Graphing Secant and Cosecant Lesson

Over the course of the previous two lessons, we learned how to sketch the graphs for sine and cosine, as well as tangent and cotangent. At this point, we should know that sine and cosine are periodic functions with a period of 2$π$. Since cosecant and secant are the reciprocals of sine and cosine, respectively, the period for both is also 2$π$. $$\csc \hspace{.1em}x=\csc\left(x + 2πn\right)$$ $$\sec \hspace{.1em}x=\sec\left(x + 2πn\right)$$ Where n is any integer and x is a member of the domain for the given function. $$\csc\left(\frac{π}{6}\right)=\csc\left(\frac{π}{6}+ 2π\right)$$ $$\csc\left(\frac{π}{6}\right)=\csc\left(\frac{π}{6}+ \frac{12π}{6}\right)$$ $$\csc\left(\frac{π}{6}\right)=\csc\left(\frac{13π}{6}\right)$$ $$2=2$$ $$\sec\left(\frac{π}{4}\right)=\sec\left(\frac{π}{4}+ 2π\right)$$ $$\sec\left(\frac{π}{4}\right)=\sec\left(\frac{π}{4}+ \frac{8π}{4}\right)$$ $$\sec\left(\frac{π}{4}\right)=\sec\left(\frac{9π}{4}\right)$$ $$\sqrt{2}=\sqrt{2}$$

Secant Function

$$\sec \hspace{.1em}x=\frac{1}{\cos \hspace{.1em}x}$$ Since we can't divide by zero, the secant function will be undefined where cosine values are zero. Recall from our last lesson, that this occurs at odd multiples of $π$/2, and just like we saw with our tangent function, we will have vertical asymptotes for those values. Let's create a basic sketch for secant over one period. We will use a table for the basic points found over the interval from [-$π$, $π$]:

y = sec x
x	y
$\large\pm\frac{\pi}{2}$	undefined
$0$	$1$
$\large\pm\frac{\pi}{6}$	$\large\frac{2\sqrt{3}}{3}$
$\large\pm\frac{\pi}{4}$	$\large\sqrt{2}$
$\large\pm\frac{\pi}{3}$	$\large2$
$\large\pm\frac{2\pi}{3}$	$\large-2$
$\large\pm\frac{3\pi}{4}$	$\large-\sqrt{2}$
$\large\pm\frac{5\pi}{6}$	$\large-\frac{2\sqrt{3}}{3}$
$\large\pmπ$	$\large-1$

Graphing the Secant Function over One Period

Desmos link for more detail As mentioned earlier, the period of secant is 2$π$ since the secant values are reciprocals of the corresponding cosine values. When the cosine value is -1, the secant value will also be -1. This occurs at $π$ + 2$π$n, where n is any integer. Similarly, when the cosine value is 1, the secant value will also be 1. This occurs at 2$π$n, where n is any integer. The range for cosine is from [-1, 1], which means that |sec x| ≥ 1 for all x in its domain. Recall that dividing 1 by a number between 0 and 1 will give you a value that is larger than 1. For example, think about the cosine of $π$/3 and the secant of $π$/3. $$\cos\left(\frac{π}{3}\right)=\frac{1}{2}$$ $$\sec\left(\frac{π}{3}\right)=\frac{1}{\cos\left(\frac{π}{3}\right)}=\frac{1}{\frac{1}{2}}=2$$ Desmos link for more detail Looking at the above graph, we can see how the graphs of y = cos x and y = sec x are related.

• x-intercepts on the cosine curve correspond to vertical asymptotes on the secant curve
• A maximum point on the cosine curve corresponds to a minimum point on a continuous portion of the secant curve
• A minimum point on the cosine curve corresponds to a maximum point on a continuous portion of the secant curve

An easy way to remember this is to say that the "hills" and "valleys" are interchanged. To be more specific, we could say that a valley or minimum point on the cosine graph, gives us a hill or relative maximum on the secant graph, and a hill or maximum point on the cosine graph, gives us a valley or relative minimum on the secant graph.

