4. Graphing Reciprocal Trigonometric Functions
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Chapter 6
4.
Graphing Reciprocal Trigonometric Functions
Reciprocal trigonometric functions, such as the secant, cosecant, and cotangent, are derived from fundamental trigonometric functions like sine, cosine, and tangent. When plotted on a graph, these reciprocal functions display distinct behaviors. For instance, the cosecant function, being the reciprocal of the sine function, has asymptotes at points where the sine function is zero. Similarly, the secant function, related to the cosine function, has its asymptotes where the cosine function zeroes out. These functions play a pivotal role in various fields, from engineering to physics. Their graphical representations can depict real-world scenarios, like the length of wires used to anchor a communication tower or the curvature of an umbrella shielding a garden. Gaining proficiency in graphing these functions equips individuals to interpret and predict diverse real-world phenomena.
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Lesson Settings & Tools
| | 13 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Graphing Reciprocal Trigonometric Functions
Slide of 13
When discussing trigonometric functions such as the sine, cosine, and tangent functions, it is natural to consider their reciprocal functions.
| Trigonometric Function | Reciprocal Trigonometric Function |
|---|---|
| y=sin x | y=1/sin x |
| y=cos x | y=1/cos x |
| y=tan x | y=1/tan x |
This lesson will explore the graphs of these reciprocal trigonometric functions.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Communication Tower
A massive communication tower is anchored to the ground with wires.
These wires are attached to the tower at a height of 5 meters above the ground. The following function models the length of a wire. y=5csc θ Here, θ is the measure of the angle formed by the wire and the ground. Graph the given function to find the length of the wire that makes an angle of π4 radians with the ground. Round the answer to the nearest integer.
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Exploring the Graphs of the Cosine and Sine Functions, and Their Quotient
In the diagram, the graphs of the functions y=sin x, y=cos x, and y= cos xsin x can be seen.
Consider the graph of y= cos xsin x. How is it different from the other two graphs? What happens in the graph of y= cos xsin x when sin x=0?
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The Cotangent Function
Let P be the point of intersection of the terminal side of an angle in standard position and the unit circle. The cotangent function, denoted by cot, is defined as the ratio of the x-coordinate to the y-coordinate of P.
cotθ=cosθ/sin θ
Since division by 0 is not defined, the graph of the parent cotangent function y=cot x has vertical asymptotes where sin x=0. This means that the graph has vertical asymptotes at every multiple of π. The graph of y=cot x can be drawn by making a table of values.
Consider now the general form of a cotangent function.
y=acotbx
Here, a and b are non-zero real numbers and x is measured in radians. The properties of the cotangent function are stated below.
| Properties of y=acotbx | |
|---|---|
| Amplitude | No amplitude |
| Number of Cycles in [0,2π] | 2|b| |
| Period | π/|b| |
| Domain | All real numbers except multiples of π|b| |
| Range | All real numbers |
Method
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Graphing a Cotangent Function
Recall the format of a cotangent function.
y= acot b θ
Here, a and b are non-zero real numbers and θ is measured in radians. The values of a and b can be used to graph the function. Consider an example function.
y= 3cot 1/2 θ
In this function, a= 3 and b= 12. To sketch one cycle of a cotangent curve, its asymptotes and three points can be used. As with other trigonometric functions, there are five elements that are equally spaced through one cycle. The asymptote-point-zero-point-asymptote pattern is helpful for graphing this function. Four steps will be followed.
1
Find the Period and Cycle
expand_more The period of a cotangent function is the quotient of π and |b|. Period π/|b| ⇒ π/| 12| =2π The period of this function is 2π. This means that one cycle goes from 0 to 2π.
2
Graph the Asymptotes
expand_more Asymptotes occur at the end of each cycle. Therefore, the given function has asymptotes at θ = 0 and θ = 2π.
3
Divide the Period and Locate Points
expand_more Divide the period into fourths and locate the three equidistant points between the asymptotes. The period for this function goes from 0 to 2π, so a table of values will be made for θ= π2, θ=π, and θ= 3π2.
| θ | 3cot (1/2θ) | y |
|---|---|---|
| π/2 | 3cot (1/2* π/2) | 3 |
| π | 3cot (1/2* π) | 0 |
| 3π/2 | 3cot (1/2* 3π/2) | - 3 |
The points found in the table are ( π2,3), (π,0), and ( 3π2,- 3).
