Exercises: Restrict domain for inverse trig functions - Common Core Math - HS Functions

A

Recall / Warm-Up

1.

A function has an inverse only if it is one-to-one. Which test
determines whether a function is one-to-one from its graph?

A.

Vertical Line Test

B.

Horizontal Line Test

C.

Zero Product Property

D.

Symmetry Test across $y = x$

2.

The sine function $y = \sin(x)$ is periodic with period $2\pi$ .
Draw a horizontal line at $y = 0.5$ across the sine graph.
How many times does this line intersect $y = \sin(x)$ on the
interval $[-4\pi, 4\pi]$ ?

A.

1

B.

2

C.

4

D.

Infinitely many

3.

For a function $f$ , the notation $f^{-1}(x)$ means the inverse
function of $f$ . Which expression is NOT the same as $\sin^{-1}(x)$ ?

A.

$\arcsin(x)$

B.

The angle in the restricted domain whose sine equals $x$

C.

$\frac{1}{\sin(x)}$

D.

The inverse function of restricted sine

B

Fluency Practice

1.

Evaluate $\arcsin\!\left(\dfrac{1}{2}\right)$ . Give your answer
in radians.

2.

Evaluate $\arccos\!\left(\dfrac{1}{2}\right)$ . Give your answer
in radians.

$Summary table showing the restricted domains and ranges for arcsin, arccos, and arctan.$

3.

Use the summary table to answer: what is the range of $\arctan(x)$ ?

A.

$[-1, 1]$

B.

$[0, \pi]$

C.

$(-\frac{\pi}{2}, \frac{\pi}{2})$

D.

$(-\infty, \infty)$

4.

Evaluate $\arctan(1)$ . Give your answer in radians.

5.

Evaluate $\arcsin\!\left(-\dfrac{\sqrt{2}}{2}\right)$ .
Give your answer in radians.

C

Varied Practice

D

Word Problems / Application

1.

A surveyor stands at point $A$ and measures that the top of a
vertical tower is visible at an angle of elevation $\theta$ .
The surveyor uses the equation $\tan(\theta) = \frac{45}{60}$
to model the situation and wants to find $\theta$ in radians.

1.

Simplify $\dfrac{45}{60}$ to lowest terms.

2.

Use $\arctan$ to find $\theta$ . Write your answer using
exact inverse-trig notation (e.g., $\arctan(\frac{3}{4})$ )
— do not evaluate to a decimal.

3.

The result $\theta = \arctan\!\left(\frac{3}{4}\right)$ lies
in which interval?

A.

$[-\frac{\pi}{2}, \frac{\pi}{2}]$

B.

$(-\frac{\pi}{2}, \frac{\pi}{2})$

C.

$[0, \pi]$

D.

$(-\infty, \infty)$

2.

A student is asked to find all solutions to $\cos(x) = -\frac{\sqrt{3}}{2}$
in the interval $[0, 2\pi)$ . The student first uses
$\arccos\!\left(-\frac{\sqrt{3}}{2}\right)$ to find the
reference solution in $[0, \pi]$ .

What is $\arccos\!\left(-\dfrac{\sqrt{3}}{2}\right)$ ?
Give your answer in radians.

E

Error Analysis

1.

A student evaluates $\arcsin\!\left(\sin\!\left(\frac{5\pi}{6}\right)\right)$ :

"Since $\sin\!\left(\frac{5\pi}{6}\right) = \frac{1}{2}$ , and
$\arcsin$ undoes sine, $\arcsin\!\left(\sin\!\left(\frac{5\pi}{6}\right)\right) = \frac{5\pi}{6}$ ."

What error did the student make, and what is the correct value?

A.

The student forgot that $\sin\!\left(\frac{5\pi}{6}\right) \neq \frac{1}{2}$ ; the correct value is $\arcsin\!\left(\sin\!\left(\frac{5\pi}{6}\right)\right) = \frac{5\pi}{6}$

B.

The student confused $\arcsin$ with $\csc$ ; the correct answer uses the reciprocal identity

C.

$\arccos$ should be used instead of $\arcsin$

D.

The student forgot that $\arcsin$ only outputs values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ ; since $\frac{5\pi}{6} \notin [-\frac{\pi}{2}, \frac{\pi}{2}]$ , the composition gives $\arcsin\!\left(\frac{1}{2}\right) = \frac{\pi}{6}$ , not $\frac{5\pi}{6}$

2.

A student is asked for the value of $\sin^{-1}\!\left(\frac{\sqrt{3}}{2}\right)$
and writes:

$\sin^{-1}\!\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{\sin\!\left(\frac{\sqrt{3}}{2}\right)}$

Explain the student's error and give the correct value of
$\sin^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)$ .

F

Challenge / Extension

1.

A student claims: "Since $\sin(x)$ repeats every $2\pi$ ,
restricting to any interval of length $\pi$ will make it
one-to-one." Is this claim correct? Justify your answer with
an example or counterexample.

2.