\chapter{Trigonometry}
\section{Graph of tangent}
We now turn to the graph of the tangent function. We will see that it is quite different from sine
and cosine. The main difference is that \(\tan t\) is not defined at all the quadrantal angles
on the \(y\)-axis. This means that the function will have vertical asymptotes.
As we did for sine and cosine, we make a table for a few important angles. This time
instead of using the quadrantal angles in \(0\leq t \leq 2\pi\), we will use the five angles
\(-\pi/2\), \(-\pi/4\), \(0\), \(\pi/4\), \(\pi/2\).
At \(-\pi/2\) and \(\pi/2\), the tangent function is undefined. The graph has vertical asymptotes there.
To find \(\tan 0\), use the point \((1,0)\) on the \(t\)-axis: \(\tan 0 = 0/1=0\).
To find \(\tan(\pi/4)\), use the \(45\)-\(45\)-\(90\) triangle: \(\tan (\pi/4)=1\).
The reference angle for \(-\pi/4\) is \(\pi/4\), and the angle is in Quadrant 4, where
the tangent is negative, so \(\tan (-\pi/4)=-1\).
We use ? to mean that
the function is undefined there. There will be vertical asymptotes where the function is undefined.
\[\begin{array}{c|c}
t & \tan t\\
\hline
-\dfrac{\pi}{2} & ? \\[2ex]
-\dfrac{\pi}{4} & -1 \\[2ex]
0 & 0 \\[2ex]
\dfrac{\pi}{4} & 1\\[2ex]
\dfrac{\pi}{2} & ?
\end{array} \hspace{5ex} \img{U4_8F1.png}{-5em}{15em}{}\]
We now draw one cycle of the graph. To draw the right side, start at \((0,0)\)
and draw a curve bending up,
going through \((\pi/4,1)\), and approaching the vertical asymptote at \(t=\pi/2\).
To draw the left side of the graph, start at \((0,0)\) and draw a curve bending down,
going through the point \((-\pi/4,-1)\) and approaching the vertical asymptote at
\(t=-\pi/2\):
\[
\img{U4_8F2.png}{}{8em}{}
\]
The origin \((0,0)\) is the point where the graph changes from bending down to bending up and
we think of it as the \textit{center} of the graph.
Note that the period of the tangent function is \(P=\pi\) (half the period of sine or cosine).
The domain of \(\tan t\) is complicated to write down, because we need to exclude infinitely many
points (all the points corresponding to quadrantal angles on the \(y\)-axis). But we can see from
the graph that the range extend to all real numbers: \(R=(-\infty,\infty)\).
So the amplitude is not defined for \(\tan t\).
As for the graphs of sine and cosine, the graph of \(\tan t\) extends indefinitely on both sides
of the \(t\)-axis, repeating the same shape. Three cycles of the graph are shown below:
\[
\img{U4_8F3.png}{}{18em}{}
\]
\subsection{The transformed tangent function}
A transformed tangent function will have the form
\[f(t) = A\tan(Bt+C)+k.\]
where \(A\) determines a vertical stretch (remember that there is no amplitude
for the tangent function), \(k\) is the vertical shift, and \(B\) determines the period
by the formula
\[P=\frac{\pi}{B}.\]
Notice the difference with the \(\sin\) and \(\cos\) functions, with \(\pi\) instead
of \(2\pi\) in the numerator.
The parent function \(\tan t\) has vertical asymptotes at \(t=-\pi/2\) and
\(t=\pi/2\). So to find the vertical asymptotes of the transformed function we solve the equations:
\[Bt+C=-\dfrac{\pi}{2}, \hspace{3ex} Bt+C=\dfrac{\pi}{2}.\]
To draw the graph of one cycle of a transformed tangent function, we divide the period into \(4\)
to find the unit for the \(t\)-axis.
\begin{example}
We want to draw the graph of the function
\[f(t)=3\tan(4t)-2.\]
To find the vertical asymptotes we solve the equations:
\[4t =-\frac{\pi}{2} \hspace{5ex} \mbox{and} \hspace{5ex} 4t = \frac{\pi}{2},\]
and we find
\[t=-\frac{\pi}{8}\hspace{5ex} \mbox{and} \hspace{5ex} t = \frac{\pi}{8}.\]
We begin to draw the graph:
\[
\img{U4_8F4.png}{}{12em}{}
\]
To find the unit for the \(t\)-axis, we divide the period by \(4\):
\[P=\frac{\pi}{4}, \hspace{5ex} \mbox{unit}=\frac{1}{4}\cdot \frac{\pi}{4}=\frac{\pi}{16}.\]
Now we locate the center of the graph, corresponding to the origin \((0,0)\)
for the parent function. The vertical stretch by \(3\) does not change \(0\): \(3(0)=0\). Then
shift down by \(2\): \(0-2=-2\). So the center of the graph is at \((0,-2)\):
\[
\img{U4_8F5.png}{}{12em}{}
\]
The other two points have \(y\)-coordinates \(-1\) and \(1\), so they get transformed to
\(3(1)-2=1\) and \(3(-1)-2=-5\):
\[
\img{U4_8F6.png}{}{12em}{}
\]
Now we can draw the graph, starting at the center and bending up towards the asymptote on the right, and
then bending down towards the asymptote on the left:
\[
\img{U4_8F7.png}{}{12em}{}
\]
\end{example}
In the next example we draw a tangent function with a horizontal shift and a reflection.