4
Draw the Graph
expand_more Finally, the points can be connected with a smooth curve to draw the graph for one cycle.
Once the graph for one cycle is drawn, it can be replicated as many times as desired to draw more cycles. Here, another cycle is graphed.
Extra
Alternative MethodAlternatively, the graph of a cotangent function can be drawn by considering the graphs of the sine and cosine functions. To show this method, the graph of the parent cotangent function will be drawn.
y=cot x
To draw the graph by using this method, five steps will be followed.
1
Graph the Sine and Cosine Functions
expand_more Start by drawing the graphs of the the sine and cosine functions on the same coordinate plane.
2
Draw the Asymptotes
expand_more 
3
Plot Two Zeros
expand_more Since the cotangent of an angle is the ratio of the cosine to the sine of the angle, the cotangent function and the cosine function have the same zeros.
cot θ =0 ⇔ cos θ =0
Therefore, the x-intercepts of the cotangent function occur when cos x=0.
4
Plot More Points on the Curve
expand_more By definition, the cotangent function is 1 when the cosine and the sine have the same value. This happens at their points of intersection. Therefore, the y-value of the cotangent function will be 1 at the x-coordinate of these points.
Similarly, the output of the cotangent function is - 1 when the sine and the cosine function have opposite values.
5
Draw the Graph
expand_more Finally, two cycles of the cotangent function can be graphed by connecting the marked points.
It is seen that the period of the cotangent function is π. Since each branch comes from positive infinity towards negative infinity, the cotangent function has no amplitude and its range is all real numbers.
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Tanabata Trees and a Cotangent Function
Tanabata is a Japanese festival that celebrates two mythical lovers separated by the Milky Way. A Tanabata tree is a type of tree on which people hang wishes written on paper during Tanabata. Ramsha is starting a Tanabata garden in her backyard. She realized that two of the Tanabata trees follow the path of one cycle of a cotangent function. 
After doing some calculations, she found the equation of the curve that these trees follow.
y=1/2 cot 2x
Draw the graph of the function for values of x between 0 and 2π.
External credits: Kumiko
Answer
Hint
Find the period and cycle, then graph the asymptotes and plot some points. Finally, sketch the curve.
Solution
To graph a cotangent function, the first step is to calculate the period. To do so, consider the general form of a cotangent function and the given function. General Form:& y= a cot bx [0.3em] Given Function:& y= 1/2 cot 2x Here, a= 12 and b= 2. The period of the function is the quotient of π and | b|. Period π/|b| ⇒ π/2 The period of the function is π2. This means that one cycle goes from 0 to π2, another cycle from π2 to π, and so on. Since the asymptotes occur at the end of each cycle they can be drawn on a coordinate plane.
Next, divide the period into fourths and locate three equidistant points between the asymptotes. Since a period goes from 0 to π2, a table of values will be made for x= π8, x= π4, and x= 3π8.
| x | 1/2cot 2x | y |
|---|---|---|
| π/8 | 1/2cot 2( π/8) | 1/2 |
| π/4 | 1/2cot 2( π/4) | 0 |
| 3π/8 | 1/2cot 2( 3π/8) | - 1/2 |
The points found in the table are ( π8, 12), ( π4,0), and ( 3π8,- 12). These three points can be plotted on the plane.
The points are then connected with a smooth curve to graph one period of the function.
Finally, the cycle can be replicated as many times as desired. In this case, the graph will be drawn for values of x between 0 and 2π.
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Exploring the Graphs of the Cosine Function and its Reciprocal Function
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The Secant Function
Let P be the point of intersection of the terminal side of an angle in standard position and the unit circle. The secant function, denoted as sec, is defined as the reciprocal of the x-coordinate of P.
secθ=1/cos θ
Since division by 0 is not defined, the graph of the parent secant function y=sec x has vertical asymptotes where cos x=0. This means that the graph has vertical asymptotes at odd multiples of x= π2. The graph of y=sec x can be drawn by making a table of values.
