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Algebra 2 View details

3. Graphing Trigonometric Functions

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eCourses /

Algebra 2 /

Chapter 6

Graphing Trigonometric Functions

This lesson focuses on teaching how to graph trigonometric functions like sine and cosine. It breaks down the process into understandable steps, emphasizing the importance of key elements such as amplitude, period, and frequency. The lesson is particularly useful for students who are studying Algebra 2 or higher-level math, as it provides a comprehensive look into the graphical representation of trigonometric equations. Understanding these graphs is crucial for various real-world applications, including engineering, physics, and even economics. By mastering these graphing techniques, you can better interpret and apply trigonometric functions in various scenarios.

Lesson Settings & Tools

	15 Theory slides
	9 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

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Many real-life problems can be solved by applying trigonometric functions, such as sine, cosine, and tangent. This lesson will define these functions, show how to graph them using their function rules, and explore some of their different real-life applications.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Modeling the Population of Rabbits and Foxes

On the weekend, Kriz and their family headed to the local zoo. Kriz really loves learning about animals, so they were sure to stop at all the cool information boards that teach interesting facts about them.

Of all the animals, Kriz likes foxes and rabbits the most. They were dying to learn more about them and discovered a table showing the state population of rabbits and foxes during the previous year.

The table on the zoo website contains the following information about the population of rabbits and foxes: m=0, r=1250, f=250; m=1, r=1000, f=243; m=2, r=817, f=225; m=3, r=750, f=200; m=4, r=817, f=175; m=5, r=1000, f=157; m=6, r=1250, f=150; m=7, r=1500, f=157; m=8, r=1683, f=175; m=9, r=1750, f=200; m=10, r=1683, f=225; m=11, r=1500, f=243

Analyzing the table more closely, Kriz arrived at some interesting conclusions. By answering the following questions, try to determine what Kriz discovered.

a What type of functions can be used to model the populations of rabbits and foxes?

b Find the appropriate function that models the population of rabbits

r (m)

as a function of the time

m

in months.

c Find the appropriate function that models the population of foxes

f (m)

as a function of the time

m

in months.

d Graph both functions. One function seems to chase the other. What can be a possible explanation for this?

Discussion

Trigonometric Functions

Trigonometric functions are functions that relate an input, which represents an acute angle of a right triangle, to a trigonometric ratio of two of the triangle's side lengths. The angle is usually measured in radians.

Trigonometric functions are also defined for angles

θ

that are not acute by using the appropriate reference angle

θ^{'} .

For input values that are not between

0

and

2 π,

the value of the coterminal angle that belongs to this interval is used instead. Therefore, trigonometric functions are periodic functions. Because of their close relation with the unit circle, trigonometric functions are also called circular functions. The domain of the sine and cosine functions is the set of all real numbers. Their range is the interval that goes from

- 1

1 .

Trigonometric Function	Domain	Range
$y = sin x$	All real numbers	$[- 1, 1]$
$y = cos x$	All real numbers	$[- 1, 1]$

The tangent, cotangent, secant, and cosecant functions are defined as rational functions that involve the sine and cosine functions. The domain of each function does not include values that would make their denominator zero.

Trigonometric Function	Ratio	Domain	Range
$y = tan x$	$tan x = \frac{sin x}{cos x}$	Real numbers except odd multiples of $\frac{π}{2}$	All real numbers
$y = cot x$	$cot x = \frac{cos x}{sin x}$	Real numbers except multiples of $π$	All real numbers
$y = sec x$	$sec x = \frac{1}{cos x}$	Real numbers except odd multiples of $\frac{π}{2}$	$(- \infty, - 1] \cup [1, \infty)$
$y = csc x$	$csc x = \frac{1}{sin x}$	Real numbers except multiples of $π$	$(- \infty, - 1] \cup [1, \infty)$

Now, one of the main trigonometric functions, the sine function, will be defined and examined more closely.

Concept

The Sine Function

Let $P$ be the point of intersection of the unit circle and terminal side of an angle in standard position. The sine function, denoted as $sin,$ can be defined as the $y -$ coordinate of the point $P .$

The graph of the sine function is called a sine curve.

The graph of the sine function y=sin(x) over the domain [-2pi,2pi].

Note that for $x$ in the interval $[- 2 π, 0]$ and in the interval $[0, 2 π]$ the graph looks exactly the same. This means that the sine function is a periodic function and its period is $2 π .$

$sin (θ + 2 π n) = sin θ$

Here, $n$ is any integer number. Consider the function $y = a sin b θ,$ where $a$ and $b$ are non-zero real numbers and $θ$ is measured in radians. With this information, the properties of the sine function can be defined.

Properties of $y = a sin b θ$
Amplitude	$∣ a ∣$
Number of cycles in $[0, 2 π]$	$∣ b ∣$
Period	$\frac{2 π}{∣ b ∣}$
Domain	All real numbers
Range	$[- ∣ a ∣, ∣ a ∣]$

Example

Vertical Displacement of a Buoy

Kriz is interested in a maritime topic and wants to become a sailor one day. Kriz often goes sailing with their father and loves watching how waves crash on the shore, how buoys bob up and down as waves go past, and how the sun slowly melts into the water on the horizon.

A buoy bobs up and down on the waves in the sea

Kriz was very surprised when they learned in a math lesson that the vertical displacement of the buoys with respect to the sea level at the nearest beach can be modeled by a trigonometric function.

y = 1.6 sin \frac{π}{2} t

Here,

y

is the vertical displacement in feet and

t

the time in seconds. What are the amplitude, period, and midline of this function?

Hint

In the general form $y = a sin b θ,$ the amplitude is $∣ a ∣$ and the period is $\frac{2 π}{∣ b ∣} .$

Solution

Start by recalling the general form of the sine function.

y = a sin b θ

This function has the following properties.

Properties of $y = a sin b θ$
Amplitude	$∣ a ∣$
Number of cycles in $[0, 2 π]$	$∣ b ∣$
Period	$\frac{2 π}{∣ b ∣}$

Now, analyze the given function and identify the values of the coefficients

a

and

b .

y = 1.6 sin \frac{π}{2} t ⇓ a = 1.6 and b = \frac{π}{2}

Since the value of

a

1.6,

the amplitude of the function is

∣ 1.6 ∣ = 1.6

feet. Recall that the midline of the parent sine function is the horizontal line

y = 0 .

