## Section16.5The Graphs of the Tangent and Cotangent Functions

##### The Graph of $y=\tan(t)$.

Exploring the effects of the quotient identity $\tan(t)=\frac{\sin(t)}{\cos(t)}$ on the behavior of the tangent function will give us a lot of insight into the graph $y=\tan(t)\text{.}$ Let's make some initial observations.

• There is an $t$-intercept everywhere $\sin(t)=0\text{.}$ This gives us the following $t$-intercepts.

\begin{equation*} ...,\,(-2\pi,0),\,(-\pi,0),\,(0,0),\,(\pi,0),\,(2\pi,0),\,... \end{equation*}
• The is a vertical asymptote at every value of $t$ where $\cos(t)=0\text{.}$ This gives us the following vertical asymptotes.

\begin{equation*} ...,\,t=-\frac{5\pi}{2},\,t=-\frac{3\pi}{2},\,t=-\frac{\pi}{2},\,t=\frac{\pi}{2},\,t=\frac{3\pi}{2},\,t=\frac{5\pi}{2},\,... \end{equation*}
• The $y$-coordinate of the point on the graph of $y=\tan(t)$ is either $1$ or $-1$ at every value of $t$ where the sine and cosine functions have equal or opposite values. This gives us the following points.

\begin{equation*} ...,\,\left(-\frac{5\pi}{4},-1\right),\,\left(-\frac{3\pi}{4},1\right),\,\left(-\frac{\pi}{4},-1\right),\,\left(\frac{\pi}{4},1\right),\,\left(\frac{3\pi}{4},-1\right),\,\left(\frac{5\pi}{4},1\right),\,... \end{equation*}

Let's go ahead and plot what we've discussed to this point. This is done in Figure 16.5.1.

The sign on the tangent value can only change when $t$ moves from one quadrant to the next. In Figure 16.5.1 this occurs at the asymptotes and the $t$-intercepts. We can use this fact to infer the behavior of the function as $t$ approaches the asymptotes from either side, giving us the function graph shown in Figure 16.5.2.

We should make that unlike the sine and cosine functions, the period of $y=\tan(t)$ is only $\pi\text{.}$

##### The Graph of $y=\cot(t)$.

Exploring the effects of the quotient identity $\cot(t)=\frac{\cos(t)}{\sin(t)}$ on the behavior of the cotangent function will give us a lot of insight into the graph $y=\cot(t)\text{.}$ Let's make some initial observations.

• There is an $t$-intercept everywhere $\cos(t)=0\text{.}$ This gives us the following $t$-intercepts.

\begin{equation*} ...,\,\left(-\frac{5\pi}{2},0\right),\,\left(-\frac{3\pi}{2},0\right),\,\left(-\frac{\pi}{2},0\right),\,\left(\frac{3\pi}{2},0\right),\,\left(\frac{5\pi}{2},0\right),\,\left(\frac{7\pi}{2},0\right),\,... \end{equation*}
• The is a vertical asymptote at every value of $t$ where $\sin(t)=0\text{.}$ This gives us the following vertical asymptotes.

\begin{equation*} ...,\,t=-2\pi,\,t=-\pi,\,t=0,\,t=\pi,\,t=2\pi,\,t=3\pi,\,... \end{equation*}
• The $y$-coordinate of the point on the graph of $y=\cot(t)$ is either $1$ or $-1$ at every value of $t$ where the sine and cosine functions have equal or opposite values. This gives us the following points.

\begin{equation*} ...,\,\left(-\frac{5\pi}{4},-1\right),\,\left(-\frac{3\pi}{4},1\right),\,\left(-\frac{\pi}{4},-1\right),\,\left(\frac{\pi}{4},1\right),\,\left(\frac{3\pi}{4},-1\right),\,\left(\frac{5\pi}{4},1\right),\,... \end{equation*}

Let's go ahead and plot what we've discussed to this point. This is done in Figure 16.5.3.