![The secant function y=sec(x) with asymptotes at -3pi/2, -pi/2, pi/2, 3p/2, and 5pi/2 over the domain [-3pi/2,5pi/2]](/mediawiki/images/7/7a/Mljsx_Concept_Secant_Function_2.svg)
Consider the general form of a secant function.
y=asec bx
Here, a and b are non-zero real numbers and x is measured in radians. The properties of the secant function are be stated in the table below.
| Properties of y=asec bx | |
|---|---|
| Amplitude | No amplitude |
| Number of Cycles in [0,2π] | |b| |
| Period | 2π/|b| |
| Domain | All real numbers except odd multiples of π2|b| |
| Range | (-∞,- |a|] ⋃ [|a|,∞) |
Method
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Graphing a Secant Function
Recall the format of a secant function.
y= asec b θ
Here, a and b are non-zero real numbers and θ is measured in radians. The values of a and b can be used to graph the function. Consider an example function.
y= 2sec 1/2 θ
In this function, a= 2 and b= 12. The asymptotes, some points, and the graph of the cosine function can be used to sketch one cycle of a secant curve. Then the cycle can be replicated as many times as desired. Five steps will be followed.
The points from the table are (- π2,2.83), ( π2,2.83), ( 3π2,- 2.83), and ( 5π2,- 2.83).
1
Find the Period
expand_more The period of a secant function is the quotient of 2π and |b|. Period 2π/|b| ⇒ 2π/| 12| =4π The period of this function is 4π.
2
Graph the Related Cosine Function
expand_more Consider the related cosine function. cc Secant & Related Cosine Function & Function [0.5em] y=2sec 1/2θ & y=2cos 1/2θ The graph of the cosine function can be drawn on a coordinate plane.
3
Graph the Asymptotes and Plot Points
expand_more By definition, the asymptotes of the secant function occur when the output of the cosine function is 0. Furthermore, the maximum and minimum points of the cosine function are also points on the secant function's graph.
4
Divide the Period and Locate Points
expand_more Here, the period is 4π and the asymptotes occur every 2π radians. Therefore, the asymptotes are located not only at the beginning and at the end of each period, but also in the middle of each period. Divide the interval between two asymptotes in fourths by the following pattern.
Notice that the middle point in the pattern, the maximum or minimum point, was already plotted in the previous step. The remaining four points can be found for the interval that goes from - π to 3π by making a table of values.
| θ | 2sec 1/2θ | y |
|---|---|---|
| - π/2 | 2sec (1/2( - π/2)) | 2sqrt(2)≈ 2.83 |
| π/2 | 2sec (1/2* π/2) | 2sqrt(2)≈ 2.83 |
| 3π/2 | 2sec (1/2* 3π/2) | - 2sqrt(2)≈ - 2.83 |
| 5π/2 | 2sec (1/2* 5π/2) | - 2sqrt(2)≈ - 2.83 |
5
Draw the Graph
expand_more Finally, the points can be connected with a smooth curve to draw the graph for one cycle.
Once the graph for one cycle is drawn, it can be replicated as many times as desired to draw more cycles. Here, one more cycle will be graphed.
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A Bowl and a Secant Function
Ramsha is thinking about a sustainable way of fertilizing her new garden. She collects her kitchen scraps in a bowl so that she can compost them and use the compost for the Tanabata garden. While mixing these scraps with soil to make some fertilizer, Ramsha noticed that the shape of the bowl matches the shape of one branch of a secant function.
After doing some calculations, she found the equation of the curve that models the bowl.
y=2 sec x
Draw the graph of the above function for values of x between - 3π2 and 5π2.
The asymptotes of the secant function occur at the zeros of the cosine function. Furthermore, the maximum and minimum points of the cosine function are also points on the curve of the secant function.
Next, more points on the curve of y=2sec x can be found by making a table of values.
External credits: Kumiko
Answer
Hint
Graph the related cosine function and recall that the asymptotes of the secant function occur at the zeros of the cosine function.
