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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

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