The sign on the cotangent value can only change when $t$ moves from one quadrant to the next. In Figure 16.5.3 this occurs at the asymptotes and the $t$-intercepts. We can use this fact to infer the behavior of the function as $t$ approaches the asymptotes from either side, giving us the function graph shown in Figure 16.5.4.

We should note that like the tangent function, the period of $y=\cot(t)$ is $\pi\text{.}$

Let's move on and discuss graphical transformations effected upon $y=\tan(t)$ and $y=\cot(t)\text{.}$

##### Period of a Tangent or Cotangent Function.

As is the case with the sine and cosine function, if $\omega$ is a nonzero constant that is not equal to $1$ or $-1\text{,}$ then the graph of $y=\tan(\omega t)$ or $y=\cot(\omega t)$ will be different than the periods of the graphs of $y=\tan(t)$ and $y=\cot(t)\text{.}$ However, in this case the new period is $\frac{\pi}{\abs{\omega}}$ (as opposed to $\frac{2\pi}{\abs{\omega}}$ which is the period of either $y=\sin(\omega t)$ or $y=\cos(\omega t)$). Let's see a couple of examples.

###### Example16.5.5.

Sketch three periods of the function $y=\tan(2t)\text{.}$

Solution

We can see that $\omega=2\text{,}$ so the period of this function is $\frac{\pi}{2}\text{.}$

When there is no horizontal shift, the tangent function has vertical asymptotes half of a period to both the left and the right of the $y$-axis. In this case, this means that both $t=-\frac{\pi}{4}$ and $t=\frac{\pi}{4}$ are vertical asymptotes. Every time we move one full period to the left or the right from a vertical asymptote] we will encounter another vertical asymptote. This means that both $t=-\frac{3\pi}{4}$ and $t=\frac{3\pi}{4}$ are vertical asymptotes. These vertical asymptotes are shown in Figure 16.5.6. Note that the scale on the $t$-axis is $\frac{\pi}{4}\text{.}$ That is because $\frac{\pi}{4}$ is one-half of a full period.

There is a $t$-intercept halfway between each pair of asymptotes. A quarter of a period to the left of each $t$intercept there is a point with a $y$-coordinate of $-1\text{.}$ A quarter of a period to the right of each $t$-intercept there is a point with a $y$-coordinate of $1\text{.}$ All of these points have been added to the graph in Figure 16.5.7.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.8.

###### Example16.5.9.

Sketch three periods of the function $y=\cot\left(\frac{\pi}{3}\right)\text{.}$

Solution

Let's begin by noting that $\frac{t}{3}=\frac{1}{3}t\text{,}$ so $\omega=\frac{1}{3}\text{.}$ The period for $y=\cot\left(\frac{pi}{3}\right)$ is derived below.

\begin{align*} \frac{\pi}{\abs{\omega}}\amp=\frac{\pi}{\frac{1}{3}}\\ \amp=\frac{\pi}{1} \cdot \frac{3}{1} \end{align*}

When there is no horizontal shift the $y$-axis is a vertical asymptote for the graph of a cotangent function. Every time we move one full period to the left of a vertical asymptote we encounter another vertical asymptote. In this case this means that $t=-3\pi\text{,}$ $t=3\pi\text{,}$ and $t=6\pi$ are also vertical asymptotes. These vertical asymptotes are shown in Figure 16.5.10. Note that the scale on the $t$-axis is $\frac{3\pi}{2}\text{.}$ That is because $\frac{3\pi}{2}$ is one-half of a period.

There is a $t$-intercept halfway between each pair of asymptotes. A quarter of a period to the left of each $t$intercept there is a point with a $y$-coordinate of $1\text{.}$ A quarter of a period to the right of each $t$-intercept there is a point with a $y$-coordinate of $-1\text{.}$ All of these points have been added to the graph in Figure 16.5.7.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.12.

