5. Transformations of Trigonometric Functions and Their Graphs
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Chapter 6
5.
Transformations of Trigonometric Functions and Their Graphs
This lesson delves into the intriguing subject of trigonometric functions, specifically focusing on how these functions can be transformed. It explains how parameters like amplitude and period can significantly alter the graphs of sine, cosine, and tangent functions. These transformations are not just theoretical; they have practical applications too. For instance, understanding these concepts can help in modeling the movement of amusement park rides or predicting natural phenomena like tides. The information is designed to be accessible, offering visual aids to help grasp these somewhat complex mathematical ideas.
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Lesson Settings & Tools
| | 14 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Transformations of Trigonometric Functions and Their Graphs
Slide of 14
This lesson will introduce the transformations of the graphs of the sine, cosine, and tangent function. These trigonometric functions will be used to model the movement of a variety of attractions in an amusement park. Their transformations will be determined by comparing the parameters of the function rules with their parent functions.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Topics Related to Transformations of Functions
- Translation of a Function
- Horizontal Stretch and Shrink of a Function
- Vertical Stretch and Shrink of a Function
- Reflection of a Function
Topics Related to Trigonometric Functions
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Stretching and Shrinking a Graph
The graphs of sine, cosine, and tangent functions are shown in the following applet. Investigate how their graphs are transformed as the values a and b change.
Think about the following questions!
- How is each graph transformed as the value of a changes?
- How is each graph transformed as the value of b changes?
Discussion
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Stretching and Shrinking Trigonometric Functions
Like other functions, the parent functions of sine, cosine, and tangent can be transformed. The graphs of y=asin (bx), y=acos (bx), and y=atan (bx) represent stretch or shrink transformations of their parent functions.
Vertical Shrink or Stretch
The graph of a parent trigonometric function can be vertically stretched or shrunk by multiplying the function rule by a constant, positive number a. If a>1, the function will be stretched. Conversely, the function will be shrunk if 0< a<1.
| Vertical Stretch or Shrink | ||
|---|---|---|
| Parent Function | a>1 | 0 |
| y=sin x | y= asin x Stretch parent function vertically by a factor of a |
y= asin x Shrink parent function vertically by a factor of a |
| y=cos x | y= acos x Stretch parent function vertically by a factor of a |
y= acos x Shrink parent function vertically by a factor of a |
| y=tan x | y= atan x Stretch parent function vertically by a factor of a |
y= atan x Shrink parent function vertically by a factor of a |
Horizontal Shrink or Stretch
The graph of a trigonometric function can be horizontally stretched or shrunk by multiplying the input of the function by a positive number b. If b>1, the graph shrinks horizontally by a factor of b. Conversely, if 0< b<1, the graph stretch horizontally by a factor of b.
| Horizontal Stretch or Shrink | ||
|---|---|---|
| Parent Function | 0 | b>1 |
| y=sin x | y=sin ( bx) Stretch parent function horizontally by a factor of b |
y=sin ( bx) Shrink parent function horizontally by a factor of b |
| y=cos x | y=cos ( bx) Stretch parent function horizontally by a factor of b |
y=cos ( bx) Shrink parent function horizontally by a factor of b |
| y=tan x | y=tan ( bx) Stretch parent function horizontally by a factor of b |
y=tan ( bx) Shrink parent function horizontally by a factor of b |
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Ferris Wheel
Dominika and her best friend Paulina go to an amusement park to have fun after school. They first decide to ride on the Ferris wheel.
The following sine function models the height h in meters of their seat above the ground after the time t in seconds.
h=20 sin (0.16 t)
In the given function rule, the value of a is greater than 1, which represents a vertical stretch by a factor of 20. The value of b is less than 1 and greater than 0, which represents a horizontal stretch by a factor of 0.16. This information can be used to transform the graph of the parent sine function to h= 20 sin ( 0.16t).
a Identify the amplitude and period of this function. Round the answer to one decimal place if necessary.
b Identify the transformations done on the parent sine function and then graph the given function.