The Secant Curve and its Characteristics

• The domain consists of all real numbers except odd multiples of $π$/2
• f(x) = sec x is discontinuous at odd multiples of $π$/2
  • The graph contains vertical asymptotes at those values
• There are no x-intercepts
• The period is 2$π$
• There are no minimum or maximum values
  • The range is: $(-\infty, -1] ∪ [1, \infty)$
  • The graph has no amplitude
• The graph is symmetric with respect to the y-axis
  • sec(-x) = sec x, for all x in the domain
  • f(x) = sec x is an even function

Graphing Variations of the Secant Function

1. Replace secant with cosine and sketch the graph using a dashed curve
   • Cosine is used as a guide
2. Sketch the vertical asymptotes
   • These occur at 1/4 and 3/4 of the way through the cosine cycle
3. Sketch the graph by drawing the u-shaped branches between consecutive vertical asymptotes
   • The "hills" and "valleys" are interchanged
4. (Optional) Use the period to find additional cycles to the left or right as needed

Example #1: Find the period, phase shift, and vertical shift. Sketch the graph. $$f(x)=\frac{1}{2}\sec\left(2x + \frac{5π}{6}\right) + 1$$ As we have done previously, let's factor to make finding the phase shift easier: $$f(x)=\frac{1}{2}\sec\left[2\left(x + \frac{5π}{12}\right)\right] + 1$$ To find the period, we follow the same procedure we did with cosine. We use the standard period of 2$π$ and then divide by the coefficient of x (in unfactored form) or the number outside of the parentheses (in factored form). Here, we divide 2$π$ by 2 to obtain a period of $π$. $$\text{Period}: π$$ $$\text{Phase Shift}: \text{left}\frac{5π}{12}$$ $$\text{Vertical Shift}: \text{Up}\hspace{.1em}1$$ Replace secant with cosine and sketch the graph. $$f(x)=\frac{1}{2}\cos\left[2\left(x + \frac{5π}{12}\right)\right] + 1$$ Since using graphing transformations is slightly faster and has been covered extensively over the course of the past two lessons, we will use that approach. You can always use the phase shift to get the first x-value of the cycle and then add quarter periods to get the other four. This is a matter of personal preference, so pick whichever method is best for you. We will begin with points from our basic cosine function (y = cos x) and think about the transformations that are applied to x and y separately. $$f(x)=\cos \hspace{.1em}x$$ $$(0, 1)$$ $$\left(\frac{π}{2}, 0\right)$$ $$\left(π, -1\right)$$ $$\left(\frac{3π}{2}, 0\right)$$ $$(2π, 1)$$

Horizontal Transformations (x)

1. Multiply by 1/2
2. Subtract away 5$π$/12

In other words, we have a horizontal shrink by a factor of 1/2 and a shift left by 5$π$/12.

Vertical Transformations (y)