##### Horizontal Shift of a Tangent or Cotangent Function.

As with any other function, the nonzero constant $h$ in the expression $y=\tan(\omega(t-h))$ or $y=\cot(\omega(t-h))$ affects a horizontal shift of $\abs{h}$ upon their parent function ($y=\tan(\omega t)$ or $y=\cot(\omega t)$). When $h$ is positive, the shift is rightward and when $h$ is negative the shift if leftward.

For example, $y=\tan\left(t-\frac{\pi}{3}\right)$ lies $\frac{\pi}{3}$ to the right of $y=\tan(t)$ whereas $y=\tan\left(t+\frac{\pi}{3}\right)$ lies $\frac{\pi}{3}$ to the left of $y=\tan(t)\text{.}$

Let's see a couple of examples of the graphing process when there is a horizontal shift.

###### Example16.5.13.

Sketch three periods of the function $y=\tan\left(t-\frac{\pi}{3}\right)\text{.}$

Solution

The function $y=\tan(t)$ has vertical asymptote at $t=-\frac{3\pi}{2}\text{,}$ $t=-\frac{\pi}{2}\text{,}$ $t=\frac{\pi}{2}\text{,}$ and $t=\frac{3\pi}{2}\text{.}$ The function $y=\tan\left(t-\frac{\pi}{3}\right)$ lies $\frac{\pi}{3}$ to the right of the function $y=\tan(t)\text{.}$ The effect this has on the location of the vertical asymptotes for a graph of $y=\tan\left(t-\frac{\pi}{3}\right)$ is calculated below

\begin{align*} t\amp=-\frac{3\pi}{2}+\frac{\pi}{3}\\ t\amp=-\frac{7\pi}{6} \end{align*}
\begin{align*} t\amp=-\frac{\pi}{2}+\frac{\pi}{3}\\ t\amp=-\frac{\pi}{6} \end{align*}
\begin{align*} t\amp=-\frac{\pi}{2}+\frac{\pi}{3}\\ t\amp=-\frac{5\pi}{6} \end{align*}
\begin{align*} t\amp=-\frac{\pi}{2}+\frac{\pi}{3}\\ t\amp=-\frac{11\pi}{6} \end{align*}

Another way we could have figured out the location of these vertical asymptotes is to have applied the shift to $t=\frac{\pi}{2}$ (resulting in $t=\frac{5\pi}{6})$ and then repeatedly adding or subtracting a full period (in this case $\pi$) to $\frac{5\pi}{6}\text{.}$

Regardless of the strategy used to identify the vertical asymptotes, in this case they are $t=-\frac{7\pi}{6}\text{,}$ $t=-\frac{\pi}{6}\text{,}$ $t=\frac{5\pi}{6}\text{,}$ and $t=\frac{11\pi}{6}\text{.}$ These asymptotes are shown in Figure 16.5.14. Note that the midpoint between each pair of consecutive asymptotes has also been identified. These are the points where the tangent curves will cross the $t$-axis.

There is a $t$-intercept halfway between each pair of asymptotes. A quarter of a period to the left of each $t$intercept there is a point with a $y$-coordinate of $-1\text{.}$ A quarter of a period to the right of each $t$-intercept there is a point with a $y$-coordinate of $1\text{.}$ All of these points have been added to the graph in Figure 16.5.15.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.16.