Since the given function has not been translated up or down, its midline is also

y = 0 .

To calculate the period, substitute

\frac{π}{2}

for

b

into the expression

\frac{2 π}{∣ b ∣}

and evaluate.

\frac{2 π}{∣ b ∣}

Substitute

$b = \frac{π}{2}$

\frac{2 π}{∣ ∣ ∣ \frac{π}{2} ∣ ∣ ∣}

AbsPos

$∣ ∣ ∣ ∣ \frac{π}{2} ∣ ∣ ∣ ∣ = \frac{π}{2}$

\frac{2 π}{\frac{π}{2}}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

\frac{4 π}{π}

ReduceFrac

$\frac{a}{b} = \frac{a / π}{b / π}$

\frac{4}{1}

DivByOne

$\frac{a}{1} = a$

4

Therefore, the period of the function is

4 .

This means that every

4

seconds the curve repeats itself, which can be seen on the graph.

Discussion

The Cosine Function

Let $P$ be the point of intersection of the unit circle and terminal side of an angle in standard position. The cosine function, denoted as $cos,$ can be defined as the $x -$ coordinate of the point $P .$

The graph of the cosine function looks as follows.

The graph of cosine function (function equation) that has the domain:(-2pi,2pi)

Note that for $x$ in the interval $[- 2 π, 0]$ and in the interval $[0, 2 π]$ the graph looks exactly the same. This means that the cosine function is a periodic function and its period is $2 π .$

$cos (θ + 2 π n) = cos θ$

Here, $n$ is any integer number. Consider the function $y = a cos b θ,$ where $a$ and $b$ are non-zero real numbers and $θ$ is measured in radians. With this information, the properties of the cosine function can be defined.

Properties of $y = a cos b θ$
Amplitude	$∣ a ∣$
Number of cycles in $[0, 2 π]$	$∣ b ∣$
Period	$\frac{2 π}{∣ b ∣}$
Domain	All real numbers
Range	$[- ∣ a ∣, ∣ a ∣]$

Example

Operation of Submarines' Radars

Kriz visited the port on a day when a special exhibition was taking place where scientists explained how they use a submarine in ocean exploration. They learned that radars are used to monitor objects under the sea. Even more interesting, in operating radars, sine and cosine functions are involved.

External credits: @freepik, @upklyak

A wave signal received by a radar can be modeled by the following equation.

y = 3.5 cos 12 t

Here,

y

is the vertical displacement from the shooting point in centimeters and

t

is time in seconds. What are the amplitude, period, and midline of this function? Write the period in exact form.

Hint

In the general form $y = a cos b θ,$ the amplitude is $∣ a ∣$ and the period is $\frac{2 π}{∣ b ∣} .$

Solution

First, recall the general form of a cosine function.

y = a cos b x

This function has the following properties.

Properties of $y = a cos b θ$
Amplitude	$∣ a ∣$
Number of cycles in $[0, 2 π]$	$∣ b ∣$
Period	$\frac{2 π}{∣ b ∣}$

Next, examine the given function and identify the values of its coefficients

a

and

b .

y = 3.5 sin 12 t ⇓ a = 3.5 and b = 12

The value of

a

3.5,

which means that the amplitude of the function is

∣ 3.5 ∣ = 3.5

centimeters. Recall that the midline of the parent cosine function is

y = 0 .

The given function has not been translated vertically, so its midline is also

y = 0 .

These two pieces of information can be shown on a graph.

To calculate the period, substitute

12

for

b

into the expression

\frac{2 π}{∣ b ∣}

and simplify.

\frac{2 π}{∣ b ∣}

Substitute

$b = 12$

\frac{2 π}{∣ 1 2 ∣}

AbsPos

$∣ 12 ∣ = 12$

\frac{2 π}{1 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{π}{6}

Therefore, the period of the function is

\frac{π}{6},

which means that every

\frac{π}{6}

seconds the graph of the function repeats itself. This is illustrated in the graph.

Pop Quiz

Identifying Amplitude and Period of Sine and Cosine Functions

Consider the equation or the graph of a sine or cosine function. Identify its amplitude and period. If needed, round the answers to two decimal places.

Different graphs and equations of sine and cosine functions are shown

Discussion

Graphing the Sine and Cosine Functions

Sine and cosine functions can be graphed by closely analyzing their function rules and the graphs of their parent functions, which are

y = sin x

and

y = cos x .

As an example, consider the following function.

f (x) = 0.5 cos 2 x + 1

In order to graph it, there are four steps to follow.

Find the Amplitude, Period, and Translation of the Function

First, recall the general form of the cosine function and identify the value of each coefficient by comparing it with the given function.

y f (x) = a cos b θ = 0.5 cos 2 x + 1

The amplitude of the cosine function is

∣ a ∣

and the period is

\frac{2 π}{∣ b ∣} .

Therefore, the amplitude of the given function is

∣ 0.5 ∣ = 0.5 .

To find its period, substitute

2

for

b

into the corresponding expression.

\frac{2 π}{∣ 2 ∣} = π

The period of the function is

π .

Finally, if

y = 0.5 cos 2 x

is considered,

1

must be added to obtain the given function.

f (x) = 0.5 cos 2 x + 1

This means that the function is translated

1

unit upward.

Draw the Midline

The midline of the parent cosine function is $y = 0 .$ However, since the considered function is translated $1$ unit upward, its midline is also translated. This means that the equation of the midline is $y = 1 .$

The midline of y=1 on a coordinate plane

Plot Some Key Points on the Graph

Key points ranging over at least one cycle of the function should now be plotted. These key points are the maximums, minimums, and intersections with the midline. The maximums of

y = cos x

occur at even multiples of

π .

x = 0, 2 π, 4 π, \dots

The period of

f

π,

which is

\frac{1}{2}

of the parent function's period

2 π .

Also,

f

has not been translated horizontally, so the maximums neither shifted to right nor to the left. Therefore, the maximums of

f

occur at the following

x -

coordinates.

x = \frac{1}{2} (0), \frac{1}{2} (2 π), \frac{1}{2} (4 π), \dots ⇕ x = 0, π, 2 π, \dots

This means that the maximums of

f

occur at multiples of

π .

Since the midline is

y = 1

and the amplitude is

0.5,

these maximums will all have a

y -

value of

1 + 0.5 = 1.5 .