1. Multiply by 1/2
2. Add 1

In other words, we have a vertical shrink by a factor of 1/2 and a shift up by 1.
Applying the transformations to each x and each y gives us: $$(0, 1) → \left(-\frac{5π}{12}, \frac{3}{2}\right)$$ $$\left(\frac{π}{2}, 0\right) → \left(-\frac{π}{6}, 1\right)$$ $$\left(π, -1\right) → \left(\frac{π}{12}, \frac{1}{2}\right)$$ $$\left(\frac{3π}{2}, 0\right) → \left(\frac{π}{3}, 1\right)$$ $$(2π, 1) → \left(\frac{7π}{12}, \frac{3}{2}\right)$$ Desmos link for more detail Now that we have the guide, we need to sketch the vertical asymptotes. On the graph of our basic cosine function (y = cos x), we have x-intercepts at $π$/2 and 3$π$/2. This occurs at 1/4 and 3/4 of the way through our cycle. Those x-values are now -$π$/6 and $π$/3. The y-values will be 1 in each case since we have a shift up by 1 unit. This means we will draw our vertical asymptotes at: $$x=-\frac{π}{6}$$ $$\text{and}$$ $$x=\frac{π}{3}$$ Now, we are ready to sketch our graph. We will add in some additional vertical asymptotes using the period. This is optional and only done to extend the graph to the left and to the right a bit. Graphing one cycle is generally good enough for most coursework as we just need to understand the concept.
Desmos link for more detail Example #2: Find the period, phase shift, and vertical shift. Sketch the graph. $$f(x)=\sec\left(\frac{x}{3}- \frac{2π}{3}\right) + 1$$ As we have done previously, let's factor to make finding the phase shift easier: $$f(x)=\sec\left[\frac{1}{3}\left(x - 2π\right)\right] + 1$$ For the period, we will divide 2$π$ by 1/3, which is the same as multiplying by 3. This will give us a period of 6$π$. $$\text{Period}: 6π$$ $$\text{Phase Shift}: \text{right}\hspace{.1em}2π$$ $$\text{Vertical Shift}: \text{Up}\hspace{.1em}1$$ Replace secant with cosine and sketch the graph. $$f(x)=\cos\left[\frac{1}{3}\left(x - 2π\right)\right] + 1$$ We will begin with points from our basic cosine function and think about the transformations that are applied to x and y separately. $$f(x)=\cos \hspace{.1em}x$$ $$(0, 1)$$ $$\left(\frac{π}{2}, 0\right)$$ $$\left(π, -1\right)$$ $$\left(\frac{3π}{2}, 0\right)$$ $$(2π, 1)$$

Horizontal Transformations (x)

1. Multiply by 3
2. Add 2$π$

In other words, we have a horizontal stretch by a factor of 3 and a shift right by 2$π$.

Vertical Transformations (y)

1. Add 1

In other words, we have a shift up by 1.
Applying the transformations to each x and each y gives us: $$(0, 1) → \left(2π, 2\right)$$ $$\left(\frac{π}{2}, 0\right) → \left(\frac{7π}{2}, 1\right)$$ $$\left(π, -1\right) → \left(5π, 0\right)$$ $$\left(\frac{3π}{2}, 0\right) → \left(\frac{13π}{2}, 1\right)$$ $$(2π, 1) → \left(8π, 2\right)$$ Desmos link for more detail Now that we have the guide, we need to sketch the vertical asymptotes. On the graph of our basic cosine function (y = cos x), we have x-intercepts at $π$/2 and 3$π$/2. This occurs at 1/4 and 3/4 of the way through our cycle. Those x-values are now 7$π$/2 and 13$π$/2. The y-values will be 1 in each case since we have a shift up by 1 unit. This means we will draw our vertical asymptotes at: $$x=\frac{7π}{2}$$ $$\text{and}$$ $$x=\frac{13π}{2}$$ Now, we are ready to sketch our graph. We will add in some additional vertical asymptotes using the period. This is optional and only done to extend the graph to the left and to the right a bit. Graphing one cycle is generally good enough for most coursework as we just need to understand the concept.
Desmos link for more detail

Cosecant Function

$$\csc \hspace{.1em}x=\frac{1}{\sin \hspace{.1em}x}$$ Since we can't divide by zero, the cosecant function will be undefined where sine values are zero. Recall from our last lesson, that this occurs at integer multiples of $π$, and just like we saw with our cotangent function, we will have vertical asymptotes for those values. Let's create a basic sketch for cosecant over one period. We will use a table for the basic points found over the interval from (-$π$, $π$):