###### Example16.5.17.

Sketch three periods of the function $y=\cot\left(2t+\frac{\pi}{2}\right)\text{.}$

Solution

Let's begin by remarking that the period of this function is $\frac{\pi}{2}\text{.}$ Before we can state the horizontal shift, we need to factor $2$ away from both $t$ and $\frac{\pi}{2}\text{.}$ This is done below.

\begin{equation*} \cot\left(2t+\frac{\pi}{2}\right)=\cot\left(2\left(t+\frac{\pi}{4}\right)\right) \end{equation*}

We can now see that there is a leftward shift of $\frac{\pi}{4}\text{.}$

The lines $t=-\frac{\pi}{2}\text{,}$ $t=0\text{,}$ $t=\frac{\pi}{2}\text{,}$ and $t=\pi$ are all vertical asymptotes for the function $y=\cot(2t)\text{.}$ The function $y=\cot\left(2t+\frac{\pi}{2}\right)$ lies $\frac{\pi}{4}$ to the left of $y=\cot(2t)\text{,}$ so it has the vertical asymptotes $t=-\frac{3\pi}{4}\text{,}$ $t=-\frac{\pi}{4}\text{,}$ $t=\frac{\pi}{4}\text{,}$ and $t=\frac{3\pi}{4}\text{.}$ These asymptotes are shown in Figure 16.5.18. Note that the midpoint between each pair of consecutive asymptotes has also been identified. These are the points where the tangent curves will cross the $t$-axis.

There is a $t$-intercept halfway between each pair of asymptotes. A quarter of a period to the left of each $t$intercept there is a point with a $y$-coordinate of $-1\text{.}$ A quarter of a period to the right of each $t$-intercept there is a point with a $y$-coordinate of $1\text{.}$ All of these points have been added to the graph in Figure 16.5.19.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.20.

##### Vertical Stretches, Compressions, Reflections, and Shifts.

Vertical stretches, compressions, reflections, and shifts work exactly the same way for the tangent and cotangent functions as they work for any other function. To wit:

• The graph of $y=A \tan(t)\text{,}$ $A \neq 0\text{,}$ $A \neq \pm 1\text{,}$ is either a vertical stretch or a vertical compression of the graph of $y=\tan(t)\text{.}$ If $\abs{A} \gt 1\text{,}$ the effect is a stretch away from the $t$-axis by a factor of $A\text{.}$ If $\abs{A} \lt 1\text{,}$ the effect is a vertical compression towards the $t$-axis by a factor of $\abs{A}\text{.}$ Additionally, if $A \lt 0\text{,}$ the curves of $y=\tan(t)$ are reflected across the $t$-axis.

• The graph of $y=\tan(t)+k\text{,}$ $k \neq 0\text{,}$ is a vertical shift of the graph of $y=\tan(t)\text{.}$ When $k \gt 0\text{,}$ the shift is upward by $k\text{.}$ When $k \lt 0\text{,}$ the shift is downward by $\abs{k}\text{.}$

Let's see an example.

###### Example16.5.21.

Sketch three periods of the function $y=-3\cot(t)+2\text{.}$

Solution

In words, the graphical transformations affected on the parent function $y=\cot(t)$ are a stretch by a factor of $3\text{,}$ a reflection across the $t$-axis, and lastly an upward shift by $2\text{.}$

When graphing $y=\cot(t)\text{,}$ between each pair of consecutive pair of asymptotes the are three key points. The points, from left-to-right, have the form

\begin{equation*} (t,1),\,\,(t,0),\,\,(t,-1). \end{equation*}

On the graph of $y=-3\cot(2)+2$ those points become

\begin{equation*} (t,3(1)+2),\,\,(t,3(0)+2),\,\,(t,3(-1)+2) \end{equation*}

which simplify to

\begin{equation*} (t,5),\,\,(t,2),\,\,(t,-2). \end{equation*}

Because there are no changes in the horizontal direction, we know that $t=-\pi\text{,}$ $t=0\text{,}$ $t=\pi\text{,}$ and $t=2\pi$ are all vertical asymptotes for the function. Those asymptotes and the aforementioned points are plotted in Figure 16.5.22.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.23.

### ExercisesExercises

Sketch three periods of each function.