Now, plot the points on the graph with the midline.

The maximums (0,1.5), (pi,1.5), (2pi,1.5) are plotted on a coordinate plane

The minimums of

f

are horizontally located between the maximums.

x = 0.5 π, 1.5 π, 2.5 π, \dots

These points have

y -

values of

1 - 0.5 = 0.5 .

The minimums (0.5pi,0.5), (1.5pi,0.5) plotted on the coordinate plane

Lastly, in-between every neighboring maximum and minimum are the intersections with the midline.

x = 0.25 π, 0.75 π, 1.25 π, \dots

Since these points lie on the midline, their

y -

coordinate is

1 .

The intersections with the midline (0.25pi,1), (0.75pi,1), (1.25pi,1) are plotted on the coordinate plane

Draw the Graph

By connecting the plotted points with a smooth curve and continuing it periodically in both directions, the graph of the function can finally be drawn.

Extra

Formulas for the Key Points

There are formulas for the key points such as $x -$ intercepts, maximum value, and minimum value of a sine function of the form $y = a sin b x .$

	Formula
$x -$ intercepts	$(0, 0),$ $(\frac{1}{2} \cdot \frac{2 π}{b}, 0),$ $(\frac{2 π}{b}, 0)$
Maximum	$a > 0$ $(\frac{1}{4} \cdot \frac{2 π}{b}, a)$	$a < 0$ $(\frac{3}{4} \cdot \frac{2 π}{b}, - a)$
Minimum	$a > 0$ $(\frac{3}{4} \cdot \frac{2 π}{b}, - a)$	$a < 0$ $(\frac{1}{4} \cdot \frac{2 π}{b}, a)$

Similarly, there are also formulas for the $x -$ intercepts, maximum, and minimum of a cosine function of the form $y = a cos b x .$

	Formula
$x -$ intercepts	$(\frac{1}{4} \cdot \frac{2 π}{b}, 0),$ $(\frac{3}{4} \cdot \frac{2 π}{b}, 0)$
Maximum	$a > 0$ $(0, a), (\frac{2 π}{b}, a)$	$a < 0$ $(\frac{1}{2} \cdot \frac{2 π}{b}, - a)$
Minimum	$a > 0$ $(\frac{1}{2} \cdot \frac{2 π}{b}, - a)$	$a < 0$ $(0, a), (\frac{2 π}{b}, a)$

These formulas can be useful when graphing a sine or a cosine function. By using them, the first five points of a function can be plotted. Then, the function can be extended along the $x -$ axis by imitating the found pattern.

Extra

Graphing Parent Sine and Cosine Functions

The graph of the parent sine function can be obtained by using a unit circle. Recall that the sine values are represented by the

y -

coordinate of a point on this circle. Therefore, as the point is rotated, its

y -

coordinates will be plotted on a coordinate plane.

The graph of the parent cosine function can be drawn in a similar manner. The values of cosine are represented by the

x -

coordinates of a point on a unit circle. Rotate the point and plot its

x -

coordinates with the respective

θ -

values on a coordinate plane.

Example

Graphing Light Waves of Different Colors

After learning how trigonometric functions are abundant in objects related to the ocean, Kriz was stoked to go to Physics class first thing Monday. There, they learned that light travels in waves and, therefore, can be modeled by sine and cosine functions. Different colors have different wavelengths, or periods, and the amplitude of the wave affects the brightness of the color.

For example, the light visible as red has the longest period, while the light visible as violet has the shortest period. Additionally, the greater the amplitude of the light wave, the brighter it looks.

a Use a sine function to graph the dimmed red light wave with a period of

660

nanometers, an amplitude of

0.3

units, and whose midline is

y = 0.5 .

b Graph the bright violet light wave modeled by the following equation.

y = 1.4 cos \frac{π}{2 0 0} x

Answer

Hint

a First, plot the midline and then identify the locations of the maximums, minimums, and the intersections with the midline.

b Identify the values of

a

and

b

and use the fact the period is

\frac{2 π}{∣ b ∣} .

Analyze the locations of maximums, minimums, and intersections with the midline of the parent function of cosine.

Solution

a The first step is to graph the midline of the function, which is said to be

y = 0.5 .

Next, some key points, like maximums, minimums, and intersections with the midline should be plotted. The parent sine function intersects the midline at each half-period.

In this case, the period is

660,

so its half-period is

\frac{6 6 0}{2} = 330

nanometers. Therefore, the

x -

coordinates of the intersections of the function and the midline occur at

x -

values that are multiples of

330 .

x = 0, 330, 660, \dots

These points lie on the midline, so their

y -

coordinate is

0.5 .

The maximums and minimums of a sine function occur once every period between two points of intersection with the midline. Analyzing the graph of the parent sine function starting from the origin, it can be seen that the maximum of the function occurs before the minimum.

Therefore, the maximum of the given function is in the middle between the intersections

(0, 0.5)

and

(330, 0.5),

while the minimum is in the middle between

(330, 0.5)

and

(660, 0.5) .

By adding and subtracting the amplitude of

0.3

to the midline, the

y -

coordinates of the maximum and minimum, respectively, can be found.

Maximum (\frac{3 3 0}{2}, 0.5 + 0.3) ⇕ (165, 0.8) Minimum (330 + \frac{3 3 0}{2}, 0.5 - 0.3) ⇕ (495, 0.2)

Now, plot both points on the coordinate plane.

Finally, connect the points with a smooth curve and continue it periodically.

b To graph the given function, start by comparing it with the general form of a cosine function to identify the values of the coefficients.

General Form : Given Funtion : y = a cos b x y = 1.4 cos \frac{π}{2 0 0} x

Since the value of

a

1.4,

the amplitude of the function is

1.4

units. The value of

b,

which is

\frac{π}{2 0 0},

can be used to find the period of the function.

Period = \frac{2 π}{∣ b ∣}

Substitute

$b = \frac{π}{2 0 0}$

Period = \frac{2 π}{∣ ∣ ∣ \frac{π}{2 0 0} ∣ ∣ ∣}

▼

Evaluate right-hand side

AbsPos

$∣ ∣ ∣ ∣ \frac{π}{2 0 0} ∣ ∣ ∣ ∣ = \frac{π}{2 0 0}$

Period = \frac{2 π}{\frac{π}{2 0 0}}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

Period = \frac{4 0 0 π}{π}

ReduceFrac

$\frac{a}{b} = \frac{a / π}{b / π}$

Period = \frac{4 0 0}{1}

DivByOne

$\frac{a}{1} = a$

Period = 400

The midline of the parent cosine function is

y = 0 .