Graphing the Cosecant Function over One Period

Desmos link for more detail As mentioned earlier, the period of cosecant is 2$π$ since the cosecant values are reciprocals of the corresponding sine values. When the sine value is -1, the cosecant value will also be -1. This occurs at $-\frac{π}{2}$ + 2$π$n, where n is any integer. Similarly, when the sine value is 1, the cosecant value will also be 1. This occurs at $\frac{π}{2}$ + 2$π$n, where n is any integer. The range for sine is from [-1, 1], which means that |csc x| ≥ 1 for all x in its domain. Recall that dividing 1 by a number between 0 and 1 will give you a value that is larger than 1. For example, think about the sine of $π$/6 and the cosecant of $π$/6. $$\sin\left(\frac{π}{6}\right)=\frac{1}{2}$$ $$\csc\left(\frac{π}{6}\right)=\frac{1}{\sin\left(\frac{π}{6}\right)}=\frac{1}{\frac{1}{2}}=2$$ Desmos link for more detail Looking at the above graph, we can see how the graphs of y = sin x and y = csc x are related.

• x-intercepts on the sine curve correspond to vertical asymptotes on the cosecant curve
• A maximum point on the sine curve corresponds to a minimum point on a continuous portion of the cosecant curve
• A minimum point on the sine curve corresponds to a maximum point on a continuous portion of the cosecant curve

An easy way to remember this is to say that the "hills" and "valleys" are interchanged. To be more specific, we could say that a valley or minimum point on the sine graph, gives us a hill or relative maximum on the cosecant graph, and a hill or maximum point on the sine graph, gives us a valley or relative minimum on the cosecant graph.

The Cosecant Curve and its Characteristics

• The domain consists of all real numbers except $π$n, where n is any integer
• f(x) = csc x is discontinuous at $π$n, where n is any integer
  • The graph contains vertical asymptotes at those values
• There are no x-intercepts
• The period is 2$π$
• There are no minimum or maximum values
  • The range is: $(-\infty, -1] ∪ [1, \infty)$
  • The graph has no amplitude
• The graph is symmetric with respect to the origin
  • csc(-x) = -csc x, for all x in the domain
  • f(x) = csc x is an odd function

Graphing Variations of the Cosecant Function

1. Replace cosecant with sine and sketch the graph using a dashed curve
   • Sine is used as a guide
2. Sketch the vertical asymptotes
   • These occur at the beginning, middle, and end of the sine cycle
3. Sketch the graph by drawing the u-shaped branches between consecutive vertical asymptotes
   • The "hills" and "valleys" are interchanged
4. (Optional) Use the period to find additional cycles to the left or right as needed

Example #3: Find the period, phase shift, and vertical shift. Sketch the graph. $$f(x)=\frac{1}{2}\csc\left(x + \frac{π}{3}\right) + 2$$ $$\text{Period}: 2π$$ $$\text{Phase Shift}: \text{left}\frac{π}{3}$$ $$\text{Vertical Shift}: \text{Up}\hspace{.1em}2$$ Replace cosecant with sine and sketch the graph. $$f(x)=\frac{1}{2}\sin\left(x + \frac{π}{3}\right) + 2$$ We will begin with points from our basic sine function and think about the transformations that are applied to x and y separately. $$f(x)=\sin \hspace{.1em}x$$ $$(0, 0)$$ $$\left(\frac{π}{2}, 1\right)$$ $$\left(π, 0\right)$$ $$\left(\frac{3π}{2}, -1\right)$$ $$(2π, 0)$$

Horizontal Transformations (x)

1. Subtract away $π$/3

In other words, we have a shift left by $π$/3.