###### 1.

$y=\cot(3t)$

Solution

The only transformation being affected upon the function $y=\cot(t)$ is a change in the period. The new period is $\frac{\pi}{3}\text{.}$ The lines $t=-\frac{\pi}{3}\text{,}$ $t=0\text{,}$ $t=\frac{\pi}{3}\text{,}$ and $t=\frac{2\pi}{3}$ will all be vertical asymptotes for the graph. Halfway between each consecutive pair of asymptotes the curve will cross the $t$-axis. One-quarter period to the left of each $t$-intercept there is a point with a $y$-coordinate of $1$ and one-quarter-period to the right of each $t$-intercept there is a point with a $y$-coordinate of $-1\text{.}$ All of this is plotted in Figure 16.5.24.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.24.

###### 2.

$y=\tan\left(t-\frac{\pi}{8}\right)$

Solution

The only transformation being affected on the graph of $y=\tan(t)$ is a rightward shift of $\frac{\pi}{8}\text{.}$ The effect this has upon the vertical asymptotes is computed below.

\begin{align*} t\amp=-\frac{3\pi}{2}+\frac{\pi}{8}\\ t\amp=-\frac{11\pi}{8} \end{align*}
\begin{align*} t\amp=-\frac{\pi}{2}+\frac{\pi}{8}\\ t\amp=-\frac{3\pi}{8} \end{align*}
\begin{align*} t\amp=\frac{\pi}{2}+\frac{\pi}{8}\\ t\amp=\frac{5\pi}{8} \end{align*}
\begin{align*} t\amp=\frac{3\pi}{2}+\frac{\pi}{8}\\ t\amp=-\frac{13\pi}{8} \end{align*}

Halfway between each consecutive pair of asymptotes the curve will cross the $t$-axis. One-quarter period to the left of each $t$-intercept there is a point with a $y$-coordinate of $-1$ and one-quarter-period to the right of each $t$-intercept there is a point with a $y$-coordinate of $1\text{.}$ All of this is plotted in Figure 16.5.26.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.27.

###### 3.

$y=\frac{2}{3}\cot(t)-2$

Solution

All of the modifications to the graph of $y=\cot(t)$ are in the vertical direction. There is a vertical compression by a factor of $\frac{2}{3}$ followed by a downward shift of 2.

When graphing $y=\cot(t)\text{,}$ between each pair of consecutive pair of asymptotes the are three key points. The points, from left-to-right, have the form

\begin{equation*} (t,1),\,\,(t,0),\,\,(t,-1). \end{equation*}

On the graph of $y=y=\frac{2}{3}\cot(t)-2$ those points become

\begin{equation*} (t,\frac{2}{3}(1)-2),\,\,(t,\frac{2}{3}(0)-2),\,\,(t,\frac{2}{3}(-1)-2) \end{equation*}

which simplify to

\begin{equation*} (t,-\frac{4}{3}),\,\,(t,-2),\,\,(t,-\frac{8}{3}). \end{equation*}

Let's go ahead and plot our asymptotes, $t=-\pi\text{,}$ $t=0\text{,}$ $t=\pi\text{,}$ and $t=2\pi\text{.}$ Let's also plot the points discussed above. This is done in Figure 16.5.28.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.29.

###### 4.

$y=-\tan\left(3\left(t+\frac{\pi}{6}\right)\right)$

Solution

We begin by observing that the period is $\frac{\pi}{3}\text{.}$

The lines $y=-\frac{3\pi}{2}\text{,}$ $y=-\frac{\pi}{2}\text{,}$ $y=\frac{\pi}{2}\text{,}$ and $y=\frac{3\pi}{2}$ are all vertical asymptotes for a graph of $y=\tan(t)\text{.}$ These lines all need to be compressed towards the $y$-axis by a factor of $\frac{1}{3}\text{.}$ The resultant vertical asymptotes for a graph $y=\tan(3t)$ are $y=-\frac{\pi}{2}\text{,}$ $y=-\frac{\pi}{6}\text{,}$ $y=\frac{\pi}{6}\text{,}$ and $y=\frac{\pi}{2}\text{.}$ These lines are all shifted to the left by $\frac{\pi}{6}$ on a graph of $y=-\tan\left(3\left(t+\frac{\pi}{6}\right)\right)\text{.}$ The effect this has upon the vertical asymptotes is calculated below.