In the equation of the function, there is no value added to or subtracted from the cosine term, which means that the function is neither translated up nor down. Therefore, its midline is also

y = 0 .

Next, the key points should be identified and plotted. Consider the parent cosine function.

As can be seen, the maximums occur at

x = - 2 π, 0, 2 π, \dots,

which are the multiples of its period

2 π .

This means that using the period of

400

of the considered function, its maximums can be found.

Maximums x = 0, 400, 800, \dots y = 1.4

Furthermore, since the equation of the midline is

y = 0

and the amplitude is

1.4,

the

y -

coordinate of the maximums is

1.4 .

The minimums of the parent function occur at

x = - π, π, \dots,

which are the values of its half-period. In this case, the half-period of the function is

200 .

Minimums x = 200, 600, 1000, \dots y = - 1.4

The

y -

coordinates of the minimums are

- 1.4 .

Lastly, in-between every neighboring maximum and minimum are the intersections with the midline.

Intersections x = 100, 300, 500, \dots y = 0

Finally, plot all the found points on the coordinate plane with the midline.

By connecting the points with a smooth curve and continuing it periodically, the graph of the given function can be obtained.

Discussion

Frequency of a Periodic Function

The frequency of a periodic function is the number of cycles in a given unit of time. The frequency of a function's graph is the reciprocal of the function's period.

$Frequency = \frac{1}{Period}$

For example, if the period of a function is

2

seconds, then the frequency is

\frac{1}{2}

cycles per second. If the period is

\frac{1}{5}

seconds, then the frequency is

5

cycles per second. More examples can be seen in the following applet.

The graph of sine functions with different periods and frequencies are shown

When frequency is calculated per second, it is measured with a unit called hertz. For instance,

10

Hz means

10

times per second.

Example

Frequencies That Animals Can Hear

Later that day, Kriz was excitingly sharing their impressions with their classmate Zain about their visit to the zoo. Kriz told Zain that they were impressed to learn that elephants can hear frequencies $20$ times lower than humans, while mice can hear astronomically high frequencies, up to $70$ - $80$ kHz.

An elephant and a mouse with sound ways of low and high frequencies

a Write a sine function in the form

y = a sin b x,

with

a

and

b

as positive real numbers, that models a sound wave with a frequency of

10

Hz and an amplitude of

1

unit that elephants can hear.

b Write a cosine function in the form

y = a cos b x,

with

a

and

b

positive real numbers, that models the sound wave with a frequency of

75

kHz and an amplitude of

0.2

units that mice can hear.

Hint

a For a sine function with the form

y = a sin b x,

the amplitude is

∣ a ∣

and

b

can be used to find the period of the function.

b Use the formula relating the frequency and the period to calculate the period of the function. Be aware that

1 kHz

is the same as

1000 Hz .

Solution

a Start by recalling the general form of a sine function.

y = a sin b x

Here,

∣ a ∣

is the amplitude while

b

is the coefficient used to find the period of the function. It is given that the amplitude of the function is

1

unit. Therefore,

∣ a ∣ = 1 .

Because

a

is a positive number, it is known that

a = 1 .

y = 1 sin b x ⇓ y = sin b x

To find the value of

b,

use the fact that the function has a frequency of

10

Hz. This means that the function has

10

cycles per

1

unit of time. Now, review the formula that relates frequency and period.

Frequency = \frac{1}{Period}

Substitute

10

for frequency and solve the equation for the period.

Frequency = \frac{1}{Period}

Substitute

$Frequency = 10$

10 = \frac{1}{Period}

MultEqn

$LHS \cdot Period = RHS \cdot Period$

10 \cdot Period = 1

DivEqn

$LHS / 10 = RHS / 10$

Period = \frac{1}{1 0}

Next, use the relationship between the coefficient

b

and the period of the sine function.

Period = \frac{2 π}{∣ b ∣}

Because

b

is a positive number, it is known that

∣ b ∣ = b .

Period = \frac{2 π}{∣ b ∣} ⟶ b > 0 Period = \frac{2 π}{b}

Since the value of the period was already found, substitute it into the equation and calculate the value of

b .

Period = \frac{2 π}{b}

Substitute

$Period = \frac{1}{1 0}$

\frac{1}{1 0} = \frac{2 π}{b}

CrossMult

Cross multiply

1 \cdot b = 10 \cdot 2 π

Multiply

b = 20 π

Now that the value of

b

is known, the equation of the sine function describing the sound wave heard by elephants can be completed.

y = sin b x ⇓ y = sin 20 π x

Kriz once heard in a documentary that elephants can communicate when they are miles away from other elephants by listening to vibrations that travel through the ground. Kriz realized that this function could potentially describe a low-frequency sound wave that elephants can hear when communicating miles apart!

b Similarly to Part A, first recall the general form of a cosine function.

y = a cos b x

This time, the amplitude is

0.2

units. Since

a

is a positive number, the value of

a

0.2 .

y = a cos b x ⇓ y = 0.2 cos b x

It is also known that the frequency of the function is

75

kHz or

75000

Hz. Use this value to find the period of the function.

Frequency = \frac{1}{Period}

Substitute

$Frequency = 75000$

75000 = \frac{1}{Period}

MultEqn

$LHS \cdot Period = RHS \cdot Period$

75000 \cdot Period = 1

DivEqn

$LHS / 75000 = RHS / 75000$

Period = \frac{1}{7 5 0 0 0}

The period of a cosine function is also given by

\frac{2 π}{∣ b ∣} .

Because

b

is positive, the period of the function can be written as

\frac{2 π}{b} .

Therefore, the value of the period can now be used to calculate

b .

Period = \frac{2 π}{b}

Substitute

$Period = \frac{1}{7 5 0 0 0}$

\frac{1}{7 5 0 0 0} = \frac{2 π}{b}

CrossMult

Cross multiply

1 \cdot b = 75000 \cdot 2 π

Multiply

b = 150000 π

Finally, substitute the value of

b

and complete the equation of the cosine function that models the sound wave heard by the mice.

y = 0.2 cos b x ⇓ y = 0.2 cos 150000 π x

Kriz imagined a hundred different possibilities of how high-frequency sound waves help mice navigate the dark forests filled with animals out to eat them!