Answer
a Amplitude: 20
Period: 39.3
Period: 39.3
b Transformations: A vertical stretch by a factor of 20 and a horizontal stretch by a factor of 0.16 of the graph of the parent sine function.
Graph:
Hint
a In a function in the form y=a sin(bx), |a| represents the amplitude and the period is given by 2π|b|.
b Identify the value of a to see if the function was vertically transformed. Then look at the value of b to see if a horizontal transformation was applied.
Solution
a Notice that the given function is in the form of y=a sin(bx).
p=2π/|b|
Substitute
b= 0.16
p=2π/| 0.16|
AbsPos
|0.16|=0.16
p=2π/0.16
UseCalc
Use a calculator
p=39.269908 ...
RoundDec
Round to 1 decimal place(s)
p≈ 39.3
b To graph the function, the types of the transformations applied to the parent sine function should be identified.
y= sin x ⇒ h= 20 sin ( 0.16t) Note that the variables need to be changed from (x,y) to (t,h). The axis names of the coordinate plane also need to be changed. Consider the table to review what the values of a and b represent.
| Transformations of h=sin t to h=a sin (bt) | |
|---|---|
| Vertical Stretch or Shrink | If a>1, h= sin t stretches vertically by a h= a sin t |
| If 0< a< 1, h= sin t shrinks vertically by a h= a sin t | |
| Horizontal Stretch or Shrink | If b>1, h= sin t shrinks horizontally by b h=sin ( bt) |
| If 0< b< 1, h= sin t stretches horizontally by b h=sin ( bt) | |
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Townscape
After going on the Ferris wheel, Dominika and Paulina go up to the big red triangular tower to see the townscape from the viewing deck.
Paulina wonders how to find the height of the tower. Dominika says that the height y, in meters, of a tower like this one can be modeled by the following tangent function, where θ is the angle indicated. y=10 tan (1.5θ)
a Identify the amplitude, period, and asymptotes of the tangent function that represents the height of the towers. Round the answer to one decimal place if necessary.
b Apply the appropriate transformations to the parent tangent function to graph the given function.
c Find the height of the tower if θ=50 ^(∘). Round the answer to two decimal places.
Answer
a Amplitude: No amplitude
Period: 2π3
b Graph:
c 37.32 meters
Hint
a Recall the formulas for the period and asymptotes of a tangent function written in the form y=a tan(bx).
b What do the values of a and b represent?
c Substitute the given angle into the given tangent function.
Solution
a The amplitude, period, and asymptotes of the given tangent function will be identified one at a time. Start by identifying whether the tangent function is given in the form y=atan(b θ).
y= atan( b θ) ⇓ y= 10 tan ( 1.5θ) Remember that a tangent function has no amplitude because it has no maximum and minimum values. Now recall the formulas for the period and asymptotes of a tangent function.
| y=a tan(b θ) | |
|---|---|
| Period | π/|b| |
| Asymptotes | π/2|b| |
Next, substitute b=1.5 into the corresponding formulas to identify the period and asymptotes. Period:& π/|1.5| ⇒ 2π/3 [0.8em] Asymptote:& π/2|1.5| ⇒ π/3 The period of the function is 2π3. Recall that the period is the distance between any two consecutive vertical asymptotes. Since there is no translation, the branch that passes through the origin has one asymptotes at π3 and the other at 2π3 units in the negative direction. Second Asymptote: π/3-2π/3=-π/3 The second asymptote of this branch is located at - π3. The next asymptotes can be found by moving 2π3 units to the left or right. Positive Asymptotes:& +2π/3 ⇒ π, 5π/3, 7π/3... [0.8em] Negative Asymptotes:& -2π/3 ⇒ -π, -5π/3, -7π/3... In other words, the asymptotes can be located at the odd multiples of π3.