Solution
To graph a secant function, the first step is to calculate the period. Consider the general form of a secant function and the given function. General Form y= asec bx [0.5em] Given Function y=2sec x ⇔ y= 2sec 1x Here, a= 2 and b= 1. The period of the function is the quotient of 2π and | b|. Period 2π/|b| ⇒ 2π/| 1|=2π Now that it is known that the period of the function is 2π, consider the related cosine function. cc Secant & Related Cosine Function & Function [0.5em] y=2sec x & y=2cos x This cosine function will now be graphed on a coordinate plane for values of x between - 3π2 and 5π2.
| x | 2sec x | y |
|---|---|---|
| - 5π/4 | 2sec ( - 5π/4) | - 2sqrt(2)≈ - 2.83 |
| - 3π/4 | 2sec ( - 3π/4) | - 2sqrt(2)≈ - 2.83 |
| - π/4 | 2sec ( - π/4) | 2sqrt(2)≈ 2.83 |
| π/4 | 2sec π/4 | 2sqrt(2)≈ 2.83 |
| 3π/4 | 2sec 3π/4 | - 2sqrt(2)≈ - 2.83 |
| 5π/4 | 2sec 5π/4 | - 2sqrt(2)≈ - 2.83 |
| 7π/4 | 2sec 7π/4 | 2sqrt(2)≈ 2.83 |
| 9π/4 | 2sec 9π/4 | 2sqrt(2)≈ 2.83 |
The points found in the table can now be plotted. Finally, each set of points can be connected with smooth curves.
The graph of the curve of the bowl has been drawn.
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Exploring the Graphs of the Sine Function and its Reciprocal Function
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The Cosecant Function
Let P be the point of intersection of the terminal side of an angle in standard position and the unit circle. The cosecant function, denoted as csc, is defined as the reciprocal of the y-coordinate of P.
cscθ=1/sin θ
Since division by 0 is not defined, the graph of the parent cosecant function y=csc x has vertical asymptotes where sin x=0. This means that the graph has vertical asymptotes at multiples of π. The graph of y=csc x can be drawn by making a table of values.
Consider the general form of a cosecant function.
y=acsc bx
Here, a and b are non-zero real numbers and x is measured in radians. The properties of the cosecant function are stated below.
| Properties of y=acscbx | |
|---|---|
| Amplitude | No amplitude |
| Number of Cycles in [0,2π] | |b| |
| Period | 2π/|b| |
| Domain | All real numbers except multiples of π|b| |
| Range | (-∞,- |a|] ⋃ [|a|,∞) |
Method
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Graphing a Cosecant Function
Recall the format of a cosecant function.
y= acsc b θ
Here, a and b are non-zero real numbers and θ is measured in radians. The values of a and b can be used to graph the function. Consider an example function.
y= 2csc 1/2 θ
In this function, a= 2 and b= 12. To sketch one cycle of a cosecant curve, its asymptotes, some points, and the graph of the sine function can be used. Then, the cycle can be replicated as many times as desired. Five steps will be followed.
![The graph of the sine function 2*sin(1/2 theta) over the domain [-pi,7pi]](/mediawiki/images/3/3e/Mljsx_Method_Graphing_a_Cosecant_Function_1.svg)
The points found in the table are (- 3π2,- 2.83), (- π2,- 2.83), ( π2,2.83), and ( 3π2,2.83).
1
Find the Period
expand_more The period of a cosecant function is the quotient of 2π and |b|. Period 2π/|b| ⇒ 2π/| 12| =4π The period of this function is 4π.
2
Graph the Related Sine Function
expand_more Consider the related sine function. cc Cosecant & Related Sine Function & Function [0.5em] y=2csc 1/2θ & y=2sin 1/2θ The graph of the sine function can be drawn on a coordinate plane.
3
Graph the Asymptotes and Plot Points
expand_more By definition, the asymptotes of the cosecant function occur when the output of the sine function is 0. Furthermore, the maximum and minimum points of the sine function are also points on the cosecant function's graph.
4
Divide the Period and Locate Points
expand_more Here, the period is 4π and the asymptotes occur every 2π radians. Therefore, the asymptotes are located not only at the beginning and at the end of each period, but also in the middle of each period. Divide the interval between two asymptotes in fourths by the following pattern.