Explore

Investigating the Graphs of the Functions Involving Sine and Cosine

It is interesting to explore the graphs of functions that are defined by applying some basic operations, like addition, multiplication, and division, to sine and cosine functions. First, try to draw the graph of

f (x) = sin x + cos x .

Now, the graph of

f (x) = sin x cos x

will be drawn.

Finally, by applying the same method, try to draw the graph of

f (x) = \frac{s i n x}{c o s x} .

What function does the obtained graph resemble?

Discussion

The Tangent Function

Let $P$ be the point of intersection of the unit circle and terminal side of an angle in standard position. The tangent function, denoted as $tan,$ can be defined as the ratio of the $y -$ coordinate to the $x -$ coordinate of the point $P .$

Recall that the

x -

and

y -

coordinates of the point correspond to the cosine and sine of the angle, respectively. Therefore, the tangent function can also be defined as the ratio of

sin θ

cos θ .

$tan θ = \frac{sin θ}{cos θ}$

The graph of the tangent function is as follows.

The graph of the tangent function y=tan(x) with asymptotes at -3pi/2, -pi/2, pi/2, 3pi/2, and 5pi/2 over the domain [-3pi/2,5pi/2]

The period of the tangent function is $π .$ Since each branch comes from negative infinity towards positive infinity, the tangent function has no amplitude and its range is all real numbers. Consider the function $y = a tan b x,$ where $a$ and $b$ are non-zero real numbers and $θ$ is measured in radians. The properties of the tangent function can be identified from the function rule.

Properties of $y = a tan b x$
Amplitude	No amplitude
Interval of One Cycle	$(- \frac{π}{2 ∣ b ∣}, \frac{π}{2 ∣ b ∣})$
Asymptotes	At the end of each cycle
Period	$\frac{π}{∣ b ∣}$
Domain	All real numbers except odd multiples of $\frac{π}{2 ∣ b ∣}$
Range	All real numbers

Method

Graphing a Tangent Function

A tangent function can be graphed by examining its function rule and determining some of its key characteristics, like period and asymptotes. Consider the following function.

y = 1.5 tan \frac{π}{2} x

In order to draw the graph, there are four steps to follow.

Find the Period of the Function

Start by comparing the function with the general form of a tangent function to identify the value of its coefficients.

General Form : Given Function : y = a tan b x y = 1.5 tan \frac{π}{2} x

The period of a tangent function is given by

\frac{π}{∣ b ∣} .

Substitute

\frac{π}{2}

for

b

and calculate the period of the given function.

\frac{π}{∣ b ∣}

Substitute

$b = \frac{π}{2}$

\frac{π}{∣ ∣ ∣ \frac{π}{2} ∣ ∣ ∣}

AbsPos

$∣ ∣ ∣ ∣ \frac{π}{2} ∣ ∣ ∣ ∣ = \frac{π}{2}$

\frac{π}{\frac{π}{2}}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

\frac{2 π}{π}

CalcQuot

Calculate quotient

2

Therefore, the period of the function is

2

units.

Draw the Asymptotes

Recall that one cycle of a tangent function occurs in the interval from

- \frac{π}{2 ∣ b ∣}

\frac{π}{2 ∣ b ∣} .

Again, substitute the value of

b

and find the interval corresponding to the considered function.

\frac{π}{2 ∣ b ∣}

Substitute

$b = \frac{π}{2}$

\frac{π}{2 ∣ ∣ ∣ \frac{π}{2} ∣ ∣ ∣}

AbsPos

$∣ ∣ ∣ ∣ \frac{π}{2} ∣ ∣ ∣ ∣ = \frac{π}{2}$

\frac{π}{2 ( \frac{π}{2} )}

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

\frac{π}{π}

QuotOne

$\frac{a}{a} = 1$

1

One cycle of the function occurs in the interval

[- 1, 1] .

Furthermore, recall that the asymptotes occur at the end of each cycle of the function. Therefore, the function has asymptotes at

x = - 1

and

x = 1,

which can be drawn on a coordinate plane.

The asymptotes y=-1 and y=1 on a coordinate plane

Plot Some Points on the Graph

Next, divide the period of the function into four equal parts and locate three points between the asymptotes. In this case, the period is $2,$ so each fourth is $0.5$ units long.

Therefore, the $x -$ coordinates of the points that will be plotted are $- \frac{1}{2},$ $0,$ and $\frac{1}{2} .$ Substitute these values for $x$ into the function rule and evaluate the corresponding $y -$ coordinates.

$y = 1.5 tan \frac{π}{2} x$
$x -$ Coordinate	Substitute	Simplify	Evaluate
$x = - \frac{1}{2}$	$y = 1.5 tan \frac{π}{2} (- \frac{1}{2})$	$y = 1.5 tan (- \frac{π}{4})$	$y = - 1.5$
$x = 0$	$y = 1.5 tan \frac{π}{2} (0)$	$y = 1.5 tan 0$	$y = 0$
$x = - \frac{1}{2}$	$y = 1.5 tan \frac{π}{2} (\frac{1}{2})$	$y = 1.5 tan \frac{π}{4}$	$y = 1.5$

Now that both coordinates of the three points are known, plot them on the coordinate plane with the asymptotes.

Three points (-0.5,-1.5), (0,0), and (0.5,1.5) are plotted on the graph

Draw the Graph

Finally, connect the three points with a smooth curve and continue the function to the left and right keeping in mind that it should get closer to the asymptotes but will never intersect them.

The graph of the function for the cycle between -1 and 1

Replicate the branch to obtain the graph for other intervals.

Extra

Key Characteristics of a Tangent Function

When drawing one period of a function of the form $y = a tan b θ,$ the following characteristics of a tangent function can be used.

	Formula
$x -$ intercept	$(0, 0)$
Asymptotes	$θ = - \frac{π}{2 ∣ b ∣},$ $θ = \frac{π}{2 ∣ b ∣}$
Halfway Points	$(- \frac{π}{4 b}, - a),$ $(\frac{π}{4 b}, a)$

Extra

Graphing the Parent Tangent Function

Recall that a tangent function can be defined as the quotient of a sine and a cosine function. Therefore, the graph of the parent tangent function can be drawn as a rational function.

tan x = \frac{sin x}{cos x}

First, draw the graphs of the sine and cosine functions.

