse swaps inputs and outputs: $f(2) = 5$ means $f^{-1}(5) = 2$ .

Correct: $2$ :
$5$ :
$\frac{1}{5}$ :
$\frac{1}{2}$ :

Part 4 · See it

See it first

Cosine restricted to [0, pi], falling from (0, 1) to (pi, -1), and its reflection across the line y = x, the graph of inverse cosine, falling from (-1, pi) to (1, 0). — Cosine kept only on $[0, \pi]$ (blue) takes each value from $-1$ to $1$ once. Reflecting it across $y = x$ gives $y = \cos^{-1} x$ (green). LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Sine, cosine and tangent repeat outputsthey have no inverse on their whole domains
Restrict each to one interval where it is one-to-one $\sin$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , $\cos$ on $[0, \pi]$ , $\tan$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Swap inputs and outputs of the restricted function $\sin^{-1}$ , $\cos^{-1}$ and $\tan^{-1}$ , whose outputs are angles in those intervals
Evaluate $\sin^{-1}(\sin x)$ you get $x$ back only when $x$ is already in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Part 6 · Key ideas

Key ideas

An inverse trigonometric function takes a ratio and returns an angle.
$\sin^{-1}$ and $\tan^{-1}$ return angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ ; $\cos^{-1}$ returns angles between $0$ and $\pi$ .
$\sin^{-1} x$ is the inverse function, not the reciprocal $\frac{1}{\sin x}$ .
For a composite like $\cos^{-1}(\cos x)$ , find the inner value first, then the angle in the right range.

Part 7 · Misconception

A common mistake

The wrong idea: $\tan^{-1}(-1) = \frac{3\pi}{4}$ , because $\tan\frac{3\pi}{4} = -1$ .

What actually happens: Many angles have tangent $-1$ , but $\tan^{-1}$ returns the one in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ : $-\frac{\pi}{4}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $y = \cos^{-1} x$

1. The graph shows $y = \cos^{-1} x$ . What is the range of $\cos^{-1}$ ?

$[0, \pi]$
$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$[-1, 1]$
$[0, 2\pi]$

Show the answer

$\cos^{-1}$ undoes cosine restricted to $[0, \pi]$ , so its outputs are the angles in $[0, \pi]$ . The graph runs from height $\pi$ at $x = -1$ down to $0$ at $x = 1$ .

Correct: $[0, \pi]$ : Right: the outputs are angles from $0$ to $\pi$ .
$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ : That is the range of $\sin^{-1}$ and (open) of $\tan^{-1}$ , not of $\cos^{-1}$ .
$[-1, 1]$ : That is the domain of $\cos^{-1}$ : the inputs are cosine values.
$[0, 2\pi]$ : Cosine is not one-to-one on $[0, 2\pi]$ ( $\cos\frac{\pi}{3} = \cos\frac{5\pi}{3}$ ), so that interval cannot be the range.

2. Using the graph or the unit circle, find the exact value of $\cos^{-1}\left(-\frac{1}{2}\right)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\cos^{-1}\left(-\frac{1}{2}\right)$ is the angle in $[0, \pi]$ whose cosine is $-\frac{1}{2}$ . That angle is in Quadrant II with reference angle $\frac{\pi}{3}$ : $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$ .

Answer: $\frac{2\,\pi}{3}$ ≈ 2.094

3. The graph of $y = \cos^{-1} x$ is symmetric about the point $\left(0, \frac{\pi}{2}\right)$ . If $\cos^{-1}(0.6) = a$ , which expression equals $\cos^{-1}(-0.6)$ ?

$\pi - a$
$-a$
$a$
$\pi + a$

Show the answer

Symmetry about $\left(0, \frac{\pi}{2}\right)$ means the heights at $x$ and $-x$ average to $\frac{\pi}{2}$ : $\cos^{-1}(-0.6) = \pi - a$ . On the unit circle, the angle with cosine $-0.6$ is the reflection of angle $a$ across the $y$ -axis, which is $\pi - a$ .