\begin{align*} t\amp=-\frac{\pi}{2}-\frac{\pi}{6}\\ t\amp=-\frac{2\pi}{3} \end{align*}
\begin{align*} t\amp=-\frac{\pi}{6}-\frac{\pi}{6}\\ t\amp=-\frac{\pi}{3} \end{align*}
\begin{align*} t\amp=\frac{\pi}{6}-\frac{\pi}{6}\\ t\amp=0 \end{align*}
\begin{align*} t\amp=\frac{\pi}{2}-\frac{\pi}{6}\\ t\amp=\frac{\pi}{3} \end{align*}

Let's go ahead and plot the vertical asymptotes. This is done in Figure 16.5.30.

There is a curve that crosses the $t$-axis halfway between each pair of consecutive asymptotes. On a graph of $y=\tan(t)$ there is a point one-quarter of a period to the left of each $t$-intercept with a $y$-coordinate of $1$ and a point one-quarter of a period to the right of each $t$-intercept with a $y$-coordinate of $1\text{.}$ Because the function we are graphing has a negative sign in front of the tangent expression, the points with a $y$-coordinate of $1$ will fall to the left of the $t$-intercepts and the points with a $y$-coordinate of $-1$ will fall to the right of the $t$-intercepts. These points have been added to the graph in Figure 16.5.31.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.32.

###### 5.

$y=\cot\left(\frac{t}{2}-\frac{\pi}{4}\right)-3$

Solution

Let's first remark that $\frac{t}{2}=\frac{1}{2}t\text{,}$ so the value of $\omega$ is $\frac{1}{2}\text{.}$ The period is calculated below.

\begin{align*} \frac{\pi}{\abs{\omega}}\amp=\frac{\pi}{\frac{1}{2}}\\ \amp=\frac{\pi}{1} \cdot \frac{2}{1}\\ \amp=2\pi \end{align*}

Before we can establish the horizontal shift, we need to factor $\frac{1}{2}$ away from both $t$ and $\frac{\pi}{4}\text{.}$ This is done below.

\begin{equation*} \cot\left(\frac{t}{2}-\frac{\pi}{4}\right)-3=\cot\left(\frac{1}{2}\left(t-\frac{\pi}{2}\right)\right)-3 \end{equation*}

We can now see that there is a rightward shift of $\frac{\pi}{2}\text{.}$

The lines $y=-2\pi\text{,}$ $y=0\text{,}$ $y=2\pi\text{,}$ and $y=4\pi$ are all vertical asymptotes on a graph of $y=\cot\left(\frac{t}{2}\right)\text{.}$ On a graph $y=\cot\left(\frac{t}{2}-\frac{\pi}{4}\right)-3$ those asymptotes all need to shifted to the right by $\frac{\pi}{2}\text{.}$ The resultant asymptotes are $y=-\frac{3\pi}{2}\text{,}$ $y=\frac{\pi}{2}\text{,}$ $y=\frac{5\pi}{2}\text{,}$ and $y=\frac{9\pi}{2}\text{.}$ These asymptotes are plotted in Figure 16.5.33.

Between the asymptotes, at regular intervals of one-quarter of a period, the $y$-coordinates of an unmodified cotangent function follow the pattern $1\text{,}$ $0\text{,}$ $-1\text{.}$ The cotangent function we are graphing has a downward shift of $3\text{,}$ so that pattern is shifted downward to $-2\text{,}$ $-3\text{,}$ $-4\text{.}$ The resultant points have been added to the graph in Figure 16.5.34.

All that's left to do is connect the points in a way that demonstrates the asymptotic behavior of the function. This is done in Figure 16.5.35.