Exercises: Explain symmetry and periodicity - Common Core Math - HS Functions

A

Recall / Warm-Up

1.

On the unit circle, the angle $x$ has terminal point $(a, b)$ .
Which expression gives $\cos(x)$ ?

A.

$b$

B.

$a$

C.

$\frac{b}{a}$

D.

$\frac{a}{b}$

2.

A function $f$ is called an even function if $f(-x) = f(x)$ for all $x$ in its domain. Which statement correctly describes an even function's graph?

A.

The graph has rotational symmetry about the origin.

B.

The graph is symmetric about the y-axis.

C.

All output values are non-negative (the graph stays above the x-axis).

D.

The graph is symmetric about the x-axis.

3.

A point travels counterclockwise around the unit circle.
After one complete revolution (adding $2\pi$ radians to the angle),
the terminal point returns to its starting position.

If $\sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$ ,
what is $\sin\!\left(\dfrac{\pi}{6} + 2\pi\right)$ ?

Enter your answer as a fraction.

B

Fluency Practice

1.

Use the even/odd properties of cosine and sine to evaluate.

Given that $\cos\!\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$ ,
find $\cos\!\left(-\dfrac{\pi}{4}\right)$ .

Express your answer in simplest radical form.

2.

Given that $\cos\!\left(\dfrac{2\pi}{3}\right) = -\dfrac{1}{2}$ ,
find $\cos\!\left(-\dfrac{2\pi}{3}\right)$ .

Express your answer as a fraction.

3.

Given that $\sin\!\left(\dfrac{5\pi}{6}\right) = \dfrac{1}{2}$ ,
find $\sin\!\left(-\dfrac{5\pi}{6}\right)$ .

Express your answer as a fraction.

4.

Which statement about the tangent function is correct?

A.

$\tan(-x) = \tan(x)$ , so tangent is an even function.

B.

$\tan(-x) = -\tan(x)$ , so tangent is an odd function.

C.

Tangent is neither even nor odd because it has vertical asymptotes.

D.

$\tan(-x) = \tan(x)$ , so tangent is an odd function.

5.

Use the periodicity of cosine to evaluate.

Given that $\cos\!\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$ ,
find $\cos\!\left(\dfrac{\pi}{3} + 6\pi\right)$ .

Express your answer as a fraction.

6.

The tangent function has period $\pi$ .

Given that $\tan\!\left(\dfrac{\pi}{4}\right) = 1$ ,
which of the following also equals $1$ ?

A.

$\tan\!\left(\dfrac{\pi}{4} + 2\pi\right)$

B.

$\tan\!\left(\dfrac{\pi}{4} + \pi\right)$

C.

$\tan\!\left(\dfrac{\pi}{4} + \dfrac{\pi}{2}\right)$

D.

$\tan\!\left(-\dfrac{\pi}{4}\right)$

C

Varied Practice

D

Word Problems / Application

1.

A Ferris wheel rotates so that a rider's height above the ground follows $h(t) = \sin(t)$ (in suitable units), where $t$ is measured in radians.

The rider is at height $\sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$
at time $t = \dfrac{\pi}{6}$ .

At time $t = \dfrac{\pi}{6} + 4\pi$ , what is the rider's height?
Express your answer as a fraction.

2.

A physics experiment records the displacement of a pendulum as $d(\theta) = \tan(\theta)$ (approximate model for small angles). Because tangent has period $\pi$ , the displacement pattern repeats every $\pi$ radians of the drive angle.

The measured displacement at $\theta = \dfrac{\pi}{3}$ is
$\tan\!\left(\dfrac{\pi}{3}\right) = \sqrt{3}$ .

Find the displacement at $\theta = \dfrac{\pi}{3} + 5\pi$ .

Express your answer in simplest radical form.

3.

A student evaluates $\sin\!\left(-\dfrac{11\pi}{6}\right)$ using a chain of symmetry and periodicity steps.