Correct: $\pi - a$ : Right: the two outputs add to $\pi$ .
$-a$ : $-a$ is negative, outside the range $[0, \pi]$ .
$a$ : $\cos a = 0.6$ , not $-0.6$ ; the inputs differ in sign, so the angles differ.
$\pi + a$ : $\pi + a$ is greater than $\pi$ , outside the range.

Table

Four wheelchair ramp designs

Each ramp rises a vertical distance over a horizontal run. The angle $\theta$ between a ramp and the ground satisfies $\tan\theta = \frac{\text{rise}}{\text{run}}$ , with $0 < \theta < \frac{\pi}{2}$ . Use a calculator in radian mode.

Rise and horizontal run of four ramp designs, in meters
Ramp	Rise (meters)	Run (meters)
K	$0.5$	$6$
L	$0.8$	$10$
M	$1.2$	$12$
N	$0.9$	$8$

4. Use a calculator to find the angle, in radians, between ramp L and the ground. Give your answer correct to three decimal places.

Calculator allowed. Give the answer correct to three decimal places, in radians.

Show the answer

Set up $\tan\theta = \frac{0.8}{10} = 0.08$ , so $\theta$ $= \tan^{-1}(0.08)$ $\approx 0.079830$ radians (calculator in radian mode), which is $0.080$ to three decimal places.

Answer: 0.080 radians

5. $f(x) = 2\sin x + 1$ with domain $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is invertible. Find $f^{-1}(x)$ for $-1 \le x \le 3$ .

Type an expression for $f^{-1}(x) =$ .

Show the answer

Write $y = 2\sin x + 1$ , swap: $x = 2\sin y + 1$ . Undo in reverse order: subtract $1$ , divide by $2$ , then apply $\sin^{-1}$ : $y = \sin^{-1}\left(\frac{x - 1}{2}\right)$ . The restricted domain $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is the range of $\sin^{-1}$ , so this is the inverse.

Answer: $f^{-1}(x) =$ $\sin^{-1}\left(\frac{x - 1}{2}\right)$

6. Find the exact value of $\tan^{-1}(-1)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\tan^{-1}$ gives the angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with the given tangent. $\tan\frac{\pi}{4} = 1$ and tangent is odd, so $\tan^{-1}(-1) = -\frac{\pi}{4}$ .

Answer: $\frac{-\pi}{4}$ ≈ -0.785

7. What is the exact value of $\cos\left(\sin^{-1}\frac{3}{5}\right)$ ?

$\frac{4}{5}$
$\frac{3}{4}$
$-\frac{4}{5}$
$\frac{5}{4}$

Show the answer

Let $\theta = \sin^{-1}\frac{3}{5}$ , so $\sin\theta = \frac{3}{5}$ with $\theta$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; here $\theta$ is in Quadrant I. Draw a right triangle with opposite side $3$ and hypotenuse $5$ : the adjacent side is $\sqrt{25 - 9} = 4$ . So $\cos\theta = \frac{4}{5}$ , positive because $\theta$ is in Quadrant I.

Correct: $\frac{4}{5}$ : Right: adjacent over hypotenuse in the $3$ - $4$ - $5$ triangle.
$\frac{3}{4}$ : That is $\tan\theta = \frac{3}{4}$ , opposite over adjacent.
$-\frac{4}{5}$ : $\sin^{-1}$ of a positive number is an angle in Quadrant I, where cosine is positive.
$\frac{5}{4}$ : That is hypotenuse over adjacent, the reciprocal of the cosine.

Part 9 · Summary

Summary

Restricted to

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

,

[0, \pi]

and

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

, sine, cosine and tangent are invertible.

\sin^{-1}

,

\cos^{-1}

and

\tan^{-1}

take a ratio and return the one angle in that interval. Inverse is not reciprocal, and

\sin^{-1}(\sin x) = x

only for

x

in the restricted interval.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of y=cos−1⁡xy = \cos^{-1} x

Four wheelchair ramp designs

Summary

What comes next

Connections

The graph of $y = \cos^{-1} x$