\begin{example}
Let \(\displaystyle f(t)=-2\tan\left(\frac{\pi}{2} t +\frac{\pi}{4}\right)\).
The period is
\begin{center}
\(P=\dfrac{\pi}{\frac{\pi}{2}}=\pi\cdot \dfrac{2}{\pi}=2\),
\end{center}
and the unit for the \(t\)-axis is:
\begin{center}
\(\mbox{unit}=\dfrac{P}{4}=\dfrac{2}{4}=\dfrac{1}{2}.\)
\end{center}
To find the vertical asymptotes, solve:
\[\begin{array}{rclcrcl}
\displaystyle \frac{\pi}{2} t +\frac{\pi}{4}&=&\displaystyle -\frac{\pi}{2} &
\hspace{5ex} \mbox{and} \hspace{5ex}
& \displaystyle \frac{\pi}{2} t +\frac{\pi}{4}&=& \displaystyle \frac{\pi}{2}\\[2ex]
\displaystyle \frac{\cancel{\pi}}{2} t +\frac{\cancel{\pi}}{4}&=& \displaystyle
-\frac{\cancel{\pi}}{2}
& \hspace{6ex} \hspace{6ex}
& \displaystyle \frac{\cancel{\pi}}{2} t +\frac{\cancel{\pi}}{4}&=& \displaystyle
\frac{\cancel{\pi}}{2}\\[2ex]
\displaystyle \frac{1}{2} t +\frac{1}{4}&=& \displaystyle
-\frac{1}{2}
& \hspace{6ex} \hspace{6ex}
& \displaystyle \frac{1}{2} t +\frac{1}{4}&=& \displaystyle
\frac{1}{2}\\[2ex]
\displaystyle \frac{1}{2} t &=& \displaystyle
-\frac{1}{2}-\frac{1}{4}
& \hspace{6ex} \hspace{6ex}
& \displaystyle \frac{1}{2} t &=& \displaystyle
\frac{1}{2}-\frac{1}{4}\\[2ex]
\displaystyle \frac{1}{2} t &=& \displaystyle
-\frac{3}{4}
& \hspace{6ex} \hspace{6ex}
& \displaystyle \frac{1}{2} t &=& \displaystyle
\frac{1}{4}\\[2ex]
x &=& \displaystyle
-\frac{3}{2}
& \hspace{6ex} \hspace{6ex}
& x &=& \displaystyle
\frac{1}{2}
\end{array}.
\]
Now we can draw the vertical asymptotes and mark the \(t\)-axis:
\[
\img{U4_8F8.png}{}{12em}{}
\]
The center of the graph is the midpoint between the asymptotes, so it's at \(t=-1/2\).
There is no vertical shift, so the center is at \((-1/2,0)\). The other two points are reflected and
stretched by \(2\):
\[
\img{U4_8F9.png}{}{12em}{}
\]
Now we can draw the graph. Because of the reflection across the \(t\)-axis, the right side will bend down towards the asymptote at \(t=1/2\), and the left side will bend up towards the asymptote
at \(t=-3/2\):
\[
\img{U4_8F10.png}{}{12em}{}
\]
\end{example}
| Graph of \(y=\tan x\) (three cycles) | \(\hspace{7ex}\) Properties of \(\tan t\) |
| \(\img{U4_8F3.png}{}{18em}{}\) |
- Period\(\displaystyle= P=\frac{\pi}{B}\)
- Amplitude is undefined.
- Range\(=(-\infty,\infty)\).
- Vertical asymptotes at \(\displaystyle t=-\frac{\pi}{2}\) and
\(\displaystyle t=\frac{\pi}{2}\).
- Center of graph at \((0,0)\).
- Other two points at \(\displaystyle \left(-\frac{\pi}{4},-1\right)\) and \(\displaystyle
\left(\frac{\pi}{4},1\right)\).
|
Steps to plot one cycle of \(f(x)=A\tan(Bx+C)+k\)
- \(\displaystyle P=\frac{\pi}{B}\).
- Vertical shift\(=k\).
- Two asymptotes: solve \(\displaystyle Bt+C=-\frac{\pi}{2}\) and
\(\displaystyle Bt+C=\frac{\pi}{2}\)
- Unit\(\displaystyle =\frac{P}{4}\).