Because the tangent function has cosine in its denominator, the asymptotes of the tangent function are located at the zeros of the cosine function.

Since the sine function is the numerator of the tangent function, the zeros of the sine function will be the zeros of the tangent function as well.

The $y -$ values of the sine and cosine functions are equal at the intersection of both graphs. This means that the $y -$ value of the tangent function will be $1$ at the $x -$ coordinates of these points.

Likewise, three more points can be plotted between the left asymptotes of each period and the zeros. This time, however, since the sine and cosine functions have opposite $x -$ values, the $y -$ value of the tangent function will be $- 1 .$

Finally, the branches are drawn by starting from the bottom of the left asymptote and moving towards the top of the right asymptote.

Example

Modeling a Distance with a Tangent Function

After school, Kriz and Zain will meet some friends to play volleyball near Zain's home. Zain lives in a modern $300 -$ feet building. As Kriz was approaching the building about $175$ feet from its base, they saw Zain going down in the elevator and waved to them.

Kriz on the way to Zain seeing Zain in the elevator; city skyline in on the background

External credits: @upklyak

a Write an equation that models Zain's distance

d

in feet from the top of the building as a function of the angle of elevation

x .

b Graph one cycle of the function from Part A ignoring the limitations of the context.

c What part of the graph makes sense considering the given context?

Answer

d = 300 - 175 tan x

c The part in Quadrant I.

Hint

a Which trigonometric function relates the legs of a right triangle with an acute angle? Set up an equation and solve it for

d .

b Use the fact that the period of a tangent function is given by

\frac{π}{∣ b ∣}

and one cycle occurs in the interval from

- \frac{π}{2 ∣ b ∣}

\frac{π}{2 ∣ b ∣} .

c Recall what

x

and

d

represent. What values cannot they have?

Solution

a Begin by analyzing the triangle formed by Kriz, Zain and the base of the building.

External credits: @upklyak

It is a right triangle in which the side that measures

175

feet is adjacent

∠ x .

The side representing the distance between Zain and the base of the building equals

300 - d

and is opposite

∠ x .

Therefore, the tangent ratio can be used to find a relationship between the sides and the angle.

tan x = \frac{opp}{adj} \Rightarrow tan x = \frac{3 0 0 - d}{1 7 5}

To find the equation that models Zain's distance

d

from the top of the building, solve this equation for

d .

tan x = \frac{3 0 0 - d}{1 7 5}

MultEqn

$LHS \cdot 175 = RHS \cdot 175$

175 tan x = 300 - d

AddEqn

$LHS + d = RHS + d$

d + 175 tan x = 300

SubEqn

$LHS - 175 tan x = RHS - 175 tan x$

d = 300 - 175 tan x

b The function found in Part A can be seen as a vertical translation of the function

y = - 175 tan x

300

units up.

d = 300 - 175 tan x

Therefore, to obtain the graph of

d,

first graph

y = - 175 tan x

and then translate it vertically. Start by comparing the function to the general form of a tangent function to identify the values of the coefficients

a

and

b .

General Form : Given Function : y = a tan b x y = - 175 tan 1 x

Recall that the period of the parent tangent function is given by

\frac{π}{∣ b ∣} .

Since

b = 1,

the period of the function can be calculated.

Period \frac{π}{∣ b ∣} \Rightarrow \frac{π}{∣ 1 ∣} = π

One cycle of a tangent function occurs in the interval from

- \frac{π}{2 ∣ b ∣}

\frac{π}{2 ∣ b ∣} .

Substitute the value of

b

and find this interval for the given function.

(- \frac{π}{2 ∣ 1 ∣}, \frac{π}{2 ∣ 1 ∣}) \Leftrightarrow (- \frac{π}{2}, \frac{π}{2})

Additionally, the asymptotes occur at the end of each cycle. This means that two asymptotes are

x = - \frac{π}{2}

and

x = \frac{π}{2} .

Graph them on a coordinate plane.

Next, three points should be plotted between the asymptotes. By dividing the interval between

- \frac{π}{2}

and

\frac{π}{2}

into four equal parts, the

x -

coordinates of three points are obtained.

x = - \frac{π}{4}, x = 0, x = \frac{π}{4}

Substitute them into the function rule for

x

and find the corresponding

y -

coordinates.

$y = - 175 tan x$
$x -$ Coordinate	Substitute	Evaluate
$x = - \frac{π}{4}$	$y = - 175 tan (- \frac{π}{4})$	$y = 175$
$x = 0$	$y = - 175 tan 0$	$y = 0$
$x = \frac{π}{4}$	$y = - 175 tan (\frac{π}{4})$	$y = - 175$

Now, plot the points on the coordinate plane with the asymptotes.

Finally, the graph of $y$ can be drawn by connecting the points with a smooth curve such that, as $x$ tends to $- \frac{π}{2}$ and $\frac{π}{2},$ the graph gets closer and closer to the asymptotes but never crosses them.

The last step is to translate the graph $300$ units up vertically to obtain the graph of $d .$

c To determine which part of the graph makes sense in the context of the given situation, recall that

d

represents the distance between Zain and the top of the building, while

x

is the measure of the angle of elevation. In real life, the values of these two variables can only be greater than or equal to

0 .

Therefore, only the part of the graph in the first quadrant makes sense considering the context.

Closure

Drawing Conclusions About the Population of Rabbits and Foxes

Earlier, it was mentioned that on the weekend Kriz went with their family to the local zoo.

Kriz with their sister and mom at the zoo entrance where they can see a giraffe and an elephant

External credits: @storyset

Of all the great animals at the zoo, Kriz likes foxes and rabbits the most. Eager to learn more information about them, Kriz found the following table that shows the state population of rabbits and foxes from the previous year.

After analyzing the table closely, Kriz arrived at some interesting conclusions. By answering the following questions, try to determine what Kriz found.

a What type of functions can be used to model the populations of rabbits and foxes?

b Find the appropriate function to model the population of rabbits

r (m)

as a function of the time

m

in months.

c Find the appropriate function to model the population of foxes

f (m)

as a function of the time

m

in months.

d Graph both functions. One function seems to chase the other. What can be a possible explanation for this?