Notice that the middle point, which is the maximum or minimum point, was already plotted in the previous step. Four more points can be found for the interval that goes from - 2π to 2π by making a table of values.
| θ | 2csc 1/2θ | y |
|---|---|---|
| - 3π/2 | 2csc (1/2( - 3π/2)) | - 2sqrt(2)≈ - 2.83 |
| - π/2 | 2csc (1/2( - π/2)) | - 2sqrt(2)≈ - 2.83 |
| π/2 | 2csc (1/2* π/2) | 2sqrt(2)≈ 2.83 |
| 3π/2 | 2csc (1/2* 3π/2) | 2sqrt(2)≈ 2.83 |
5
Draw the Graph
expand_more Finally, each set of points can be connected with a smooth curve to draw the graph for one cycle.
Once the graph for one cycle is drawn, it can be replicated as many times as desired to draw more cycles. Another cycle is graphed below.
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An Umbrella and a Cosecant Function
To protect her new garden, Ramsha decides to set up an umbrella to cover the small plants when heavy rains are forecast. She realizes that the umbrella has the shape of one branch of a cosecant function.
After doing some calculations, she found the equation of the curve that models the umbrella.
y=1/2 csc x
Draw the graph of the function for values of x between - 2π and 2π.
The asymptotes of the cosecant function occur at the zeros of the sine function. Furthermore, the maximum and minimum points of the sine function are also points on the curve of the cosecant function.
Next, more points on the curve of y= 12csc x will be found by making a table of values.
External credits: Kumiko
Answer
Hint
Graph the related sine function and recall that the asymptotes of the cosecant function occur at the zeros of the sine function.
Solution
To graph a cosecant function, the first step is to calculate the period. Consider the general form of a cosecant function and the given function. General Form y= acsc bx [0.5em] Given Function y=1/2csc x ⇔ y= 1/2csc 1x Here, a= 12 and b= 1. The period of the function is the quotient of 2π and | b|. Period 2π/|b| ⇒ 2π/| 1|=2π The period of the function is 2π. Now consider the related sine function. cc Cosecant & Related Sine Function & Function [0.5em] y=1/2csc x & y=1/2sin x This sine function will now be graphed on a coordinate plane for values of x between - 2π and 2π.
| x | 1/2csc x | y |
|---|---|---|
| - 5π/3 | 1/2csc ( - 5π/3) | sqrt(3)/3≈ 0.58 |
| - 4π/3 | 1/2csc ( - 4π/3) | sqrt(3)/3≈ 0.58 |
| - 2π/3 | 1/2csc ( - 2π/3) | - sqrt(3)/3≈ - 0.58 |
| - π/3 | 1/2csc ( - π/3) | - sqrt(3)/3≈ - 0.58 |
| π/3 | 1/2csc π/3 | sqrt(3)/3≈ 0.58 |
| 2π/3 | 1/2csc 2π/3 | sqrt(3)/3≈ 0.58 |
| 4π/3 | 1/2csc 4π/3 | - sqrt(3)/3≈ - 0.58 |
| 5π/3 | 1/2csc 5π/3 | - sqrt(3)/3≈ - 0.58 |
The points found in the table can now be plotted. Finally, connect the sets of points with smooth curves.
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Finding the Period
Find the period of the following functions. Round the answers to two decimal places.
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Drawing and Using the Graph of a Cosecant Function
The challenge presented at the beginning can be solved with the topics covered in this lesson. It was given that a massive communication tower is anchored to the ground with wires.
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Hint
Graph the cosecant function and pay close attention to the y-value for θ= π4.
