8.6: Graphs of the Remaining Trigonometric Functions

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Note to the Instructor (click to expand): The graphs of the rest of the trigonometric functions are covered in this section. Try your best to avoid treating this material like an afterthought. The graphs of the secant and cosecant spring up in odd places in Calculus, so the students need to be aware that this material, while held off for the end of the chapter, is critical to understand.

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

Graph the cosecant, secant, and cotangent functions, along with transformations of these functions.
Determine the equation of a trigonometric function given its graph.

A Summary of the Graphs of the Fundamental Trigonometric Functions

Before we go too far, let's summarize what we know about the fundamental trigonometric functions.

Theorem: Properties of the Graph of the Sine Function

Theorem: Properties of the Graph of the Cosine Function

Theorem: Properties of the Graph of the Tangent Function

The principal cycle of the tangent is $\left[ 0,\pi \right]$, even though the tangent is undefined halfway through the cycle. It's just the cycle we use as the "standard" for our base graph, and it is okay that the tangent is undefined at some point within the cycle. The principal cycle is not the domain.

An Interlude—Vertical Asymptotes

We now turn our attention to the graphs of the remaining trigonometric functions—$\csc\left( x \right)$, $\sec\left( x \right)$, and $\cot\left( x \right)$. In each of these cases, we must understand the conceptual approach to graphing we developed in when we introduced the graphs of the fundamental trigonometric functions.

Recall, if given a fraction where the numerator is held constant and the denominator is approaching zero (but not equal to zero), then the value of the fraction is approaching either $+\infty$ or $-\infty$. To drive the point home, let's demonstrate this behavior both numerically and graphically.

For the numerical demonstration, you can create any rational function you want where the numerator is a constant and the denominator is an algebraic expression. All we require is that the denominator approaches zero as the values of $x$ approach some value. For example, suppose we choose the numerator to be $7$ and the denominator to be $6x^2 + 11x + 3$. Then the function we are working with is\[R(x)=\dfrac{7}{6x^2 + 11x + 3}. \nonumber\]We now want the values of $x$ to approach some number that causes division by zero. To reveal what causes division by zero for this function, we factor the denominator as follows:\[R(x) = \dfrac{7}{\left( 3x + 1 \right)\left( 2x + 3 \right)}. \nonumber\]We can see this function would have division by zero at $x = -\frac{1}{3}$ and $x = -\frac{3}{2}$.

Let's build two tables of values—the first to see what happens to the function as $x \to -\frac{3}{2}^-$, and the second to see what happens as $x \to -\frac{3}{2}^+$. For convenience, we will rewrite $-\frac{3}{2}$ in its decimal form, $-1.5$.\[\begin{array}{|c|c|c|} \hline x & \text{Denominator} = (3x+1)(2x+3) & R(x) \\[6pt] \hline -2 & 5 & 1.4 \\[6pt] \hline -1.6 & 0.76 & 9.2105263 \\[6pt] \hline -1.55 & 0.365 & 19.178082 \\[6pt] \hline -1.51 & 0.0706 & 99.150142 \\[6pt] \hline -1.505 & 0.03515 & 199.14651 \\[6pt] \hline -1.501 & 0.007006 & 999.14359 \\[6pt] \hline \end{array} \quad \begin{array}{|c|c|c|} \hline x & \text{Denominator} = (3x+1)(2x+3) & R(x) \\[6pt] \hline -1 & -2 & -3.5 \\[6pt] \hline -1.4 & -0.64 & -10.9375 \\[6pt] \hline -1.45 & -0.335 & -20.895522 \\[6pt] \hline -1.49 & -0.0694 & -100.86455 \\[6pt] \hline -1.495 & -0.03485 & -200.86083 \\[6pt] \hline -1.499 & -0.006994 & -1000.8579 \\[6pt] \hline \end{array} \nonumber\]In the first table, as $x$ is approaching $-1.5$ from the left, the denominator of $R(x)$ is approaching zero from the right. Simultaneously, the function values are becoming larger and larger positive numbers. That is,\[\text{As} \, x \to -\dfrac{3}{2}^-, \, R(x) \to \infty. \nonumber\]Similarly, the second table demonstrates that, as $x$ is approaching $-1.5$ from the right, the denominator of $R(x)$ approaches zero from the left. This causes the function values to become larger and larger negative numbers. We would say,\[\text{As} \, x \to -\dfrac{3}{2}^+, \, R(x) \to -\infty. \nonumber\]

For the graphical demonstration, we quickly sketch the graph of $R(x)$ around $x = -\frac{3}{2}$.