Answer

a Sine or cosine functions.

r (m) = - 500 sin \frac{π}{6} m + 1250

f (m) = 50 cos \frac{π}{6} m + 200

d Graph of

r (m)

Graph of $f (m)$ :

Explanation: A predator-prey relationship between foxes and rabbits.

Hint

a First, analyze the values in the table and try to find patterns. Then, plot the points and examine the shape of the curve that connects them.

b Use the graph and the values from the table to determine the amplitude, midline, and period of the function.

c The equation of the midline indicates the number of units by which the function has to be translated vertically.

d Connect the given and plotted points with a smooth curve. Analyze what happens to the population of foxes when the population of rabbits increases and decreases, and vice versa.

Solution

a Begin by analyzing the values of the rabbit population given in the table. Are there any patterns?

It can be seen that some values, like

1250,

1000,

817,

1500,

and

1683,

repeat in the table twice. This means that the function

r (m)

has some kind of local symmetries. Now, plot all the points on a coordinate plane and examine the shape of the graph.

The points remind the shape of a sine or a cosine graph. Similarly, analyze and try to find any patterns in the values of the fox population.

Again, some values in the table repeat, which implies that the function has some kind of symmetry. The number of foxes starts at a maximum value of

250,

then decreases to a minimum of

150,

followed by almost returning to the maximum by the end of the year. By plotting the points, the shape of the function can be seen.

Both functions have a general shape of a sine or a cosine function.

b Start by examining the shape of the graph in order to determine some of its characteristics, like midline, period, and amplitude.

On the graph almost one full cycle of the function is shown and it can be assumed that it will repeat periodically moving to the right along the

x -

axis. By analyzing the graph's pattern, it can be concluded that for

x = 12,

the function will have a value of

1250,

the same value as for

x = 0 .

Therefore, the function has a period of

12

months.

Period 12 months

Additionally, the maximum value is

1750,

while the minimum is

750 .

By calculating the mean of these values, the amplitude can be found.

Amplitude \frac{1 7 5 0 - 7 5 0}{2} = 500

Next, by subtracting the value of the amplitude from the maximum, or by adding its value to the minimum, the equation of the midline can be determined.

Midline y = 1250

Show all of these characteristics on the graph and try to identify which function, sine or cosine, can represent it.

The above graph resembles the graph of a sine function. However, after intersecting the midline at

m = 0,

the graph goes down to its minimum instead of going up to its maximum. This leads to the conclusion that the graph is a reflection of a sine function over the midline.

General Form : Reflection : y = - a sin b m y = - a sin b m

In this general form,

a

represents the amplitude and

b

represents the coefficient related to the period by the following formula.

Period = \frac{2 π}{∣ b ∣}

Substitute the known value of the period and calculate the value of

b .

Period = \frac{2 π}{∣ b ∣}

▼

Substitute

12

for

Period

and evaluate

Substitute

$Period = 12$

12 = \frac{2 π}{∣ b ∣}

MultEqn

$LHS \cdot ∣ b ∣ = RHS \cdot ∣ b ∣$

12 ∣ b ∣ = 2 π

DivEqn

$LHS / 12 = RHS / 12$

∣ b ∣ = \frac{2 π}{1 2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

∣ b ∣ = \frac{π}{6}

Note that coefficients

a

and

b

can have both positive and negative values. For simplicity, only the positive values will be considered.

a = 500 b = \frac{π}{6}

Now that the values of the both coefficients are known, substitute them into the equation.

y = - a sin b m ⇓ y = - 500 sin \frac{π}{6} m

Furthermore, the midline at

y = 1250

indicates that the function has been translated

1250

units up. This fact can be shown in the function rule by adding

1250

to the right-hand side.

r (m) = - 500 sin \frac{π}{6} m + 1250

The function that models the population of rabbits during the previous year was obtained using the graph and the values from the table, making it possible to determine the needed characteristics.

c Similar to Part B, the function that models the population of foxes can be written. Again, begin by analyzing the shape of the graph and try to draw some conclusions about it.

Since almost one full cycle of the function is shown, it can be assumed that it will repeat periodically moving to the right along the

x -

axis. Also, it can be concluded that for

x = 12,

the function will have the value

250,

the same value as for

x = 0 .

This means that the function has a period of

12

months.

Period 12 months

The maximum value is

y = 250,

while the minimum is

150 .

Calculate the mean of these values to find the amplitude of the function.

Amplitude \frac{2 5 0 - 1 5 0}{2} = 50

Now, by subtracting the value of the amplitude from the maximum, or by adding its value to the minimum, the equation of the midline can be written.

Midline y = 200

Show all of these characteristics on the graph and try to identify which function, sine or cosine, can represent it.

The diagram reminds the graph of a cosine function. Recall the general form of a cosine function.

y = a cos b m

Here,

a

represents the amplitude and

b

represents the coefficient related to the period by the following formula.

Period = \frac{2 π}{∣ b ∣}

Since for this function period is also

12

months, the value of

∣ b ∣

\frac{π}{6} .

Taking into account only the positive values of

a

and

b,

substitute them into the equation.

y = a cos b m ⇓ y = 50 cos \frac{π}{6} m

Furthermore, the midline at

y = 200

indicates that the function has been translated

200

units up. This fact can be shown in the function rule by adding

200

to the right-hand side.

f (m) = 50 cos \frac{π}{6} m + 200

The function that models the population of foxes during the previous year was obtained using the graph and the values from the table, making it possible to determine the needed characteristics.

d To graph

r (m),

connect the given points plotted in Part A with a smooth curve.

Similarly, by connecting the points given for the function $f (m),$ its graph can be drawn.

The fox population seems to be behind the rabbit population by three months. In other words, the fox population chases the rabbit population. This can be explained by the predator-prey relationship that exists between foxes and rabbits, as rabbits are a source of food for foxes. Where there is a big population of rabbits, the fox population grows.

Many rabbits \to Foxes ↑

At the same time, when there are a lot of foxes, more rabbits are being hunted, so the rabbit population decreases.

Many foxes \to Rabbits ↓

When the rabbit population gets quite small, foxes have a limited food supply and, therefore, their population decreases.

Few rabbits \to Foxes ↓

Once the fox population is small enough, the rabbit population can recover, and this cycle continues.

Few foxes \to Rabbits ↑

Level 1

Level 2

Level 3

Consider the following trigonometric function.

y = 7.1 sin 11 t

What is the amplitude of the function

y ?