Solution
To find the length of the wire that makes an angle of π4 radians with the ground, graph the cosecant function. The first step is to calculate the period of the function. Consider the general form of a cosecant function and the given function. General Form y= acsc bθ [0.5em] Given Function y=5csc θ ⇔ y= 5csc 1θ Here, a= 5 and b= 1. The period of the function is the quotient of 2π and | b|. Period 2π/|b| ⇒ 2π/| 1|=2π The period of the function is 2π. Now consider the related sine function. cc Cosecant & Related Sine Function & Function [0.5em] y=5csc θ & y=5sin θ This sine function can be graphed on a coordinate plane.
| θ | 5 csc θ | y |
|---|---|---|
| - 3π/4 | 5csc ( - 3π/4) | - 5sqrt(2)≈ - 7.07 |
| - π/4 | 5csc ( - π/4) | - 5sqrt(2)≈ - 7.07 |
| π/4 | 5csc π/4 | 5sqrt(2)≈ 7.07 |
| 3π/4 | 5csc 3π/4 | 5sqrt(2)≈ 7.07 |
Finally, the points found in the table are be plotted and each set of points connected with smooth curves.
Now that the function has been graphed, the point at θ = π4 can be located and its y-coordinate identified.
The exact y-coordinate cannot be found in the graph, but its nearest integer can be identified as 7. Therefore, the length of the wire that makes an angle of π4 radians with the ground is about 7 meters.
Graphing Reciprocal Trigonometric Functions
Exercises
Find the period of each reciprocal trigonometric function. Write the answers in exact form.
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The period of a secant function with the form y=asec bx is the quotient of 2π and |b|. Secant Function & Period y = a sec bx & 2π/| b| In our equation we have that b=- 1. Let's substitute this value in the above formula to find the period of y=8sec (- x).
The period of a cotangent function with the form y=acot bx is the quotient of π and |b|. Cotangent Function & Period y=acot bx & π/| b| In our equation we have that b=- 5. Let's substitute this value in the above formula to find the period of y=- 9cot (- 5x).
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Which of the following graphs is the graph of y= 14cot 2x?
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To determine which of the given options corresponds to the graph of y= 14cot 2x, we will draw the graph by ourselves and compare it with the given choices. To graph a cotangent function, the first step is to calculate its period. To do so, consider the general form of a cotangent function. General Form:& y= a cot bx [0.3em] Given Function:& y= 1/4 cot 2x Here, a= 14 and b= 2. The period of the function is the quotient of π and the absolute value of b. Period π/|b| ⇒ π/| 2|=π/2 The period of the function is π2. This means that one cycle goes from 0 to π2, another cycle from π2 to π, and so on. Since the asymptotes occur at the end of each cycle we can draw them on a coordinate plane.
Next, we can divide the period into fourths and locate three equidistant points between the asymptotes. Since a period goes from 0 to π2, we will make a table of values for x= π8, x= π4, and x= 3π8.
| x | 1/4cot 2x | y=1/4cot 2x |
|---|---|---|
| π/8 | 1/4cot 2( π/8) | 1/4 |
| π/4 | 1/4cot 2( π/4) | 0 |
| 3π/8 | 1/4cot 2( 3π/8) | - 1/4 |
The points we found in the table are ( π8, 14), ( π4,0), and ( 3π8,- 14). Let's plot these three points on our the plane.
We can connect these points to graph one period of the given function.
Finally, let's draw all the cycles for values of x between - π and π.
This graph matches choice B.
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Which of the following graphs is the graph of y= 12sec x?
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To determine which of the given options corresponds to the graph of y= 12sec x, we will draw the graph by ourselves and compare it with the given choices. To graph a secant function, the first step is to calculate the period. Let's consider the general form of a secant function and the given function. General Form y= asec bx [0.5em] Given Function y=1/2sec x ⇔ y= 1/2sec 1x Here, we see that a= 12 and b= 1. We know that the period of the function is the quotient of 2π and the absolute value of b. Period 2π/|b| ⇒ 2π/| 1|=2π Now that we know that the period of the function is 2π, let's consider the related cosine function. cc Secant & Related Cosine Function & Function [0.5em] y=1/2sec x & y=1/2cos x We can now graph this cosine function on a coordinate plane for values of x between - π2 and 3π2.
The asymptotes of the secant function occur at the zeros of the cosine function. Furthermore, the maximum and minimum points of the cosine function are also points on the curve of the secant function.