- Mark the \(t\)-axis with multiples of the unit.
- Draw two vertical asymptotes.
- Center of graph is the midpoint between asymptotes, shifted vertically by \(k\).
- Plot other two key points using vertical stretch \(A\).
|
Problems
\problem
Let \(\displaystyle f(x)=2\tan\left(2x-\frac{\pi}{4}\right)+1\).
\begin{enumerate}
\item
Find the period for the function.
\item
Find the unit for the \(x\)-axis.
\item Find two vertical asymptotes.
\item Sketch one cycle.
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item The period is \(P=\pi/2\).
\item
The unit is
\[\mbox{unit}=\frac{P}{4}=\frac{1}{4}\cdot \frac{\pi}{2}=\frac{\pi}{8}.\]
\item
Solve the equations:
\[\begin{array}{rclcrcl}
\displaystyle 2x-\frac{\pi}{4}& = & \displaystyle -\frac{\pi}{2} &
\hspace{5ex} \mbox{and} \hspace{5ex}
& \displaystyle 2x-\frac{\pi}{4}&=& \displaystyle \frac{\pi}{2}\\[1ex]
\displaystyle 2x &=& \displaystyle -\frac{\pi}{2}+\frac{\pi}{4}
& \hspace{6ex} \hspace{6ex}
& \displaystyle 2 x&=& \displaystyle
\frac{\pi}{2}+\frac{\pi}{4}\\[1ex]
\displaystyle 2x &=& \displaystyle -\frac{\pi}{4}
& \hspace{6ex} \hspace{6ex}
& \displaystyle 2 x&=& \displaystyle \frac{3\pi}{4}\\[1ex]
\displaystyle x &=& \displaystyle -\frac{\pi}{8}
& \hspace{6ex} \hspace{6ex}
& \displaystyle x&=& \displaystyle \frac{3\pi}{8}
\end{array}.
\]
So there are vertical asymptotes at \(x=-\pi/8\) and \(x=3\pi/8\).
\item
The horizontal shift is
\[\begin{array}{rcl}
\displaystyle 2x-\frac{\pi}{4}& = &0\\[1ex]
\displaystyle 2x& = & \displaystyle \frac{\pi}{4}\\[1ex]
\displaystyle x& = & \displaystyle \frac{\pi}{8}
\end{array}
\]
and the vertical shift is \(k=1\),
so the center is at \((\pi/8,1)\). The vertical stretch is \(2\), so the other two points
have \(y\)-coordinates \(2+1=3\) and \(-2+1=-1\).
\[\img{U4_8F11.png}{}{15em}{}\]
\end{enumerate}
\end{sol}
\mproblem
Let \(\displaystyle f(x)=3\tan\left(4x+\frac{\pi}{2}\right)-1\).
\begin{enumerate}
\item
Find the period for the function.
\item
Find the unit for the \(x\)-axis.
\item Find two vertical asymptotes.
\item Sketch one cycle.
\end{enumerate}
\problem
Let \(\displaystyle f(x)=-\tan\left(5x\right)+2\).
\begin{enumerate}
\item
Find the period for the function.
\item
Find the unit for the \(x\)-axis.
\item Find two vertical asymptotes.
\item Sketch one cycle.
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item The period is \(P=\pi/5\).
\item
The unit is
\[\mbox{unit}=\frac{P}{4}=\frac{1}{4}\cdot \frac{\pi}{5}=\frac{\pi}{20}.\]
\item
Solve the equations:
\[\begin{array}{rclcrcl}
\displaystyle 5x& = & \displaystyle -\frac{\pi}{2} &
\hspace{5ex} \mbox{and} \hspace{5ex}
& \displaystyle 5x &=& \displaystyle \frac{\pi}{2}\\[2ex]
\displaystyle x &=& \displaystyle \frac{\pi}{10}
& \hspace{6ex} \hspace{6ex}
& \displaystyle x&=& \displaystyle -\frac{\pi}{10}
\end{array}.
\]
So there are vertical asymptotes at \(x=-\pi/10\) and \(x=\pi/10\).
\item
There is a reflection across the \(x\)-axis, with no vertical stretch.
There is no horizontal shift,
and the vertical shift is \(k=2\),
so the center is at \((0,2)\). The other two points
have \(y\)-coordinates \(3\) and \(1\).
\[\img{U4_8F12.png}{}{15em}{}\]
\end{enumerate}
\end{sol}
\mproblem
Let \(\displaystyle f(x)=-2\tan\left(3x\right)-1\).
\begin{enumerate}
\item
Find the period for the function.
\item
Find the unit for the \(x\)-axis.
\item Find two vertical asymptotes.
\item Sketch one cycle.
\end{enumerate}