5.5.1 Figure.png — **Figure $\PageIndex{1}$**

As you can see from the graph, the $y$-values of the function (also called the function values) become more and more positive as the $x$-values get closer and closer to $-1.5$ from the left. Likewise, the function values become increasingly more negative as the $x$-values approach $-1.5$ from the right.

The moral of the story can be stated as follows:

If the numerator is a nonzero constant and the denominator is tending to (but not becoming) zero, the ratio\[\dfrac{\text{numerator}}{\text{denominator}} \nonumber\]approaches either $+\infty$ or $-\infty$. The sign can be determined by comparing the signs of the numerator and denominator as the denominator approaches zero. Graphically, this means there is a vertical asymptote at this value of $x$.

Graphing the Cosecant Function

We know that $\csc\left( x \right) = \frac{1}{\sin\left( x \right)}$. Therefore, by the preceding discussion, as the denominator tends to zero, the overall ratio (which is the cosecant) will tend to $+\infty$ or $-\infty$. The zeros of the sine function occur at $x = k \pi$, where $k \in \mathbb{Z}$ (see Figure $\PageIndex{2}$). Let's start by focusing on the zero at $x = 0$.

5.5.2 Figure.png — **Figure $\PageIndex{2}$:** *$y = \sin\left( x \right)$*

From Figure $\PageIndex{2}$, we can see as $x \to 0^-$, the $y$-values of $\sin\left( x \right)$ are approaching $0^-$. Therefore,\[\text{as} \, x \to 0^-, \, \dfrac{1}{\sin\left( x \right)} \to \dfrac{1}{0^-} \to -\infty. \nonumber\]It is negative infinity because the numerator is positive, but the denominator is negative (albeit, very close to zero). Similarly, as $x \to 0^+$, the $y$-values of $\sin\left( x \right)$ are approaching $0^+$. Therefore,\[\text{as} \, x \to 0^+, \, \dfrac{1}{\sin\left( x \right)} \to \dfrac{1}{0^+} \to +\infty = \infty. \nonumber\]Again, we know the result is positive infinity because the numerator is positive and the denominator is positive. Sketching this behavior, we get the following:

5.5.4 Figure.png — **Figure $\PageIndex{3}$:** *Beginning the sketch of $y = \csc\left( x \right)$*

It's hard to see in Figure $\PageIndex{3}$, but we drew a vertical asymptote at $x = 0$ (the $y$-axis). A similar investigation shows that\[\text{as} \, x \to \pi^-, \, \dfrac{1}{\sin\left( x \right)} \to \dfrac{1}{0^+} \to \infty \nonumber\]and\[\text{as} \, x \to \pi^+, \, \dfrac{1}{\sin\left( x \right)} \to \dfrac{1}{0^-} \to -\infty. \nonumber\]Using this to fill in a little more of the graph, we get the following:

5.5.5 Figure.png — **Figure $\PageIndex{4}$:** *Getting closer to the graph of $y = \csc\left( x \right)$*

From this, we expect the cosecant function to have vertical asymptotes at these zeros of $\sin\left( x \right)$ (due to the division by zero). Furthermore, we expect the cosecant to be positive when its reciprocal (the sine) is positive, and negative when the reciprocal is negative. This gets us very close to the base graph of the cosecant.

5.5.6 Figure.png — **Figure $\PageIndex{5}$:** *Almost there*

To finish out the graph of the base cosecant function, we will build a table of values for the interval $\left[ \frac{\pi}{6}, \frac{5\pi}{6} \right]$ and use this to extrapolate the features for the rest of the graph.\[\begin{array}{|c|c|} \hline x & \csc\left( x \right) \\[6pt] \hline \dfrac{\pi}{6} & 2 \\[6pt] \hline \dfrac{\pi}{4} & \sqrt{2} \\[6pt] \hline \dfrac{\pi}{3} & \dfrac{2}{\sqrt{3}} \\[6pt] \hline \dfrac{\pi}{2} & 1 \\[6pt] \hline \dfrac{2\pi}{3} & \dfrac{2}{\sqrt{3}} \\[6pt] \hline \dfrac{3\pi}{4} & \sqrt{2} \\[6pt] \hline \dfrac{5\pi}{6} & 2 \\[6pt] \hline \end{array} \nonumber\]Plotting these values, we get the following:

5.5.7 Figure.png — **Figure $\PageIndex{6}$:** *So close!*

We can now see the pattern for the entire graph of the cosecant. Filling in the blanks, we get our final graph.

5.5.8 Figure.png — **Figure $\PageIndex{7}$:** *The base graph of the cosecant, $y = \csc\left( x \right)$*

I have left a very lightly sketched sine function in this graph. This is because when you are graphing the cosecant, you will inevitably need to start by graphing its reciprocal.

Just as we did with the fundamental trigonometric functions, we now summarize the properties of the cosecant function (with the help of its graph as a visual aid).

Theorem: Properties of the Graph of the Cosecant Function

Graphing the Secant Function

The derivation of the graph of the secant function is similar to that of the cosecant function. It is left for you as an exercise.

Theorem: Properties of the Graph of the Secant Function

As with the graph of the cosecant, we usually lightly graph the reciprocal of the secant (which is the cosine) to help us in graphing the secant.

Graphing the Cotangent Function

As with the cosecant and the secant, to graph the cotangent function, we will use its reciprocal. That is, because $\cot\left( x \right) = \frac{1}{\tan\left( x \right)}$, we will first graph the tangent and discuss what happens as the tangent function tends to zero.

5.5.9 Figure Fixed.png — **Figure $\PageIndex{8}$:** *$y = \tan\left( x \right)$*

Figure $\PageIndex{8}$ shows the base graph of the tangent function. Using the same arguments as before, we can get the behavior of the reciprocal function, $\frac{1}{\tan\left( x \right)} = \cot\left( x \right)$, near the zeros of the tangent.

5.5.10 Figure.png — **Figure $\PageIndex{9}$:** *Almost done graphing the base graph of the cotangent*

Notice in Figure $\PageIndex{9}$ that the vertical asymptotes are not in the same places as the vertical asymptotes for the tangent. This is because the tangent is undefined when the cosine function is zero (at $x = \frac{\pi}{2} + k \pi$) and the cotangent is undefined when the sine is zero (at $x = k \pi$). Evaluating the cotangent function for a few special values between $x=0$ and $x = \pi$ should give us the pattern of those missing middle regions.\[\begin{array}{|c|c|} \hline x & \cot\left( x \right) \\[6pt] \hline \dfrac{\pi}{4} & 1 \\[6pt] \hline \dfrac{\pi}{2} & 0 \\[6pt] \hline \dfrac{3\pi}{4} & -1 \\[6pt] \hline \end{array} \nonumber\]Plotting these points, filling out the rest of the curve with the same pattern, and erasing the underlying tangent graph, we get the following:

5.5 Base Cotangent.png — **Figure $\PageIndex{10}$:** *The base graph of the cotangent, $y = \cot\left( x \right)$*

As before, we list all of the properties in a convenient theorem.

Theorem: Properties of the Graph of the Cotangent Function

Transformations of the Reciprocal Functions

Graphing transformations of the reciprocal functions is the same as graphing the transformations of the fundamental trigonometric functions. In fact, to graph transformations of the cosecant and secant functions, we first fully graph the transformations of their related fundamental functions. After that, graphing the transformed cosecant and secant functions is a somewhat trivial move. Let's see this in action.

Example $\PageIndex{1}$

Graph a single cycle of each function. State all key characteristics of the graph (including domain, range, period, phase shift, etc.).

$f(x) = -1.2 + 2.8\sec\left( -3x - \frac{\pi}{2} \right)$
$g(x) = 2 \cot\left( 2x - 2 \right) + 1$

Solutions

Just as we have done previously, we rewrite it in the better form\[f(x) = 2.8 \sec\left( -3x - \frac{\pi}{2} \right) - 1.2 \nonumber\]Using the fact that secant is an even function, we clean this up a little more to the following:\[f(x) = 2.8\sec\left( -\left( 3x + \frac{\pi}{2} \right) \right) - 1.2 = 2.8 \sec\left( 3x + \frac{\pi}{2} \right) - 1.2 \nonumber\]Now that we are finished "preconditioning" the function, it's time we discussed how we are going to graph it. With the secant and cosecant, we perform the intermediate step of graphing their reciprocal functions, the cosine and sine, respectively. So, let's graph\[y = 2.8\cos\left( 3x + \frac{\pi}{2} \right) - 1.2. \nonumber\]We list out all the essential information.\[\begin{array}{rclcrcl} \hline A & = & 2.8 & \implies & \text{Amplitude} & = & 2.8 \\[6pt] \hline B & = & 3 & \implies & \text{Period} & = & \dfrac{\text{Natural Period}}{B} = \dfrac{2 \pi}{3} \\[6pt] \hline \quad & \quad & \quad & \implies & \text{Step Size} & = & \dfrac{\text{Period}}{4} = \dfrac{2 \pi/3}{4} = \dfrac{\pi}{6} \\[6pt] \hline C & = & \dfrac{\pi}{2} & \implies & \text{Phase Shift} & = & - \dfrac{C}{B} = - \dfrac{\pi/2}{3} = -\dfrac{\pi}{6} \\[6pt] \hline D & = & -1.2 & \implies & \text{Vertical Shift} & = & -1.2 \\[6pt] \hline \end{array} \nonumber\]As usual, we graph and label the midline, label the initial number as the phase shift, and step to the other key numbers using the step size.
$5.5.1 Example.a.1.png$
Figure $\PageIndex{11}$

Note that the distance between the first and last key number is\[\left| \dfrac{3\pi}{6} - \left( -\dfrac{\pi}{6} \right) \right| = \dfrac{2\pi}{3} = \text{Period}, \nonumber\]so we have done something right! We then graph the non-reflected cosine, draw in and label the $y$-axis, and use that to place the $x$-axis.
$5.5.1 Example.a.2.png$
Figure $\PageIndex{12}$

Now comes the new part. Wherever the cosine function crosses the midline, the original function is undefined. This gives us our vertical asymptotes at $x = 0$, $x = \pm\frac{2\pi}{6} = \frac{\pi}{3}$, $x = \pm \frac{4\pi}{6} = \frac{2 \pi}{3}$, and so on. Graphing these vertical asymptotes allows us to graph the secant.
$5.5.1 Example.a.3.png$
Figure $\PageIndex{13}$

Notice that I didn't remove the sketch of the cosine. This is perfectly fine (and some instructors expect you to leave it). Finally, we should state as much about this function as we can. The domain (by looking at our spectacular graph) is all $x \ne \frac{\pi}{3}k$, where $k \in \mathbb{Z}$. The range (again, from the graph) is $\left( -\infty, -4 \right] \cup \left[ 1.6, \infty \right)$. The only thing that we cannot state at this point are the zeros of this function (luckily, I didn't ask for them).
Before we start, I need to be clear and state that the secant and cosecant functions are the only ones where we must first graph their reciprocal functions. The base graph for the cotangent is simple enough to handle without bothering with the tangent.

Now, there isn't any preconditioning necessary for this function, so we go ahead and jump right into the other stages. Just like the tangent, the cotangent does not have an amplitude and its natural period is $\pi$.\[\begin{array}{rclcrcl} \hline B & = & 2 & \implies & \text{Period} & = & \dfrac{\text{Natural Period}}{B} = \dfrac{\pi}{2} \\[6pt] \hline \quad & \quad & \quad & \implies & \text{Step Size} & = & \dfrac{\text{Period}}{4} = \dfrac{\pi/2}{4} = \dfrac{\pi}{8} \\[6pt] \hline C & = & -2 & \implies & \text{Phase Shift} & = & - \dfrac{C}{B} = - \dfrac{-2}{2} = 1 \\[6pt] \hline D & = & 1 & \implies & \text{Vertical Shift} & = & 1 \\[6pt] \hline \end{array} \nonumber\]Graphing and labeling the midline, plotting the phase shift as the first key number, and stepping to each subsequent key number using the step size, we get the following:
$5.5.1 Example.b.1 Fix.png$
Figure $\PageIndex{14}$

Because the phase shift was a rational number and the step size was an irrational number, we could not easily add them. Leaving them in the form above is preferred. If you look at the base graph for the cotangent, its principal cycle occurs from $x = 0$ to $x = \pi$. This cycle starts and ends with a vertical asymptote, decreases from left to right, and crosses the midline halfway through. That's precisely what we are going to do here.
$5.5.1 Example.b.2.png$
Figure $\PageIndex{15}$

Now we place the $y$-axis one unit to the left of the phase shift and the $x$-axis one unit down from the midline. How do I know how far one unit is? It's more of an art of approximation.
$5.5.1 Example.b.4.png$
Figure $\PageIndex{16}$

Finally, we should label the two points on the curve at $x=1+\frac{\pi}{8}$ and $x = 1 + \frac{3\pi}{8}$. The good news is that the $y$-values of these points vary from the midline by $|A|$. Thus, the corresponding $y$-values are $y=1+2=3$ and $y = 1-2 =-1$.
$5.5.1 Example.b.5.png$
Figure $\PageIndex{17}$

The domain of the function is all $x \ne 1 + \frac{\pi}{2}k$, where $k \in \mathbb{Z}$. The range is all real numbers, $\mathbb{R}$.

Checkpoint $\PageIndex{1}$

Graph the function. Clearly state all key information (including domain and range).\[y = 3 - 4 \csc\left( -\pi x + \dfrac{2}{3}\pi \right) \nonumber\]

Answer: Amplitude of the related sine function: $4$
Period: $2$
Step Size: $\frac{1}{2}$
Phase Shift: $\frac{2}{3}$
Vertical Shift: $3$
Domain: All $x \ne \frac{2}{3} + k$, where $k \in \mathbb{Z}$
Range: $\left( -\infty, 1 \right] \cup \left[ 7, \infty \right)$

$5.5.1 Checkpoint.png$
Figure $\PageIndex{18}$

Finding Equations Given Graphs

Finding the equation of a reciprocal function given its graph requires a little bit more work than with the fundamental trigonometric functions.

Example $\PageIndex{2}$

Find the equation of the graphed function.

$5.5.2 Example.a.1.png$
Figure $\PageIndex{19}$
$5.5.2 Example.b.1.png$
Figure $\PageIndex{20}$

Solutions

As with the sine and cosine functions, graphs of the cosecant and secant are so similar that you could use either as an equation for the given graph (they would just be shifts of one another). A good rule of thumb when building trigonometric equations from graphs is to read the graph from left to right. The piece of information (whether a point or an equation of an asymptote) you have been given will set the beginning of the cycle for the trigonometric function. Therefore, in this case, we will assume the cycle of the trigonometric function starts at $x = \frac{3}{4}$. With the cosecant and secant, it's beneficial to draw a sinusoid "helper" function. Looking at the given graph, we can see that the midline should be at $y = -\frac{1}{2}$. Drawing this in allows us to sketch our "helper" function.
$5.5.2 Example.a.2.png$
Figure $\PageIndex{21}$

We now find the equation of the "helper" function. It is definitely a sine function starting at $x = \frac{3}{4}$, so the phase shift is\[-\dfrac{C}{B} = \dfrac{3}{4}.\nonumber\]The midline is $y = -\frac{1}{2}$, so the vertical shift is\[D = -\dfrac{1}{2}.\nonumber\]Additionally, the amplitude is the distance between the midline and the maximum or minimum values of the sinusoid. Since the maximum is $y = 0$ and the midline is $y = -\frac{1}{2}$, the amplitude is $\frac{1}{2}$. Moreover, since this sine function is not reflected,\[A = \dfrac{1}{2}. \nonumber\]The only thing left to find is the period. This would give us an equation to find $B$, which in turn would allow us to find $C$. The cycle begins at $x = \frac{3}{4}$ and three-quarters of the way through the cycle, we are given the point $\left( \frac{21}{4},-1 \right)$. That is, the distance between $\frac{3}{4}$ and $\frac{21}{4}$ is three-quarters of the period.\[\begin{array}{rrcl} & \dfrac{3}{4} \, \text{of the Period} & = & \left| \dfrac{3}{4} - \dfrac{21}{4} \right| \\[6pt] & & = & \left| -\dfrac{18}{4} \right| \\[6pt] & & = & \dfrac{9}{2} \\[6pt] \end{array} \nonumber\]Therefore,\[\begin{array}{rcl} \text{Period} & = & \left( \dfrac{9}{2} \right)\left( \dfrac{4}{3} \right) \\[6pt] & = & 6 \\[6pt] \end{array} \nonumber\]Since the period is $\frac{2\pi}{B}$,\[\dfrac{2 \pi}{B} = 6 \implies \dfrac{2 \pi}{6} = B \implies \dfrac{\pi}{3} = B. \nonumber\]Earlier we stated that $-\frac{C}{B} = \frac{3}{4}$, so\[-\dfrac{C}{\pi/3} = \dfrac{3}{4} \implies C = - \dfrac{\pi}{4}. \nonumber\]Thus, the "helper" function is\[y = A \, \sin\left( Bx + C \right) + D = \dfrac{1}{2} \, \sin\left( \dfrac{\pi}{3}x - \dfrac{\pi}{4} \right) - \dfrac{1}{2}. \nonumber\]Finally, the actual function is the reciprocal of the sine. That is, the function in the initial graph is\[y = \dfrac{1}{2} \, \csc\left( \dfrac{\pi}{3}x - \dfrac{\pi}{4} \right) - \dfrac{1}{2}. \nonumber\]
Again, you could write the equation for this in terms of the (reflected) tangent function; however, we will write our equation in terms of the cotangent. Sometimes, you have to "eyeball" the features of a graph. In this example, the midline looks to be $y = 4$, so we will say $D = 4$. The phase shift could be $-\frac{\pi}{6}$ or $+\frac{\pi}{6}$, but the extra point they gave us is in the interval $\left( -\frac{\pi}{6}, \frac{\pi}{6} \right)$, so we will choose our phase shift to be $-\frac{\pi}{6}$. This means\[-\dfrac{C}{B} = -\dfrac{\pi}{6}. \nonumber\]Furthermore, the period is easy to spot on this graph. From vertical asymptote to vertical asymptote, the distance is $\frac{2\pi}{6} = \frac{\pi}{3}$. Therefore, the period is $\frac{\pi}{3}$ and, as such,\[\dfrac{\pi}{B} = \dfrac{\pi}{3} \implies B = 3. \nonumber\]With this, we can find $C$.\[-\dfrac{C}{B} = -\dfrac{\pi}{6} \implies \dfrac{C}{3} = \dfrac{\pi}{6} \implies C = \dfrac{\pi}{2}. \nonumber\]We are close to the finish line. Our function is\[y = A \, \cot\left( 3x + \dfrac{\pi}{2} \right) + 4. \nonumber\]Now we use the given data point to help us find the value of $A$.\[\begin{array}{rrcl} & 4.1 & = & A \, \cot\left( 3\left( -\dfrac{\pi}{12} \right) + \dfrac{\pi}{2} \right) + 4 \\[6pt] \implies & 0.1 & = & A \, \cot\left( -\dfrac{\pi}{4} + \dfrac{\pi}{2} \right) \\[6pt] \implies & 0.1 & = & A \, \cot\left( \dfrac{\pi}{4} \right) \\[6pt] \implies & 0.1 & = & A \, \left( 1 \right) \\[6pt] \implies & 0.1 & = & A \\[6pt] \end{array} \nonumber\]Thus, the function is\[y = 0.1 \, \cot\left( 3x + \dfrac{\pi}{2} \right) + 4. \nonumber\]