Write the midline of the function

y .

What is the period of the function

y ?

To find the amplitude of the given function, let's recall the general form of the sine function. y=asin bx This function has the following properties.

Properties of y=asinbθ
Amplitude	\|a\|
Number of cycles in [0,2π]	\|b\|
Period	2π/\|b\|

Now we will examine the given function to identify the values of its coefficients a and b. y = 7.1 sin 11 t ⇓ a= 7.1 and b= 11 We can see that the value of a is 7.1, which means that the function has an amplitude of |7.1|=7.1.

Recall that the midline of the parent sine function changes when the function is vertically translated. Let's recall how a vertical translation of the sine function looks. y=asin bx + h The constant h above indicates that the function is vertically translated h units. Now let's compare it to our given function. y = 7.1 sin 11t + 0 Looking at the given function, we can see that it has not been translated vertically, so its midline is the same as the parent sine function, y=0.

To calculate the period, we need to substitute 11 for b into the expression 2π|b| and simplify. Let's do it!

The period of the function is 2π11. This means that the graph of the function repeats itself every 2π11 units.

Use a reference angle to find the exact value of the following trigonometric function.

sin 58 5^{\circ}

We want to find the exact value of sin 585^(∘). To do so, let's start by recalling some trigonometric values for special angles.

Trigonometric Values for Special Angles
Sine	Cosine	Tangent	Cosecant	Secant	Cotangent
sin 30^(∘)=1/2	cos 30^(∘)=sqrt(3)/2	tan 30^(∘)=sqrt(3)/3	csc 30^(∘)=2	sec 30^(∘)=2sqrt(3)/3	cot 30^(∘)=sqrt(3)
sin 45^(∘)=sqrt(2)/2	cos 45^(∘)=sqrt(2)/2	tan 45^(∘)=1	csc 45^(∘)=sqrt(2)	sec 45^(∘)=sqrt(2)	cot 45^(∘)=1
sin 60^(∘)=sqrt(3)/2	cos 60^(∘)=1/2	tan 60^(∘)=sqrt(3)	csc 60^(∘)=2sqrt(3)/3	sec 60^(∘)=2	cot 60^(∘)=sqrt(3)/3

As we can see, the value we need is not on this table. To find the value, let's graph θ =585^(∘) in standard position so that we can find its reference angle. Be aware that the measure of our angle is greater than 360^(∘), so the angle is greater than a full turn.

If the measure of θ is greater than 360^(∘) or less than 0^(∘), we use a coterminal angle with a positive measure between 0^(∘) and 360^(∘) to find the reference angle. To do so, we will subtract 360^(∘) from the measure of the given angle.

Now we can find the reference angle θ ' of θ =585^(∘). This way we can use the values from our table. Note that the terminal side of 225^(∘) lies in Quadrant III. Therefore, to find its reference angle θ', we will subtract 180^(∘) from 225^(∘).

We found that θ '=45^(∘). Next, we will recall the signs of the six trigonometric functions in the different quadrants of the coordinate plane.

In Quadrant III — the quadrant where the terminal side of the angle is located — the sine is negative. With this information, we can write an equation relating the sine of the angle and the sine of its reference angle. sin 585^(∘) = - sin 45^(∘) Using our table, we can see that sin 45^(∘)= sqrt(2)2. sin 585^(∘) = - sin 45^(∘) ⇒ sin 585^(∘) =- sqrt(2)/2

The following point belongs to the unit circle.

P (\frac{2}{4}, \frac{- 1 4}{4})

The terminal side of an angle

θ

in standard position passes through

P .

Find

cos θ .

Find

sin θ .

A unit circle is a circle of radius 1 centered at the origin on the coordinate plane. The cosine of an angle in standard position whose terminal side intersects the unit circle at point P is the first coordinate of P.

We are told that the terminal side of angle θ in standard position goes through P and that P lies on the unit circle. Therefore, the terminal side of the angle intersects the unit circle at P( sqrt(2)4, -sqrt(14)4). Using this information, we can identify the cosine of θ. Let's do it! cos θ= sqrt(2)/4

We will proceed in the same fashon as we did in Part A. However, the sine of an angle in standard position whose terminal side intersects the unit circle at point P is the second coordinate of P. Let's identify the sine of θ! sin θ= -sqrt(14)/4

Let's draw a graph to represent the given situation. Be aware that P is located in Quadrant IV.

Consider the following graph of a sine function $y (x) .$

The values of $x$ are angles in radians.

What is the value of

y (π) ?

What is the value of

y (- π) ?

What is the value of

y (\frac{3 π}{2}) ?

We want to use the graph of the sine function to find the value of y(x) for x =π radians. Let's find the point on the curve with an x-coordinate of π and look for its corresponding y-coordinate.

We can see that the value of y(x) for x = π radians is 2.

This time we will look at the graph of y(x) to find the value of y(x) for x =-π radians. Let's find the point on the curve with an x-coordinate of -π and look for its corresponding y-coordinate.

We can see that the value of y(x) for x = -π radians is also 2.

Finally, we will examine the graph of y(x) to find the value of y(x) for x = 3π2 radians. Let's find the point on the curve with an x-coordinate of 3π2 and look for its corresponding y-coordinate.

We can see that the value of y(x) for x = 3π2 radians is 0. We did it!

Consider the following graph of a sine function.

What is the number of cycles in the interval from

0

2 π ?

Find the amplitude of the function.

Find the period of the function.

We want to use the given graph to determine the number of cycles that the sine function has in the interval from 0 to 2π. To determine the number of cycles, we need to identify when the shape of the graph starts repeating. Let's do it!

Looking at the graph, we can see that the graph repeats two times for the values between 0 and 2π. Therefore, there are two cycles in this interval.

Now we need to find the amplitude of the function. Let's review the definition of this term before we look at the graph.

Amplitude |- The amplitude is half the difference of the maximum and minimum values of a periodic function.

Let's identify the minimum and the maximum of the function in the graph!

We can see that the maximum and minimum values are 3 and - 3, respectively. Now we can find the amplitude!

The period is the horizontal length of one cycle. In Part A we found that there are two cycles in the interval from 0 to 2π. Therefore, we can find the period by dividing 2π by 2. Let's do it! Period = 2π/2 ⇒ Period = π

Graphing Trigonometric Functions

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