1.

Step 1 — Apply the odd property of sine:

$\sin\!\left(-\dfrac{11\pi}{6}\right) = -\sin\!\left(\dfrac{11\pi}{6}\right)$ .

Now evaluate $\sin\!\left(\dfrac{11\pi}{6}\right)$ using the unit circle.
The angle $\dfrac{11\pi}{6}$ is in QIV with reference angle $\dfrac{\pi}{6}$ .

What is $\sin\!\left(\dfrac{11\pi}{6}\right)$ ?
Express your answer as a fraction.

2.

Step 2 — Complete the evaluation:

From Step 1, $\sin\!\left(\dfrac{11\pi}{6}\right) = -\dfrac{1}{2}$ .

Now use the odd property result from Step 1:
$\sin\!\left(-\dfrac{11\pi}{6}\right) = -\sin\!\left(\dfrac{11\pi}{6}\right)$ .

What is $\sin\!\left(-\dfrac{11\pi}{6}\right)$ ?
Express your answer as a fraction.

E

Error Analysis

1.

Alex is asked: "Use the unit circle to explain why cosine is an odd function."

Alex writes:
"Angle $x$ lands at $(a, b)$ and angle $-x$ lands at $(a, -b)$ .
The y-coordinates are opposite: $b$ and $-b$ . Since the outputs change sign,
$\cos(-x) = -\cos(x)$ , so cosine is odd."

What error did Alex make?

A.

Alex confused x- and y-coordinates. Cosine equals the x-coordinate ( $a$ ), not the y-coordinate ( $b$ ). Because both points share the same x-coordinate $a$ , we have $\cos(-x) = a = \cos(x)$ — cosine is even.

B.

Alex is correct; cosine outputs can be negative, which makes it odd.

C.

Alex should have used the y-axis reflection, not the x-axis, to prove the even/odd property.

D.

Alex should have tested a specific value, like $x = \pi/3$ , rather than using general coordinates $(a, b)$ .

2.

Jordan claims: "I can show that tangent has period $2\pi$ because
both sine and cosine have period $2\pi$ . Since $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$
and both repeat every $2\pi$ , tangent must also repeat every $2\pi$ .
That makes $2\pi$ the period of tangent."

Identify the flaw in Jordan's reasoning and state the correct period.

A.

Jordan is correct. Tangent has period $2\pi$ because it is built from sine and cosine, which both have period $2\pi$ .

B.

Jordan's logic is partially right — tangent does repeat every $2\pi$ — but the period is the smallest such value. Because both $\sin(x + \pi) = -\sin(x)$ and $\cos(x + \pi) = -\cos(x)$ , the negatives cancel in the ratio, giving $\tan(x + \pi) = \tan(x)$ . So the period is $\pi$ .

C.

Jordan is wrong because tangent does not repeat at $2\pi$ ; it only repeats at multiples of $\pi$ .

D.

Jordan is wrong because tangent is undefined at odd multiples of $\pi/2$ , so it cannot have a period.

F

Challenge / Extension

1.

Evaluate $\cos\!\left(-\dfrac{13\pi}{4}\right)$ by combining the
even property of cosine with periodicity.

Hint: Use $\cos(-x) = \cos(x)$ first, then subtract multiples
of $2\pi$ to find a reference angle.

Express your answer in simplest radical form.

2.

Consider the function $f(x) = \sin^2(x)$ .

(a) Is $f$ even, odd, or neither? Justify using the definition
$f(-x) =$ ? and the odd property of sine.
(b) What is the period of $f$ ? Justify your answer.

(Hint for part b: use the identity
$\sin^2(x) = \dfrac{1 - \cos(2x)}{2}$ .)

Symmetry and Periodicity of Trigonometric Functions

Recall / Warm-Up

Fluency Practice

Varied Practice

Word Problems / Application

Error Analysis

Challenge / Extension