Vertical Transformations (y)

1. Multiply by 1/2
2. Add 2

In other words, we have a vertical shrink by a factor of 1/2 and a shift up by 2.
Applying the transformations to each x and each y gives us: $$(0, 0) → \left(-\frac{π}{3}, 2\right)$$ $$\left(\frac{π}{2}, 1\right) → \left(\frac{π}{6}, \frac{5}{2}\right)$$ $$\left(π, 0\right) → \left(\frac{2π}{3}, 2\right)$$ $$\left(\frac{3π}{2}, -1\right) → \left(\frac{7π}{6}, \frac{3}{2}\right)$$ $$(2π, 0) → \left(\frac{5π}{3}, 2\right)$$ Desmos link for more detail Now that we have the guide, we need to sketch the vertical asymptotes. On the graph of our basic sine function (y = sin x), we have x-intercepts at 0, $π$, and 2$π$. The beginning, middle, and end of the sine cycle. Those x-values are now -$π$/3, 2$π$/3, and 5$π$/3. The y-values will be 2 in each case since we have a shift up by 2 units. This means we will draw our vertical asymptotes at: $$x=-\frac{π}{3}$$ $$\text{and}$$ $$x=\frac{2π}{3}$$ $$\text{and}$$ $$x=\frac{5π}{3}$$ Now, we are ready to sketch our graph. We will add in some additional vertical asymptotes using the period. This is optional and only done to extend the graph to the left and to the right a bit. Graphing one cycle is generally good enough for most coursework as we just need to understand the concept.
Desmos link for more detail Example #4: Find the period, phase shift, and vertical shift. Sketch the graph. $$f(x)=2\csc\left(\frac{x}{2}- \frac{5π}{6}\right) - 1$$ As we have done previously, let's factor to make finding the phase shift easier: $$f(x)=2\csc\left[\frac{1}{2}\left(x - \frac{5π}{3}\right)\right] - 1$$ $$\text{Period}: 4π$$ $$\text{Phase Shift}: \text{right}\frac{5π}{3}$$ $$\text{Vertical Shift}: \text{Down}\hspace{.1em}1$$ Replace cosecant with sine and sketch the graph. $$f(x)=2\sin\left[\frac{1}{2}\left(x - \frac{5π}{3}\right)\right] - 1$$ We will begin with points from our basic sine function and think about the transformations that are applied to x and y separately. $$f(x)=\sin \hspace{.1em}x$$ $$(0, 0)$$ $$\left(\frac{π}{2}, 1\right)$$ $$\left(π, 0\right)$$ $$\left(\frac{3π}{2}, -1\right)$$ $$(2π, 0)$$

Horizontal Transformations (x)

1. Multiply by 2
2. Add 5$π$/3

In other words, we have a horizontal stretch by a factor of 2 and a shift right by 5$π$/3.

Vertical Transformations (y)

1. Multiply by 2
2. Subtract away 1

In other words, we have a vertical stretch by a factor of 2 and a shift down by 1.
Applying the transformations to each x and each y gives us: $$(0, 0) → \left(\frac{5π}{3}, -1\right)$$ $$\left(\frac{π}{2}, 1\right) → \left(\frac{8π}{3}, 1\right)$$ $$\left(π, 0\right) → \left(\frac{11π}{3}, -1\right)$$ $$\left(\frac{3π}{2}, -1\right) → \left(\frac{14π}{3}, -3\right)$$ $$(2π, 0) → \left(\frac{17π}{3}, -1\right)$$ Desmos link for more detail Now that we have the guide, we need to sketch the vertical asymptotes. On the graph of our basic sine function (y = sin x), we have x-intercepts at 0, $π$, and 2$π$. The beginning, middle, and end of the sine cycle. Those x-values are now 5$π$/3, 11$π$/3, and 17$π$/3. The y-values will be -1 in each case since we have a shift down by 1 unit. This means we will draw our vertical asymptotes at: $$x=\frac{5π}{3}$$ $$\text{and}$$ $$x=\frac{11π}{3}$$ $$\text{and}$$ $$x=\frac{17π}{3}$$ Now, we are ready to sketch our graph. We will add in some additional vertical asymptotes using the period. This is optional and only done to extend the graph to the left and to the right a bit. Graphing one cycle is generally good enough for most coursework as we just need to understand the concept.
Desmos link for more detail

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Skills Check: