7.5 Graphs of Other Trigonometric Functions

Vanessa Coffelt

7.5 Graphs of Other Trigonometric Functions

7.5.1 Graphs of the Secant and Cosecant Functions

As mentioned at the end of Section 7.4, one way to proceed with our analysis of the circular functions is to use what we know about the functions $\sin(t)$ and $\cos(t)$ to rewrite the four additional circular functions in terms of sine and cosine with help from Theorem 7.8. We use this approach to analyze $F(t) = \sec(t)$ .

Rewriting $F(t) = \sec(t)= \frac{1}{\cos(t)}$ , we first note that $F(t)$ is undefined whenever $\cos(t) = 0$ . Thanks to Example 7.2.4 number 3, we know $\cos(t) = 0$ whenever $t = \frac{\pi}{2} + \pi k$ for integers $k$ .

This gives us one way to describe the domain of $F$ : $\{ t \, | \, t \neq \frac{\pi}{2} + \pi k, \text{ for integers } k \}$ . To get a better feel for the set of real numbers we’re dealing with, we write out and graph the domain on the number line.

Running through a few values of $k$ , we find some of the values excluded from the domain: $t \neq \pm \frac{\pi}{2}, \, \pm \frac{3\pi}{2}, \, \pm \frac{5\pi}{2}$ . Using these we can graph the domain on the number line below.

A number line with marks every pi, starting at negative five pi over two and ending at five pi over two. There is a line above the number line with a hole at every odd multiple of pi over 2. — Graphical Representations of the Domain Restrictions of the Secant function.

Expressing this set using interval notation is a bit of a challenge, owing to the infinitely many intervals present. As a first attempt, we have: $\ldots \cup \left( -\frac{5\pi}{2}, -\frac{3\pi}{2}\right) \cup \left( -\frac{3\pi}{2}, -\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \cup \left(\frac{3\pi}{2}, \frac{5\pi}{2}\right) \cup \ldots$ , where, as usual, the periods of ellipsis indicate the pattern continues indefinitely. Hence, for now, it suffices to know that the domain of $F(t) = \sec(t)$ excludes the odd multiples of $\frac{\pi}{2}$ .

To find the range of $F$ , we find it helpful once again to view $F(t) = \sec(t) = \frac{1}{\cos(t)}$ . We know the range of $\cos(t)$ is $[-1,1]$ and $F(t) = \sec(t) = \frac{1}{\cos(t)}$ is undefined when $\cos(t) = 0$ , so we split our discussion into two cases: when $0 < \cos(t) \leq 1$ and when $-1 \leq \cos(t) < 0$ .

If $0 < \cos(t) \leq 1$ , then we can divide the inequality $\cos(t) \leq 1$ by $\cos(t)$ to obtain $\sec(t) = \frac{1}{\cos(t)} \geq 1$ . Moreover, we see as $\cos(t) \rightarrow 0^{+}$ , $\sec(t) \rightarrow \infty$ . If, on the other hand, if $-1 \leq \cos(t) < 0$ , then dividing by $\cos(t)$ causes a reversal of the inequality so that $\sec(t) = \frac{1}{\cos(t)} \leq -1$ . In this case, as $\cos(t) \rightarrow 0^{-}$ , $\sec(t) \rightarrow -\infty$ . As $\cos(t)$ admits all of the values in $[-1,1]$ , the function $F(t) = \sec(t)$ admits all of the values in $(-\infty, -1] \cup [1,\infty)$ .

Because $\cos(t)$ is periodic with period $2\pi$ , it shouldn’t be too surprising to find that $\sec(t)$ is also. Indeed, provided $\sec(\alpha)$ and $\sec(\beta)$ are defined, $\sec(\alpha) = \sec(\beta)$ if and only if $\cos(\alpha) = \cos(\beta)$ . Said differently, $\sec(t)$ `inherits’ its period from $\cos(t)$ .

We now turn our attention to graphing $F(t) = \sec(t)$ . Using the table of values we tabulated when graphing $y = \cos(t)$ in Section 7.3, we can generate points on the graph of $y = \sec(t)$ by taking reciprocals.

Using the techniques developed in Section 3.2, we can more closely analyze the behavior of $F$ near the values excluded from its domain. We find as $t \rightarrow \frac{\pi}{2}^{-}$ , $\cos(t) \rightarrow 0^{+}$ , so $\sec(t) \rightarrow \infty$ . Similarly, we get as $t \rightarrow \frac{\pi}{2}^{+}$ , $\sec(t) \rightarrow -\infty$ ; as $t \rightarrow \frac{3\pi}{2}^{-}$ , $\sec(t) \rightarrow -\infty$ ; and as $t \rightarrow \frac{3\pi}{2}^{+}$ , $\sec(t) \rightarrow \infty$ . This means the lines $t = \frac{\pi}{2}$ and $t = \frac{3\pi}{2}$ are vertical asymptotes to the graph of $y = \sec(t)$ .

Below on the right we graph a fundamental cycle of $y = \sec(t)$ with the graph of the fundamental cycle of $y = \cos(t)$ dotted for reference.

$\begin{array}{|r||r|r|r|} \hline t & \cos(t) & \sec(t) & (t,\sec(t)) \\ \hline 0 & 1 & 1 & (0,1) \\ [2pt] \hline \frac{\pi}{4} & \frac{\sqrt{2}}{2} & \sqrt{2} & \left(\frac{\pi}{4}, \sqrt{2} \right) \\ [2pt] \hline \frac{\pi}{2} & 0 & \text{undefined} & \\ [2pt] \hline \frac{3\pi}{4} & -\frac{\sqrt{2}}{2} & -\sqrt{2} & \left(\frac{3\pi}{4}, -\sqrt{2} \right) \\ [2pt] \hline \pi & -1 & -1 & (\pi, -1) \\ [2pt] \hline \frac{5\pi}{4} & -\frac{\sqrt{2}}{2} & -\sqrt{2} & \left(\frac{5\pi}{4}, -\sqrt{2} \right) \\ [2pt] \hline \frac{3\pi}{2} & 0 & \text{undefined} & \\ [2pt] \hline \frac{7\pi}{4} & \frac{\sqrt{2}}{2} & \sqrt{2} & \left(\frac{7\pi}{4}, \sqrt{2} \right) \\ [2pt] \hline 2\pi & 1 & 1& (2\pi, 1) \\ [2pt] \hline \end{array}$

The graph of the fundamental cycle of the secant function. The graph starts at (0,1) and curves upward. The graph has a vertical asymptote at pi over two. After pi over two the graph increases from negative infinity and increases to negative one at pi and then decreases until the next vertical asymptote at three pi over two. After three pi over two the graph decreases from positive infinity down to 1 at two pi. There is a dashed line representing the cosine function between the pieces of the secant function. — The ‘fundamental cycle’ of y = sec(t)

To get a graph of the entire secant function, we paste copies of the fundamental cycle end to end to produce the graph below. The graph suggests that $F(t) =\sec(t)$ is even. Indeed, because $\cos(t)$ is even, that is, $\cos(-t) = \cos(t)$ , we have $\sec(-t) = \frac{1}{\cos(-t)} = \frac{1}{\cos(t)} = \sec(t)$ . Hence, along with its period, the secant function inherits its symmetry from the cosine function.

A secant function on a cartesian plane. The graph extends in both as t approaches negative infinity and as t approaches positive infinity. The cosine graph is a dashed wave that slowly bounces between negative one and positive one inside the secant function. The fundamental cycle of the secant graph from the origin to when t is 2 pi is highlighted. And the remaining portion of the graph is a repetition of the one cycle of the secant graph. — The graph of y = sec (t)

As one would expect, to graph $G(t) = \csc(t)$ we begin with $y = \sin(t)$ and take reciprocals of the corresponding $y$ -values. Here, we encounter issues at $t = 0$ , $t = \pi$ , $t = 2\pi$ , and, in general, at all whole number multiples of $\pi$ , so the domain of $G$ is $\{ t \, | \, t \neq \pi k, \text{for integers } k \}$ . Not surprisingly, these values produce vertical asymptotes.

Proceeding as above, we produce the graph of the fundamental cycle of $y = \csc(t)$ below along with the dotted graph of $y=\sin(t)$ for reference.

$\begin{array}{|r||r|r|r|} \hline x & \sin(x) & \csc(x) & (x,\csc(x)) \\ \hline 0 & 0 & \text{undefined} & \\ [2pt] \hline \frac{\pi}{4} & \frac{\sqrt{2}}{2} & \sqrt{2} & \left(\frac{\pi}{4}, \sqrt{2} \right) \\ [2pt] \hline \frac{\pi}{2} & 1 & 1 & \left(\frac{\pi}{2}, 1 \right) \\ [2pt] \hline \frac{3\pi}{4} & \frac{\sqrt{2}}{2} & \sqrt{2} & \left(\frac{3\pi}{4}, \sqrt{2} \right) \\ [2pt] \hline \pi & 0 & \text{undefined} & \\ [2pt] \hline \frac{5\pi}{4} & -\frac{\sqrt{2}}{2} & -\sqrt{2} & \left(\frac{5\pi}{4}, -\sqrt{2} \right) \\ [2pt] \hline \frac{3\pi}{2} & -1 & -1 & \left(\frac{3\pi}{2},-1 \right)\\ [2pt] \hline \frac{7\pi}{4} & -\frac{\sqrt{2}}{2} & -\sqrt{2} & \left(\frac{7\pi}{4}, -\sqrt{2} \right) \\ [2pt] \hline 2\pi & 0 & \text{undefined} & \\ [2pt] \hline \end{array}$

The graph of the fundamental cycle of the secant function. The graph starts with a vertical asymptote at 0. After 0 the graph decreases from positive infinity and decreases to one at pi over two and then back up as we approach pi. The graph has a vertical asymptote at pi. After pi the graph increases from negative infinity to negative one at three pi over two and then decreases again as we approach the next vertical asymptote at two pi. There is a dashed line representing the sine function between the pieces of the cosecant function. — The ‘fundamental cycle’ of y = csc (t)

Pasting copies of the fundamental period of $y = \csc(t)$ end to end produces the graph below. Due to the fact that the graphs of $y = \sin(t)$ and $y = \cos(t)$ are merely phase shifts of each other, it is not too surprising to find the graphs of $y = \csc(t)$ and $y = \sec(t)$ are as well.

A cosecant function on a cartesian plane. The graph extends in both as t approaches negative infinity and as t approaches positive infinity. The sine graph is a dashed wave that slowly bounces between negative one and positive one inside the cosecant function. The fundamental cycle of the cosecant graph from the origin to when t is 2 pi is highlighted. And the remaining portion of the graph is a repetition of the one cycle of the cosecant graph. — The graph of y = csc (t)

As with the graph of secant, the graph of cosecant suggests symmetry. Indeed, as the sine function is odd, that is $\sin(-t) = -\sin(t)$ , so too is the cosecant function: $\csc(-t) = \frac{1}{\sin(-t)} = -\frac{1}{\sin(t)} = -\csc(t)$ . Hence, the graph of $G(t) = \csc(t)$ is symmetric about the origin.

Note that, on the intervals between the vertical asymptotes, both $F(t) = \sec(t)$ and $G(t) = \csc(t)$ are continuous and smooth. In other words, they are continuous and smooth on their domains.^[1]

The following theorem summarizes the properties of the secant and cosecant functions. Note that all of these properties are direct results of them being reciprocals of the cosine and sine functions, respectively.

Theorem 7.11 Properties of the Secant and Cosecant Functions

The function
- has domain $\left\{ t \, | \, t \neq \frac{\pi}{2} + \pi k, \, k \, \text{ is an integer} \right\}$
- has range $(-\infty, -1] \cup [1, \infty)$
- is continuous and smooth on its domain
- is even
- has period $2 \pi$
The function
- has domain $\left\{ t \, | \, t \neq \pi k, \, k \, \text{ is an integer} \right\}$
- has range $(-\infty, -1] \cup [1, \infty)$
- is continuous and smooth on its domain
- is odd
- has period $2\pi$

In the next example, we discuss graphing more general secant and cosecant curves. We make heavy use of the fact they are reciprocals of sine and cosine functions and apply what we learned in Section 7.3.

Example 7.5.1

Example 7.5.1.1

Graph one cycle of the following functions. State the period of each.

$f(t) = 1 - 2 \sec(2t)$

Solution:

Graph one cycle of $f(t) = 1 - 2 \sec(2t)$ .

To graph $f(t) = 1 - 2 \sec(2t)$ , we follow the same procedure as in Example 7.3.2. That is, we use the concept of frequency and phase shift to identify quarter marks, then substitute these values into the function to obtain the corresponding points.

If we think about a related cosine curve, $y = 1 - 2\cos(2t) = -2\cos(2t) + 1$ , we know from Section 7.3, that the frequency is $B = 2$ , so the period is $T = \frac{2\pi}{2} = \pi$ .

$C = 0$ , so there is no phase shift.

Hence, the new quarter marks for this curve are $t=0$ , $t=\frac{\pi}{4}$ , $t=\frac{\pi}{2}$ , $t=\frac{3\pi}{4}$ , and $t=\pi$ .

These same $t$ -values are the new quarter marks for $f(t) = 1 - 2 \sec(2t)$ , because we obtained the fundamental cycle of the secant curve from the fundamental cycle of the cosine curve.

Substituting these $t$ values $f(t)$ , we get the table below on the left. Note that if $f(t)$ exists, we have a point on the graph; otherwise, we have found a vertical asymptote.^[2]

We graph one cycle of $f(t) = 1 - 2 \sec(2t)$ below on the right along with the associated cosine curve, $y = 1 - 2 \cos(2t)$ which is dotted, and confirm the period is $\pi - 0 = \pi$ .

$\begin{array}{|r||r|r|} \hline t & f(t) & (t,f(t)) \\ \hline 0 & - 1 & (0,-1) \\ [2pt] \hline \frac{\pi}{4} & \text{undefined} & \\ [2pt] \hline \frac{\pi}{2} & 3 & \left(\frac{\pi}{2}, 3 \right) \\ [2pt] \hline \frac{3\pi}{4} & \text{undefined} & \\ [2pt] \hline \pi & -1 & (\pi, -1) \\ [2pt] \hline \end{array}$

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Example 7.5.1.2

Graph one cycle of the following functions. State the period of each.

$g(t) = \dfrac{\csc(- \pi t - \pi) - 5}{3}$

Solution:

Graph one cycle $g(t) = \dfrac{\csc(- \pi t - \pi) - 5}{3}$ .

As with the previous example, we start graphing $g(t) = \frac{\csc(-\pi t- \pi ) - 5}{3}$ by first finding the quarter marks of the associated sine curve: $y =\frac{\sin(- \pi t - \pi ) - 5}{3} = \frac{1}{3} \sin( -\pi t - \pi) - \frac{5}{3}$ .

The coefficient of $t$ is negative, thus we make use of the odd property of sine to rewrite the function as: $y = \frac{1}{3} \sin( - \pi t - \pi ) - \frac{5}{3} = \frac{1}{3} \sin( -(\pi t + \pi) ) - \frac{5}{3} = -\frac{1}{3} \sin(\pi t + \pi) - \frac{5}{3}$ .

We find the frequency is $B = \pi$ , so the period is $T = \frac{2\pi}{\pi} = 2$ .

$C = \pi$ , so the phase shift is $-\frac{\pi}{\pi} = -1$ .

Hence the fundamental cycle $[0, 2\pi]$ is shifted to the interval $[-1,1]$ with quarter marks $t = -1$ , $t = -\frac{1}{2}$ , $t=0$ , $t = \frac{1}{2}$ and $t=1$ .

Substituting these $t$ -values into $g(t)$ , we generate the graph below on the right confirm the period is $1 - (-1) = 2$ . The associated sine curve, $y = \frac{\sin(- \pi t- \pi) - 5}{3}$ , is dotted in as a reference.

$\begin{array}{|r||r|r|} \hline t & g(t) & (t,g(t)) \\ \hline -1 & \text{undefined} & \\ [2pt] \hline -\frac{1}{2} & -2 & \left(-\frac{1}{2}, -2\right) \\ [2pt] \hline 0 & \text{undefined} & \\ [2pt] \hline \frac{1}{2} & -\frac{4}{3} & \left(\frac{1}{2}, -\frac{4}{3} \right) \\ [2pt] \hline 1 & \text{undefined} & \\ [2pt] \hline \end{array}$

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As suggested in Example 7.5.1, the concepts of frequency, period, phase shift, and baseline are alive and well with graphs of the secant and cosecant functions. The secant and cosecant curves are unbounded, therefore we do not have the concept of `amplitude’ for these curves. That being said, the amplitudes of the corresponding cosine and sine curves do play a role here – they measure how wide the gap is between the baseline and the curve.

We gather these observations in the following result whose proof is a consequence of Theorem 7.7 and is relegated to Exercise 19.

Theorem 7.12

For $B > 0$ , the graphs of

$F(t) = A \sec( B t + C) +D \quad \text{and} \quad G(t) = A \csc( B t + C) + D$

have frequency $B$
have period $T = \dfrac{2\pi}{B}$
have phase shift $-\dfrac{C}{B}$
have `baseline’ $D$ and have a vertical gap $|A|$ units between the the baseline and the graph^[3]

We put Theorem 7.12 to good use in the next example.

Example 7.5.2

Example 7.5.2.1

Below is the graph of one cycle of a secant (cosecant) function, $y = f(t)$ .

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Write $f(t)$ in the form $F(t) = A \sec( B t + C) +D$ for $B > 0$ .

Solution:

Write $f(t)$ in the form $F(t) = A \sec( B t + C) +D$ for $B > 0$ .

We first note the period: $T = \frac{5\pi}{6} - \left(-\frac{\pi}{6}\right) = \pi$ and $T = \frac{2\pi}{B} = \pi$ , so we get $B = 2$ .

To find $C$ , we need to first determine the phase shift. Recall that what is graphed here is only one cycle of the function, so by copying and pasting one more cycle, we identify what looks like a fundamental cycle of the secant function to us^[4] as highlighted below on the left.

We get the phase shift is $\frac{7\pi}{12}$ so solving $-\frac{C}{2} = \frac{7\pi}{12}$ , we get $C = - \frac{7\pi}{6}$ .

To find the baseline, $D$ , we take a cue from our work in Example 7.3.3 in Section 7.3. We find the average of the local minimums and maximums to be $\frac{-2+0}{2} = -1$ , so $D = -1$ . As there is a $1$ unit gap between the baseline and the graph of the function, we have $A = 1$ . Alternatively, we can sketch the corresponding cosine curve (dotted in the figure below) and determine $D$ and $A$ that way.

We find our final answer to be $f(t) = \sec\left( 2 t - \frac{7\pi}{6} \right) -1$ .

As usual, we check our answer by graphing.

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Example 7.5.2.2

Below is the graph of one cycle of a secant (cosecant) function, $y = f(t)$ .

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Write $f(t)$ in the form $G(t) = A \csc( B t + C) +D$ for $B > 0$ .

Solution:

Write $f(t)$ in the form $G(t) = A \csc( B t + C) +D$ for $B > 0$ .

The secant and cosecant curves are phase shifts of each other, therefore we could find a formula for $f(t)$ in terms of cosecants by shifting our formula $F(t) = \sec\left( 2 t - \frac{7\pi}{6} \right) -1$ . We leave this to the reader.^[5]

Working `from scratch,’ we would find $T = \pi$ , $B= 2$ , $D=-1$ , and $A=1$ the same as above.^[6] To determine the phase shift, we refer to the figure below.

The phase shift is $\frac{\pi}{3}$ , thus we solve $-\frac{C}{2} = \frac{\pi}{3}$ to get $C = -\frac{2\pi}{3}$ .

Putting all our work together, we get our final answer: $f(t) = \csc\left(2t - \frac{2\pi}{3} \right) - 1$ .

Again, our best check here is to graph.

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We cannot stress enough that our answers to Example 7.5.2 are one of many. For example, in Exercise 18, we ask you to rework this example choosing $A<0$ . It is well worth the time to think about what relationships exist between the different answers, however.

7.5.2 Graphs of the Tangent and Cotangent Functions

Next, we turn our attention to the tangent and cotangent functions. Viewing $J(t) = \tan(t) = \frac{\sin(t)}{\cos(t)}$ , we find the domain of $J$ excludes all values where $\cos(t) = 0.$

Hence, the domain of $J$ is $\{ t \, | \, t \neq \frac{\pi}{2} + \pi k, \text{for integers }k \}$ .

Using this information along with the common values we given in Section 7.4, we create the table of values below.

$\begin{array}{|r||r|r|} \hline t & \tan(t) & (t,\tan(t)) \\ \hline 0 & 0 & (0, 0) \\ [2pt] \hline \frac{\pi}{4} & 1 & \left(\frac{\pi}{4},1 \right) \\ [2pt] \hline \frac{\pi}{2} & \text{undefined} & \\ [2pt] \hline \frac{3\pi}{4} & -1 & \left(\frac{3\pi}{4}, -1\right) \\ [2pt] \hline \pi & 0 & (\pi, 0) \\ [2pt] \hline \frac{5\pi}{4} & 1 & \left(\frac{5\pi}{4}, 1 \right) \\ [2pt] \hline \frac{3\pi}{2} & \text{undefined} & \\ [2pt] \hline \frac{7\pi}{4} & -1 & \left(\frac{7\pi}{4}, -1 \right) \\ [2pt] \hline 2\pi & 0 & (2\pi, 0) \\ [2pt] \hline \end{array}$

A graph of the fundamental cycle of the tangent function. The graph starts at the origin. The graph increases to the vertical asymptote at pi over two After pi over two the graph rises from negative infinity to zero at pi and then increases to the next vertical asymptote at three pi over two. After three pi over two the graph rises from negative infinity to zero at two pi — The graph of y=tan(t) over [0, 2 pi]

After the usual `copy and paste’ procedure, we create the graph of $y = \tan(t)$ below:

A tangent function on a cartesian plane. The graph extends in both as t approaches negative infinity and as t approaches positive infinity. One cycle complete and continuous cycle of the tangent graph from the negative pi over two to pi over two is highlighted. And the remaining portion of the graph is a repetition of the one cycle of the tangent graph. — The graph of y = tan(t)

The graph of $y = \tan(t)$ suggests symmetry through the origin. Indeed, tangent is odd because sine is odd and cosine is even:

$\begin{array}{rcl} \tan(-t) &=& \frac{\sin(-t)}{\cos(-t)} \\[4pt] &=& \frac{-\sin(t)}{\cos(t)} \\[4pt] &=& -\tan(t) \end{array}$

We also see the graph suggests the range of $J(t) = \tan(t)$ is all real numbers, $(-\infty, \infty)$ . We present one proof of this fact in Exercise 18.

Moreover, as noted in Section 7.4, the period of the tangent function is $\pi$ , and we see that reflected in the graph. This means we can choose any interval of length $\pi$ to serve as our `fundamental cycle.’

We choose the cycle traced out over the (open) interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ as highlighted above. In addition to the asymptotes at the endpoints $t = \pm \frac{\pi}{2}$ , we use the `quarter marks’ $t = \pm \frac{\pi}{4}$ and $t = 0$ .

It should be no surprise that $K(t) = \cot(t)$ behaves similarly to $J(t)=\tan(t)$ . As $\cot(t) = \frac{\cos(t)}{\sin(t)}$ , the domain of $K$ excludes the values where $\sin(t) = 0$ : $\{ t \, | \, t \neq \pi k, \text{for integers } k \}$ .

After analyzing the behavior of $K$ near the values excluded from its domain along with plotting points, we graph $y = \cot(t)$ over the interval $[0,2\pi]$ below on the right.

$\begin{array}{|r||r|r|} \hline t & \cot(t) & (t, \cot(t)) \\ \hline 0 & \text{undefined} & \\ [2pt] \hline \frac{\pi}{4} & 1 & \left(\frac{\pi}{4},1 \right) \\ [2pt] \hline \frac{\pi}{2} & 0 & \left(\frac{\pi}{2},0 \right) \\ [2pt] \hline \frac{3\pi}{4} & -1 & \left(\frac{3\pi}{4}, -1\right) \\ [2pt] \hline \pi & \text{undefined} & \\ [2pt] \hline \frac{5\pi}{4} & 1 & \left(\frac{5\pi}{4}, 1 \right) \\ [2pt] \hline \frac{3\pi}{2} & 0 & \left(\frac{3\pi}{2}, 0 \right) \\ [2pt] \hline \frac{7\pi}{4} & -1 & \left(\frac{7\pi}{4}, -1 \right) \\ [2pt] \hline 2\pi & \text{undefined} & \\ [2pt] \hline \end{array}$

A graph of the fundamental cycle of the cotangent function. The graph starts with a vertical asymptote at t equals 0. The graph decreases from the vertical asymptote to zero at pi over two. And then continues downward to the next vertical asymptote at pi. After pi the graph again decreases from positive infinity to zero at three pi over two and then decreases to the next vertical asymptote two pi. — The graph of y=cot(t) over [0, 2pi]

As usual, pasting copies end to end produces the graph of $K(t) = \cot(t)$ below.

A cotangent function on a cartesian plane. The graph extends in both as t approaches negative infinity and as t approaches positive infinity. One cycle complete and continuous cycle of the cotangent graph from 0 to pi is highlighted. And the remaining portion of the graph is a repetition of the one cycle of the cotangent graph. — The graph of y = cot (t)

As with $J(t) = \tan(t)$ , the graph of $K(t) = \cot(t)$ suggests $K$ is odd, a fact we leave to the reader to prove in Exercise 22. Also, we see that the period of cotangent (like tangent) is $\pi$ and the range is $(-\infty, \infty).$

We take as one fundamental cycle the graph as traced out over the interval $(0,\pi)$ , highlighted above, with quarter marks: $t= 0$ , $t=\frac{\pi}{4}$ , $t=\frac{\pi}{2}$ , $t=\frac{3\pi}{4}$ and $t=\pi .$

The properties of the tangent and cotangent functions are summarized below. As with Theorem 7.11, each of the results in Theorem 7.13 can be traced back to properties of the cosine and sine functions and the definition of the tangent and cotangent functions as quotients thereof.

Theorem 7.13 Properties of the Tangent and Cotangent Functions

The function
- has domain $\left\{ t \, | \, t \neq \frac{\pi}{2} + \pi k, \, k \, \text{ is an integer} \right\}$
- has range $(-\infty, \infty)$
- is continuous and smooth on its domain
- is odd
- has period $\pi$
The function
- has domain $\left\{ t \, | \, t \neq \pi k, \, k \, \text{ is an integer} \right\}$
- has range $(-\infty, \infty)$
- is continuous and smooth on its domain
- is odd
- has period $\pi$

Unlike the secant and cosecant functions, the tangent and cotangent functions have different periods than sine and cosine. Moreover, in the case of the tangent function, the fundamental cycle we’ve chosen starts at $-\frac{\pi}{2}$ instead of $0$ . Nevertheless, we can use the same notions of period and phase shift to graph transformed versions of tangent and cotangent functions, because these results ultimately trace back to applying Theorem 1.12. We state a version of Theorem 7.7 for tangent and cotangent functions below.

Theorem 7.14

For $B > 0$ , the functions

$J(t) = A \tan(B t + C) + D \quad \text{and} \quad K(t) = A \cot(B t + C) + D$

have frequency $B$
have period $T = \dfrac{\pi}{B}$
have vertical shift or `baseline’ $D$
The phase shift for $y = J(t)$ is $-\dfrac{C}{B} - \dfrac{\pi}{2B}$
The phase shift for $y = K(t)$ is $-\dfrac{C}{B}$

The proof of Theorem 7.14 is left to the reader in Exercise 20.

We put Theorem 7.14 to good use in the following example.

Example 7.5.3

Example 7.5.3.1

Graph one cycle of the following functions. Find the period.

$f(t) = 1 - \tan\left(\frac{t}{2} - \pi \right)$

Solution:

Graph one cycle of $f(t) = 1 - \tan\left(\frac{t}{2} - \pi \right)$ .

Rewriting $f(t)$ so it fits the form in Theorem 7.14, we get $f(t) = - \tan\left(\frac{1}{2} t + (- \pi) \right) + 1$ .

With $B = \frac{1}{2}$ , we find the period $T = \frac{\pi}{1/2} = 2 \pi$ . As $C = -\pi$ , the phase shift is $-\frac{(-\pi)}{1/2} - \frac{\pi}{2 (1/2)} = \pi .$

Hence, one cycle of $f(t)$ starts at $t=\pi$ and finishes at $t = \pi + 2\pi = 3\pi$ . Our quarter marks are $\frac{2\pi}{4} = \frac{\pi}{2}$ units apart and are $t = \pi$ , $t = \frac{3\pi}{2}$ , $t = 2\pi$ , $t = \frac{5\pi}{2}$ , and, finally, $t = 3\pi .$

Substituting these $t$ -values into $f(t)$ , we find points on the graph and the vertical asymptotes.^[7]

$\begin{array}{|r||r|r|} \hline t & f(t) & (t,f(t)) \\ \hline \pi & \text{undefined} & \\ [2pt] \hline \frac{3\pi}{2} & 2 & \left(\frac{3\pi}{2}, 2 \right) \\ [2pt] \hline 2 \pi & 1 & (2\pi,1) \\ [2pt] \hline \frac{5\pi}{2} & 0 & \left(\frac{5\pi}{2}, 0 \right) \\ [2pt] \hline 3 \pi & \text{undefined} & \\ [2pt] \hline \end{array}$

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We confirm that the period is $3\pi - \pi = 2\pi$ .

Example 7.5.3.2

Graph one cycle of the following functions. Find the period.

$g(t) = 2\cot\left(2\pi - \pi t \right) - 1$

Solution:

Graph one cycle of $g(t) = 2\cot\left(2\pi - \pi t \right) - 1$ .

To put $g(t)$ into the form prescribed by Theorem 7.14, we make use of the odd property of cotangent:

$\begin{array}{rcl} g(t) &=& 2\cot\left(2\pi - \pi t \right) - 1 \\ & =& 2\cot( -[\pi t- 2\pi]) - 1 \\ &=& -2 \cot(\pi t- 2\pi) - 1 \\ &=& -2 \cot(\pi t + (- 2\pi)) - 1 \end{array}$

We identify $B = \pi$ so the period is $T = \frac{\pi}{\pi} = 1$ . Because $C = -2\pi$ , the phase shift is $-\frac{-2\pi}{\pi} = 2$ . Hence, one cycle of $g(t)$ starts at $t = 2$ and ends at $t = 2+1 = 3$ .

Our quarter marks are $\frac{1}{4}$ units apart and are $t = 2$ , $t = \frac{9}{4}$ , $t = \frac{5}{2}$ , $t = \frac{11}{4}$ , and $t = 3$ . We generate the next graph.

$\begin{array}{|r||r|r|} \hline t & g(t) & (t,g(t)) \\ \hline 2 & \text{undefined} & \\ [2pt] \hline \frac{9}{4} & -3 & \left(\frac{9}{4}, -3 \right) \\ [2pt] \hline \frac{5}{2} & -1 & \left( \frac{5}{2}, -1 \right) \\ [2pt] \hline \frac{11}{4} & 1 & \left(\frac{11}{4}, 1 \right) \\ [2pt] \hline 3 & \text{undefined} & \\ [2pt] \hline \end{array}$

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We confirm the period is $3-2 = 1$ .

Example 7.5.4

Example 7.5.4.1

Below is the graph of one cycle of a tangent (cotangent) function, $y = f(t)$ .

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Write $f(t)$ in the form $J(t) = A \tan( B t + C) +D$ for $B > 0$ .

Solution:

Write $f(t)$ in the form $J(t) = A \tan( B t + C) +D$ for $B > 0$ .

We first find the period $T = 10-(-2) = 12$ . Per Theorem 7.14, we know $\frac{\pi}{B} = 12$ , or $B = \frac{\pi}{12}$ .

Next, we look for the phase shift. We notice the cycle graphed for us is decreasing instead of the usual increasing we expect for a standard tangent cycle. When this sort of thing happened in Examples 7.3.3 and 7.5.2, we pasted another cycle of the function and used that to help identify the phase shift in order to keep the value of $A> 0$ . Here, no amount of `copying and pasting’ will produce an increasing cycle (do you see why?), so we know $A<0$ and use $-2$ , as the phase shift.

The formula given in Theorem 7.14 tells us $-\frac{C}{B} - \frac{\pi}{2 B} = -2$ so substituting $B = \frac{\pi}{12}$ gives $C = -\frac{\pi}{3}$ .

Next, we see the baseline here is still the $t$ -axis, so $D=0$ .

This means all that’s left to find is $A$ . We have already established that $A<0$ to account for the reflection across the $t$ -axis. Moreover, the $y$ -values of the points off of the baseline are $3$ units from the baseline, indicating a vertical stretch by a factor of $3$ .

Hence, $A = -3$ and $f(t) = -3 \tan\left( \frac{\pi}{12} t - \frac{\pi}{3} \right)$ .

As usual, the ultimate check is to graph, which we will leave to the reader.

Example 7.5.4.2

Below is the graph of one cycle of a tangent (cotangent) function, $y = f(t)$ .

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Write $f(t)$ in the form $K(t) = A \cot(B t + C) +D$ for $B > 0$ .

Solution:

Write $f(t)$ in the form $K(t) = A \cot(B t + C) +D$ for $B > 0$ .

We find $T = 12$ , $B = \frac{\pi}{12}$ , and $D = 0$ as above. As the fundamental cycle of cotangent is decreasing, we know $A>0$ and identify the phase shift as $-2$ .

Using Theorem 7.14, we know $-\frac{C}{B} = -2$ so substituting $B = \frac{\pi}{12}$ , we get $C = \frac{\pi}{6}$ .

As above, the vertical stretch is by a factor of $3$ , so we take $A = 3$ for our final answer: $f(t) = 3 \cot \left( \frac{\pi}{12} t + \frac{\pi}{6}\right)$ .

We leave it to the reader to check our answer by graphing.

Once again, our answers to Example 7.5.4 are one of many, and we invite the reader to think about what all of the solutions would have in common. We close this section with an application.

Example 7.5.5

Example 7.5.5.1

Let $\theta$ be the angle of inclination from an observation point on the ground 42 feet away from the launch site of a model rocket. Assuming the rocket is launched directly upwards:

Write a formula for $f(\theta)$ , the distance from the rocket to the ground (in feet) as a function of $\theta$ . Compute and interpret $f\left( \frac{\pi}{3} \right)$ .

Solution:

We begin by sketching the scenario below. Given the rocket is launched `directly upwards,’ we may assume the rocket is launched at a $90^{\circ}$ angle which provides us with a right triangle.

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Write a formula for $f(\theta)$ , the distance from the rocket to the ground (in feet) as a function of $\theta$ . Compute and interpret $f\left( \frac{\pi}{3} \right)$ .

From the remarks preceding Theorem 7.10, we know the definitions of the circular functions agree with those specified for acute angles in right triangles as described in Definition 7.6 in Section 7.2.1.

Hence, $\tan(\theta) = \frac{f(\theta)}{42}$ , so $f(\theta) = 42 \tan(\theta)$ .

We find $f\left( \frac{\pi}{3} \right) = 42 \tan\left( \frac{\pi}{3} \right) = 30 \sqrt{3}$ . This means when the angle of inclination is $\frac{\pi}{3}$ or $60^{\circ}$ , the rocket is or $30 \sqrt{3} \approx 73$ feet off of the ground.

Example 7.5.5.2

Let $\theta$ be the angle of inclination from an observation point on the ground 42 feet away from the launch site of a model rocket. Assuming the rocket is launched directly upwards:

Write a formula for $g(\theta)$ , the distance from the rocket to the observation point on the ground (in feet) as a function of $\theta$ . Compute and interpret $g\left( \frac{\pi}{3} \right)$ .

Solution:

We begin by sketching the scenario below. Given the rocket is launched `directly upwards,’ we may assume the rocket is launched at a $90^{\circ}$ angle which provides us with a right triangle.

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Write a formula for $g(\theta)$ , the distance from the rocket to the observation point on the ground (in feet) as a function of $\theta$ . Compute and interpret $g\left( \frac{\pi}{3} \right)$ .

Again, working with the triangle, we find $\sec(\theta) = \frac{g(\theta)}{42}$ so that $g(\theta) = 42 \sec(\theta)$ .

We find $g\left( \frac{\pi}{3} \right) = 42 \sec\left( \frac{\pi}{3} \right) = 84$ , so when the angle of inclination is $60^{\circ}$ , the rocket is $84$ feet from the observation point on the ground.

Example 7.5.5.3

Let $\theta$ be the angle of inclination from an observation point on the ground 42 feet away from the launch site of a model rocket. Assuming the rocket is launched directly upwards:

Write and interpret the behavior of $f(\theta)$ and $g(\theta)$ as $\theta \rightarrow \frac{\pi}{2}^{-}$ .

Solution:

We begin by sketching the scenario below. Given the rocket is launched `directly upwards,’ we may assume the rocket is launched at a $90^{\circ}$ angle which provides us with a right triangle.

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Write and interpret the behavior of $f(\theta)$ and $g(\theta)$ as $\theta \rightarrow \frac{\pi}{2}^{-}$ .

As $\theta \rightarrow \frac{\pi}{2}^{-}$ , both $f(\theta) \rightarrow \infty$ and $g(\theta) \rightarrow \infty$ (a fact we could verify graphically, if needs be.)

This means as the angle of inclination approaches $\frac{\pi}{2}$ or $90^{\circ}$ , the distances from the rocket to the ground and from to the rocket to the observation point increase without bound. Barring the effects of drift or the curvature of space, this matches our intuition.

7.5.3 Section Exercises

In Exercises 1 – 12, graph one cycle of the given function. State the period of the function.

$y = \tan \left(t - \dfrac{\pi}{3} \right)$
$y = 2\tan \left( \dfrac{1}{4}t \right) - 3$
$y = \dfrac{1}{3}\tan(-2t - \pi) + 1$
$y = \sec \left( t - \dfrac{\pi}{2} \right)$
$y = -\csc \left( t + \dfrac{\pi}{3} \right)$
$y = -\dfrac{1}{3} \sec \left( \dfrac{1}{2}t + \dfrac{\pi}{3} \right)$
$y = \csc (2t - \pi)$
$y = \sec(3t - 2\pi) + 4$
$y = \csc \left( -t - \dfrac{\pi}{4} \right) - 2$
$y = \cot \left( t + \dfrac{\pi}{6} \right)$
$y = -11\cot \left( \dfrac{1}{5} t \right)$
$y = \dfrac{1}{3} \cot \left( 2t + \dfrac{3\pi}{2} \right) + 1$

In Exercises 13 – 14, the graph of a (co)secant function is given. Write a formula for the function in the form $F(t) = A \sec(B t + C) + D$ and $G(t) = A \csc(B t + C) + D$ . Select $B$ so $B > 0$ . Check your answer by graphing.

Asymptotes: $t = \pm \frac{\pi}{2}$ , $t=\pm \frac{3\pi}{2}$ , $\dots$
Asymptotes: $t = \pm 1$ , $t = \pm 3$ , $t = \pm 5$ , $\ldots$

In Exercises 15 – 16, the graph of a (co)tangent function given. Find a formula the function in the form $J(t) = A \tan(B t + C) + D$ and $K(t) = A \cot(B t + C) + D$ . Select $B$ so $B > 0$ . Check your answer by graphing.

Asymptotes: $t=-\frac{3 \pi}{4}$ , $t=\frac{\pi}{4}$ , $t = \frac{5\pi}{4}$ , $\dots$
Asymptotes: $t = \pm 2$ , $t = \pm 6$ , $t = \pm 10$ , $\ldots$
(a) Use the conversion formulas listed in Theorem 7.6 to create conversion formulas between secant and cosecant functions.
(b) Use a conversion formula to rewrite our first answer to Example 7.5.2, $f(t) = \sec\left( 2 t - \frac{7\pi}{6} \right) -1$ , in terms of cosecants.
Rework Example 7.5.2 and find answers with $A<0$ .
Prove Theorem 7.12 using Theorem 7.7.
Prove Theorem 7.14 using Theorem 1.12.
In this Exercise, we argue the range of the tangent function is . Let be a fixed, but arbitrary positive real number.
1. Show there is an acute angle $\theta$ with $\tan(\theta) = M$ . (Hint: think right triangles.)
2. Using the symmetry of the Unit Circle, explain why there are angles $\theta$ with $\tan(\theta) = -M$ .
3. Determine angles with $\tan(\theta) = 0$ .
4. Combine the three parts above to conclude the range of the tangent function is $(-\infty, \infty)$ .
Prove $\cot(t)$ is odd. (Hint: mimic the proof given in the text that $\tan(t)$ is odd.)

Section 7.5 Exercise Answers can be found in the Appendix … Coming soon

Just like the rational functions in Chapter 3 are continuous and smooth on their domains because polynomials are continuous and smooth everywhere, the secant and cosecant functions are continuous and smooth on their domains as a result of the cosine and sine functions are continuous and smooth everywhere. ↵
As with the examples in Section 7.3, note that we can partially check our answer because the argument of the secant function should simplify to the `original' quarter marks - the quadrantal angles. ↵
In other words, the range of these functions is $(-\infty, D-|A|] \cup [D+|A|, \infty)$ . ↵
Assuming $A>0$ , that is. ↵
See Exercise 17. ↵
Again, assuming we want $A > 0$ . ↵
Here, as with all tangent functions, we can partially check our new quarter marks by noting the argument of the tangent function simplifies, in each case, to one of the original quarter marks of the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2} \right)$ . ↵