b To graph the given tangent function, recall what the values of a and b represent for y=a tan(b θ).
| Transformations of y=tan θ to y=a tan (b θ) | |
|---|---|
| Vertical Stretch or Shrink | If a>1, y= tan θ stretches vertically by a y= a tan θ |
| If 0< a< 1, y= tan θ shrinks vertically by a y= a tan θ | |
| Horizontal Stretch or Shrink | If b>1, y= tan θ shrinks horizontally by b y=tan ( bθ) |
| If 0< b< 1, h= sin θ stretches horizontally by b h=sin ( bθ) | |
b This time the height of the tower will be found for the given angle θ. To do so, θ=50 ^(∘) will be substituted into the given tangent function, which will then be solved for y.
y=10 tan(1.5 θ)
Substitute
θ= 50 ^(∘)
y=10 tan(1.5 ( 50^(∘)))
Multiply
Multiply
y=10 tan(75^(∘))
UseCalc
Use a calculator
y=37.320508 ...
RoundDec
Round to 2 decimal place(s)
y=37.32
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Translating a Trigonometric Function Graph
Observe how the graphs of sine, cosine, or tangent functions are translated horizontally and vertically by changing the values of h and k.
Consider the following questions!
- How does each graph change when the value of h changes?
- How does each graph change when the value of k changes?
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Translations of Trigonometric Functions
Trigonometric functions can be translated vertically or horizontally like the other functions. Next, these translations will be examined one at a time.
Vertical Translation
Let f(x) be a parent trigonometric function. Then, f(x)+k will translate the parent function vertically. If k>0, then the graph moves k units up. However, if k<0, then the graph moves k units down.
| Vertical Translations | ||
|---|---|---|
| Parent Function | k>0 | k<0 |
| y=sin x | y=sin x+ k, Translation k units up |
y=sin x -k, Translation k units down |
| y=cos x | y=cos x+ k, Translation k units up |
y=cos x -k, Translation k units down |
| y=tan x | y=tan x+ k, Translation k units up |
y=tan x -k, Translation k units down |
Horizontal Translation, or Phase Shift
A horizontal translation of a periodic function is called a phase shift. The graph of f(x−h) represents a horizontal translation of f(x) by h units. For example, consider the parent functions of sine, cosine, and tangent functions. If h>0, the parent trigonometric function will be shifted to the right h units, while if h<0, the function will be shifted to the right h units.
| Horizontal Translations | ||
|---|---|---|
| Parent Function | h>0 | h<0 |
| y=sin x | y=sin (x- h) Translation h units to the right |
y=sin (x +h) Translation h units to the left |
| y=cos x | y=cos (x- h) Translation h units to the right |
y=cos (x +h) Translation h units to the left |
| y=tan x | y=tan (x- h) Translation h units to the right |
y=tan (x +h) Translation h units to the left |
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Carousel
Now Dominika and Paulina are waiting to ride the carousel in the amusement park. The girls enjoy the up and down movement of wooden horses.
Paulina thinks that the movement of a rider on a horse looks like a cosine function. While talking to the operator of the ride, the girls learn that each horse has a special route. They choose horses whose routes can be represented by the following functions. Dominika: & y= cos x Paulina: & y= cos (x+π)-1
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b Select the translations which should be applied to the parent cosine function in order to get the function that represents the route of Paulina's horse.
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c Which of the following is the graph of y=cos(x+π)-1?
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Hint
a In a function in the form y=acosb(x-h)+k, the amplitude is given by |a| and the period by 2πb.
b What do the values of h and k represent?
c Translate the parent function using the values of h and k to draw the graph of the given function.
Solution
a Start by recalling how to identify the amplitude and period of a cosine function in the case that it is written in the form y=a cos b(x-h)+k.
The functions have an amplitude of 1 and a period of 2π each.
Amplitude: |a| [0.5em] Period: 2π/|b| Using this information, rewrite the given functions to identify the values of a and b.
| Write in the Form y=a cos b(x-h)+k | Amplitude: |a| | Period: 2π/|b| | |
|---|---|---|---|
| y= cos x | y= 1 cos 1(x- 0)+ 0 | | 1|= 1 | 2π/| 1|=2 π |
| y=cos(x+π)-1 | y= 1 cos 1(x-( -π))+( -1) | | 1|= 1 | 2π/| 1|=2 π |
b In a function in the form y=acosb(x-h)+k, h represents a horizontal translation and k a vertical translation. To identify the transformations applied to the parent cosine function, consider the rewritten form of the function representing the route of Paulina's horse.
y=acosb(x-h)+k ⇓ y= 1 cos 1(x-( -π))+( -1) Note that h= -π and k= -1. Because h<0, the parent function has been translated π units to the left. Next, since k<0, the graph has been translated 1 unit down.
c Now the parent cosine function will be translated π units to the left and 1 unit down to get the function y=cos(x+π)-1.
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Roller Coaster
After riding on the carousel, Dominika and Paulina are looking for something more exciting when they hear the screams coming from the roller coaster. They watch the roller coaster for a while, then see the section where people are most excited and scream the loudest.
The track in this section is quite steep. The girls do some quick math and determine that the path of the roller coaster can be modeled by the following tangent function. y=tan (x-3π)+1
a Identify the period and asymptotes of the given function that represents the most exciting section of the roller coaster.
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b Select the translations to apply to the parent tangent function in order to get the function that represents the most exciting section of the roller coaster.
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c Which of the following is the graph of y=tan(x-3π)+1?
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Hint
a What are the formulas for the period and asymptote of a tangent function?
b Write the given function rule in the form y=atan b(x-h)+k.
c Translate the parent function using the values of h and k found in the previous part.
Solution
a Begin by writing the function in the form y=atan b(x-h)+k. Then match the variables with their corresponding values to identify the period and asymptotes of the given function.
y= a tan b(x- h)+ k ⇓ y= 1 tan 1(x- 3π)+ 1 Now recall the formulas for the period and asymptotes of a tangent function. Period: & π/| b| [0.8em] Asymptote:& π/|2 b| Since b= 1, substitute this value into the formulas. Period: π/| 1| =π [0.8em] Asymptote: π/|2( 1)|=π/2 The period of the given tangent function is π and its asymptotes are at the odd multiples of π2.
b Consider the revised form of the function written in Part A.
y= a tan b(x- h)+ k ⇓ y= 1 tan 1(x- 3π)+ 1 Here, h= 3π and k= 1 represent the horizontal and vertical translations, respectively. Consider the following table to review how these values translate the parent tangent function.
| Translations of y=tan θ to y= tan (θ -h)+k | |
|---|---|
| Vertical Translation | If k>0, y=tan θ moves $ k$ units up y= tan (x)+ k |
| If k<0, y=tan θ moves $ k$ units down y= tan (x) -k | |
| Horizontal Translation | If h>0, y=tan θ moves $ h$ units to the right y= tan (x- h) |
| If h<0, y=tan θ moves $ h$ units to the left y= tan (x + h) | |
From the table, it can be concluded that the function y=tan (x- 3π)+ 1 results from shifting the parent tangent function 3π units to the right and 1 unit up.
c Now the given function will be graphed as a translation of the parent tangent function 3π units to the right and 1 unit up. Recall that the tangent function has no amplitude because it has no maximum and minimum values. Use the period π and asymptotes at odd multiples of π2 to graph the given function.
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Math Game
After the roller coaster, Dominika and Paulina decided to take a rest. They see a math game where each winner is awarded a teddy bear and think that it may be fun to take a look at the question.
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Hint
Identify the parent function and apply one transformation at a time.
Solution
To find the correct function rule to win the game, start by recalling the general form of a transformed sine function. y= a sin b(x- h)+ k Now, recall what the variables a, b, h, and k represent.
| Transformations of y=sin x | |
|---|---|
| Vertical Stretch or Shrink | y= a sin x If a>1, y= sin x stretches vertically by a |
| y= a sin x If 0< a< 1, y= sin x shrinks vertically by a | |
| Horizontal Stretch or Shrink | y=sin ( bx) If b>1 y= sin x shrinks horizontally by b |
| y=sin ( bx) If 0< b< 1 y= sin x stretches horizontally by b | |
| Vertical Translation | y= sin (x)+ k If k>0, y= sin x moves $ k$ units up |
| y= sin (x)+ k If k<0, y= sin x moves $ k$ units down y= sin (x) -k | |
| Horizontal Translation | y= sin (x- h) If h>0, y= sin x moves $ h$ units to the right |
| y= sin (x- h) If h<0, y= sin x moves $ h$ units to the left | |
y=a sin b(x-h)+k
SubstituteValues
Substitute values
y= 4 sin 0.5(x-( - 2π))+ 3
SubNeg
a-(- b)=a+b
y=4 sin 0.5(x+2π)+3
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Reflections of Parent Trigonometric Functions
Examine the following graph pairs of sine, cosine, and tangent functions in the applet.
- What effect does the negative sign have in these graphs?
- Is it a reflection? If yes, what is the line of reflection?
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Reflecting Trigonometric Functions
Trigonometric functions can be reflected across the line called a line of reflection. This reflection can be achieved in two ways, by multiplying the function rule by -1 or by multiplying the inputs by -1. Consider some parent trigonometric functions.
c
Function & Reflection Across the x -axis y=sin x & y=-sin x y=cos x & y=-cos x y=tan x & y=-tan x
When the function rule is multiplied by -1, the graph of the function is flipped over the x-axis because the y-coordinate of each point is changed to its additive inverse. 
Recall that the graphs of the functions y=a sin x, y=a cos x, and y=a tan x represent a stretch or a shrink by a factor |a|. In the case that a<0, the graphs represent both a stretch or a shrink by a factor of |a| and a reflection due to the negative sign. Now consider the second case.
c
Function & Reflection Across the y -axis y=sin x & y=sin -x y=cos x & y=cos -x y=tan x & y=tan -x
In this case, since every input is multiplied by -1, the x-coordinate of each point is changed to its additive inverse and the graphs are flipped over the y-axis.
Note that for transformed trigonometric functions, the line of reflection becomes the midline, or y=k, for the following trigonometric functions. c
Function & Reflection Line y= a sin b(x- h)+ k & y= k y= a cos b(x- h)+ k & y= k y= a tan b(x- h)+ k & y= k
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Let's Meet Again
Dominika and Paulina have had fun during the day in the amusement park. They decide to come again together next week. To choose an appropriate day, they check the weather for the next two weeks.
a Write a sine function to model the temperature for the next two weeks.
b Explain the types of transformations applied to the parent sine function to get the function written in Part A.
Answer
a Example Function:y=-25 sin π7+70
b See solution.
Hint
a Plot the days and their corresponding temperatures as points on a coordinate plane. Then, use the general form of a transformed sine function, y=a sin b(x-h)+k, to write the function rule.
b Recall what the variables in the general sine function form represent.
Solution
a To write the sine function, start by drawing the function. To do so, plot the days and the corresponding temperatures as points in the coordinate plane.
The sine function can now be written by using all of this information. Start by recalling the general transformed form of a sine function. y= a sin b(x- h)+ k In this form, |a| is the amplitude. According to graph, the amplitude is 25. |a|=25 ⇓ a=25 or a=- 25 To determine the sign of a, compare its graph with the graph of parent sine function.
p=2π/|b|
Substitute
p= 14
14=2π/|b|
▼
Solve for b
MultEqn
LHS * |b|=RHS* |b|
14|b|=2π
DivEqn
.LHS /14.=.RHS /14.
|b|=2π/14
AbsPos
|b|=b
b=2π/14
ReduceFrac
a/b=.a /2./.b /2.
b= π/7
b To determine the types of transformations applied to the parent sine function, start by recalling what the corresponding variables in the general form of the transformed sine function represent.
| Transformations of y=sin x | |
|---|---|
| Vertical Stretch or Shrink | y= a sin x If a>1, y= sin x stretches vertically by a |
| y= a sin x If 0< a< 1, y= sin x shrinks vertically by a | |
| Horizontal Stretch or Shrink | y=sin ( bx) If b>1, y= sin x shrinks horizontally by b |
| y=sin ( bx) If 0< b< 1, y= sin x stretches horizontally by b | |
| Vertical Translation | y= sin (x)+ k If k>0, y= sin x moves $ k$ units up |
| y= sin (x) -k If k<0, y= sin x moves $ k$ units down | |
| Horizontal Translation | y= sin (x- h) If h>0, y= sin x moves $ h$ units to the right |
| y= sin (x+ h) If h<0, y= sin x moves $ h$ units to the left | |
| Reflection | y= - a sin x If a<0, y= sin x reflects in the midline y= k |
Now recall the function rule written in Part A again. y= -25 sin π/7x+ 70 According to the table, it can be said that the function rule represents a vertical stretch by a factor of 25, a horizontal stretch by a factor of π7, and a vertical translation of parent sine function 70 units up. Also, since a= -25 is less than 0, there is a reflection in the midline y= 70.
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Transformations of Cotangent, Secant, and Cosecant Functions
This lesson reviewed the transformations of sine, cosine, and tangent functions. However, the same rules can be applied to investigate the transformations of the parent functions of the remaining trigonometric functions, cotangent, secant, and cosecant. Recall the parent cotangent, secant, and cosecant function graphs.
These graphs can be stretched or shrunk vertically by a factor of a, stretched or shrunk horizontally by a factor of b, translated h units to the right and k units up. They can also be reflected if a<0.
| Parent Function | Transformed Form of the Function |
|---|---|
| y=cot x | y= a cot b(x- h)+ k |
| y=sec x | y= a sec b(x- h)+ k |
| y=csc x | y= a csc b(x- h)+ k |
Transformations of Trigonometric Functions and Their Graphs
Exercise 2.1
The following function models the blood pressure P (in millimeters of mercury) of a person at rest at time t (in seconds). P=120-15 cos 5π/2 t
Which of the following graphs is the graph of the given cosine function?
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We are given a function that models the blood pressure P of a person at rest at time t. P=120-15 cos 5π/2 t The blood pressure P is given in millimeters of mercury and t is time in seconds. We want to graph this function. We will start by comparing it with the general form of a transformed cosine function. General Form y= a cos b(t- h)+ k [0.5em] Given Function P= -15 cos 5π/2 (t- 0)+ 120 As we can see, a= - 15, b= 5π2, h= 0, and k= 120. Now let's recall how these variables affect the graph of the parent cosine function.
| Transformations of y=cos x | |
|---|---|
| Vertical Stretch or Shrink | y= a cos x If a>1, y= cos x stretches vertically by a |
| y= a cos x If 0< a< 1, y= cos x shrinks vertically by a | |
| Horizontal Stretch or Shrink | y=cos ( bx) If b>1, y= cos x shrinks horizontally by b |
| y=cos ( bx) If 0< b< 1, y= cos x stretches horizontally by b | |
| Vertical Translation | y= cos (x)+ k If k>0, y= cos x moves $ k$ units up |
| y= cos (x) -k If k<0, y= cos x moves $ k$ units down | |
| Horizontal Translation | y= cos (x- h) If h>0, y= cos x moves $ h$ units to the right |
| y= cos (x+ h) If h<0, y= cos x moves $ h$ units to the left | |
| Reflection | y= - a cos x If a<0, y= cos x reflects in the midline y= k |
Now we will transform the parent cosine function step by step. First, we will stretch the parent cosine function vertically by a factor of 15 and shrink it horizontally by a factor of 5π2, or 2.5 π. Let's calculate the amplitude and period of the cosine function. Amplitude: & | a| ⇒ | 15|=15 [0..5em] Period: & 2π/| b| ⇒ 2π/| 2.5 π| =0.8 Now we will draw the transformed function by using the amplitude and period that we found.
Great! Notice that a= - 15. Since it is less than 0, it also represents a reflection in the x-axis or across the midline y=0. Let's now reflect our graph in the x-axis!
We will next translate this function 120 units up. Since h= 0, there is no phase shift.
Finally, we will set the starting point of the graph at t=0 to find the desired graph because the time cannot be negative.
The graph matches option C.
We know that one cycle represents one heartbeat. To find the number of heartbeats per minute, we will examine the graph. From the graph we can tell that one cycle takes 0.8 seconds.
We want to determine the pulse rate of the person in heartbeats per minute. Because there are 60 seconds in 1 minute, we need to find the number of heartbeats in 60 seconds. Since one heartbeat takes 0.8 seconds, let's divide 60 by 0.8. 60/0.8=75 There are 75 heartbeats in 60 seconds. This means that the pulse rate is 75 heartbeats per minute.
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Consider the following graph of the function y=5 cos 2x.
Which of the following graphs is the graph of the function f(x)=5 sec 2x? Use the given graph to determine the correct option.
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We are asked to use the given graph of the cosine function y=5cos 2x to graph the cosecant function f(x)= 5sec 2x. Notice that these functions are reciprocal functions. 5 sec (2x)=1/5 cos (2x) This means that the asymptotes of f(x)=5sec 2x occur when 5cos 2x=0. In other words, we can draw these asymptotes at the points where the graph of y=5cos 2x intersects the x-axis.
Now, we will find the period of f(x)= 5sec 2x. Note that the period of f(x)=5sec 2x and the period of y= 5cos 2x are the same because they are reciprocals. Since our function is in the form y= a cos b, we can substitute b= 2 in the following formula to find its period. Period: 2π/| b| = 2π/| 2| ⇒ π The graph of y=5cos 2 x has a period of π.
The period of the graph of f(x)=5sec 2x will also be π. Now we will plot the points to show the local minimums and maximums of y, which also belong to the graph of f. The secant function's graph will increase when the cosine function's graph decreases and the secant function's graph will decrease when the cosine function's graph increases around these points.
To graph the left portion of the function, we come down from the left asymptote, pass through the point, and move up towards the right asymptote. Then for the right portion of the graph, we will follow similar steps and come up from the left asymptote, pass through the point, and move towards the right asymptote.
This graph matches option C.
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Tadeo thinks that it is not possible to write a secant function that has the same graph as y=csc x. Determine whether he is correct or not.
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We will determine whether it is possible to write a secant function that has the same graph as y=csc x. To do so, let's first recall what we know about the secant function. Below we have the graph of the function.
We know that y=sec x is a reciprocal function of y=cos x. Similarly, the cosecant function y=csc x is a reciprocal of the sine function y=sin x.
| Function | Reciprocal |
|---|---|
| y=sin x | y=csc x |
| y=cos x | y=sec x |
Notice that the graph of y=cos x can be translated π2 units to the right to create the same graph as y=sin x.
This means that the graph of y=sec x can be translated π2 units to the right to create the same graph as y=csc x.
By doing this, we can write the cosecant function as a horizontal translation of the secant function. csc x= sec (x - π/2) Therefore, Tadeo is not correct. Notice that this is just one possible solution. We can actually write several types of translations between secant and cosecant functions that result in the same graph.
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Transformations of Trigonometric Functions and Their Graphs
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