Checkpoint $\PageIndex{2}$

Find the equation of the graphed function.

5.5.2 Checkpoint.png — **Figure $\PageIndex{22}$**

Answer: $y = 2\sec\left(3x+\frac{\pi}{5}\right)+1$

Success in Mathematics: Math Anxiety (Part 3)

For some math students, anxiety changes their ability to show what they know. They have physical and emotional symptoms. You are responsible for recognizing if math anxiety is interfering with your success. Do you need to make some changes to reduce your math anxiety? Do you need help to do so?

Researchers have identified various events that seem to result in math anxiety. Perhaps you had negative experiences learning math. For you, learning math meant feeling dumb or embarrassed. Perhaps a teacher or parent who struggled with math has passed on their fears to you. Or you believe that your gender or ethnic heritage means you cannot do the math.

Some students learn to cope on their own with their math anxiety. They use relaxation techniques. They practice problems over and over. At test time, their anxiety does not block them from doing the work. Others find help from counselors at their college.

Math anxiety is common among college students. It is your responsibility to think about it and, if needed, get help so that you can cope with your fears.

Math anxiety affects students in physical ways such as increasing blood pressure and heart rate, breathing faster, excess sweating, dry mouth, restlessness, or nausea. Emotionally, it can cause fear, anger, nervousness, worry, and feelings of hopelessness. Students may have trouble concentrating, sitting still, reading, or keeping their legs from bouncing. They may feel desperate - like they need to run away or cry.

Here are some questions about this topic that may help you succeed in this class (and possibly many other classes as well):

Using a scale of 1 to 10, with 10 describing overwhelming anxiety that makes it extremely difficult to do math, rate your math anxiety. How does it affect you?
Most students have some anxiety about their classes. They worry about giving speeches, taking tests, writing a good paper, or completing all their work. Worrying can be motivating. It can help us get going and to get things done. Or, worrying can be a bad thing. It paralyzes us. In general, how does worrying affect you?

Applications of Mathematics: Poker

So, where does mathematics apply in poker? Poker players can use mathematics in many different situations; however, it is most commonly used when a player is on a draw, such as a flush or straight draw. If an opponent makes a bet and you are on a draw, you are faced with the decision of whether or not you should call to try and complete the draw by the next card or fold and let your opponent take the pot. In this situation, a player with good knowledge of poker mathematics will always know whether or not to call, while a player who does not know mathematics will be unsure.

Players unfamiliar with poker odds will make an educated guess on whether or not they should call. If the bet is large, they may feel it is too expensive to try and catch the right card, but if the bet is small, they will be more inclined to call. On the other hand, a player that can use mathematics correctly will be able to work out the pot odds they are getting on the hand and act accordingly. Pot odds consider the amount your opponent has bet in relation to the pot size and the likelihood of completing your draw to inform you about whether or not you should call or fold.

Can you think of three other games that rely explicitly or implicitly on mathematics? Explain.
Oddly enough, most of us are superstitious about numbers. Some people even play their "favorite" numbers when they play the lottery. Suppose you pick your favorite numbers, and I pick the numbers 1, 2, 3, 4, 5, and 6. Who has better odds of winning? Why are you still likely to win more if the odds are the same?

This page titled 8.6: Graphs of the Remaining Trigonometric Functions was last modified on Tue, 08 Jul 2025 17:45:28 GMT and is shared under a CC BY-SA 12 license and was authored, remixed, and/or curated by Roy Simpson.

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