Next, we can find more points on the curve of y=2sec x by making a table of values.
| x | 1/2sec x | y |
|---|---|---|
| - π/4 | 1/2sec ( - π/4) | sqrt(2)/2≈ 0.7 |
| π/4 | 1/2sec π/4 | sqrt(2)/2≈ 0.7 |
| 3π/4 | 1/2sec 3π/4 | - sqrt(2)/2≈ - 0.7 |
| 5π/4 | 1/2sec 5π/4 | - sqrt(2)/2≈ - 0.7 |
We can plot the points we found in the table. Finally, we can connect each set of points with smooth curves.
This graph corresponds to choice A.
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Which of the following graphs is the graph of y=- 14csc x?
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To determine which of the given options corresponds to the graph of y=- 14csc x, we will draw the graph by ourselves and compare it with the given choices. To graph a cosecant function, the first step is to calculate the period. Let's consider the general form of a cosecant function and the given function. General Form y= acsc bx [0.5em] Given Function y=- 1/4csc x ⇔ y=- 1/4csc 1x Here, a= - 14 and b= 1. The period of the function is the quotient of 2π and the absolute value of b. Period 2π/|b| ⇒ 2π/| 1|=2π The period of the function is 2π. Let's now consider the related sine function. cc Cosecant & Related Sine Function & Function [0.5em] y=- 1/4csc x & y=- 1/4sin x We can graph this sine function on a coordinate plane for values of x between - π and π.
The asymptotes of the cosecant function occur at the zeros of this sine function. Furthermore, the maximum and minimum points of the sine function are also points on the curve of the cosecant function. Let's show this in our diagram.
Next, we can find more points on the curve of y=- 14csc x by making a table of values.
| x | - 1/4csc x | y |
|---|---|---|
| - 3π/4 | - 1/4csc ( - 3π/4) | sqrt(2)/4≈ 0.35 |
| - π/4 | - 1/4csc ( - π/4) | sqrt(2)/4≈ 0.35 |
| π/4 | - 1/4csc π/4 | - sqrt(2)/4≈ - 0.35 |
| 3π/4 | - 1/4csc 3π/4 | - sqrt(2)/4≈ - 0.35 |
We can now plot the points we found in the table. Finally, we can connect the sets of points with smooth curves.
This graph matches choice D.
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Ramsha wants to find the least positive solution to the following equation.
2csc x =2
Solve the equation by graphing and find the value of x that Ramsha is looking for. Write the answer in radians rounded to three significant figures.
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To solve the equation graphically, we will graph the cosecant function whose function rule is represented by the left-hand side of the equation. The first step is to calculate the period of the function. Consider the general form of a cosecant function and our function. General Form y= acsc bx [0.5em] Our Function y=2csc x ⇔ y= 2csc 1x Here, a= 2 and b= 1. The period of the function is the quotient of 2π and | b|. Period 2π/|b| ⇒ 2π/| 1|=2π The period of the function is 2π. Let's now consider the related sine function. cc Cosecant & Related Sine Function & Function [0.5em] y=2csc x & y=2sin x We can graph this sine function on a coordinate plane. Since we are looking for a positive value of x, we will not consider the negative values of the x-axis.
The asymptotes of the cosecant function occur at the zeros of the sine function. Furthermore, the maximum and minimum points of the sine function are also points on the curve of the cosecant function.
Next, we can find more points on the curve of y=2csc θ by making a table of values.
| θ | 2 csc θ | y |
|---|---|---|
| π/4 | 2csc π/4 | 2sqrt(2)≈ 2.8 |
| 3π/4 | 2csc 3π/4 | 2sqrt(2)≈ 2.8 |
| 5π/4 | 2csc 5π/4 | - 2sqrt(2)≈ - 2.8 |
| 7π/4 | 2csc 7π/4 | - 2sqrt(2)≈ - 2.8 |
Finally, let's plot the points we found in the table. We will also connect each set of points with smooth curves.
Now that we have graphed the function, we can locate the point whose x-coordinate is the least positive number and whose y-coordinate is 2.
We can see that the value of x that Ramsha is looking for is π2. This number, rounded to three significant figures, is 1.57. π/2 = 1.570796 ... ⇔ π/2 ≈ 1.57
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Graphing Reciprocal Trigonometric Functions
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