\chapter{Trigonometry}
\section{Graphs of sine and cosine}
In this unit we study the graphs of the sine and cosine functions. We will often use \(x\) or
\(t\) for the input variable, such as \(\sin x\) or \(\sin t\). So \(x\) or \(t\) are angles,
and it is always understood that they are measured in radians. We will never draw graphs
of trig functions with angles in degrees.
\textbf{Note:} If the variable \(x\) is used as the input of \(\sin x\), (and so \(x\) represents an angle),
be careful not to confuse it
with the letter \(x\) used in Unit 4.4 for the \(x,y,r\)
definition, that was the \(x\) coordinate of a point on the terminal side of the angle.
The quadrantal angles will be especially useful for the graphs. Recall that
\[90^\circ = \frac{\pi}{2}, \hspace{2ex} 180^\circ =\pi, \hspace{2ex} 270^\circ = \frac{3\pi}{2},
\hspace{2ex} 360^\circ = 2\pi.\]
We will think of the sine and cosine functions as new parent functions to add to our list from
Module 1. Then we will apply the simple transformations to get more functions.
We begin with the sine function.
\subsection{The graph of \(\sin t\)}
The first thing to notice is that the sine function is defined for every value of the input
variable, because using the \(x,y,r\) definition from Unit 4.4,
\[\sin t = \frac{y}{r},\]
where \(y\) is the \(y\)-coordinate of any point on the terminal side of \(\theta\), and \(r\) is
the distance from the origin. Since \(r\) is always positive, the sine function is always defined.
So we conclude that the domain of the sine function is \((-\infty, \infty)\).
To understand the graph of \(\sin t\), we make a table for the first five quadrantal angles
\(0,\pi/2, \pi, 3\pi/2, 2\pi\):
\[\begin{array}{c|c}
t & \sin t\\
\hline
0 & 0 \\[2ex]
\displaystyle \frac{\pi}{2} & 1 \\[2ex]
\pi & 0 \\[2ex]
\displaystyle \frac{3\pi}{2} & -1\\[2ex]
2\pi & 0
\end{array} \hspace{2ex} \img{U4_7F1.png}{-6em}{15em}{}\]
Plotting the five points and joining them with a smooth curve, we find \textit{one cycle} of the
graph of \(\sin t\):
\[\img{U4_7F2.png}{}{20em}{}\]
Note from the graph that the range is \(R=[-1,1]\).
The horizontal line that cuts the graph in
the middle is called the \textit{midline}. So the midline for \(\sin t\) is the \(t\)-axis,
or \(y=0\). We will usually draw the midline as a dashed line on our graphs.
The height of one "peak" above the midline is called the \textit{Amplitude}.
So the amplitude
of the sine function is \(A=1\).
The length of one cycle is called the \textit{Period}, and denoted by \(P\).
So the period of the sine function is \(P=2\pi\).
Note that the five points we sketched are either on the midline, or at the highest or lowest
point of
one cycle.
The highest point of a cycle is called a \textit{maximum}, and the lowest point a \textit{minimum}.
So for the sine function the five key points are:
\[\mbox{At midline} \hspace{5ex} \mbox{Maximum} \hspace{5ex} \mbox{At midline}
\hspace{5ex} \mbox{Minimum} \hspace{5ex} \mbox{At midline}.\]
\[\fbox{\(\img{U4_7F23.png}{}{20em}{}\)}\]
Of course the graph will extend indefinitely to the left and right, because the domain
is \((-\infty,\infty)\). But the rest of the graph is just a repetition of more cycles,
identical to the first one. The picture below shows one more cycle in the negative direction:
\[\img{U4_7F3.png}{}{20em}{}\]
\subsection{The graph of \(cos(t)\)}
We now turn to \(\cos t\).
\textbf{The graph of \(\cos t\)}
As for the sine function, \(\cos t\) is defined for all \(t\), and the domain is \((-\infty,\infty)\).
As before, we make a table for the first five quadrantal angles:
\[\begin{array}{c|c}
t & \cos t\\
\hline
0 & 1 \\[2ex]
\displaystyle \frac{\pi}{2} & 0 \\[2ex]
\pi & -1 \\[2ex]
\displaystyle \frac{3\pi}{2} & 0\\[2ex]
2\pi & 1
\end{array} \hspace{5ex} \img{U4_7F1.png}{-6em}{15em}{}\]
One cycle of \(\cos t\) is shown in the picture:
\[\img{U4_7F4.png}{}{20em}{}\]
The period and amplitude of \(\cos t\) are the same as for \(\sin t\):
\\
\(P=2\pi\), \(\mbox{Amplitude}=1\).
The five points we sketched are once again on the midline and at the highest and lowest
point of one cycle, but this time the first one is a maximum:
\[\mbox{Maximum}, \hspace{5ex} \mbox{At midline}, \hspace{5ex} \mbox{Minimum},
\hspace{5ex} \mbox{At midline}, \hspace{5ex} \mbox{Maximum}.\]
\[\fbox{\(\img{U4_7F24.png}{}{20em}{}\)}\]
Two cycles of the graph are shown below:
\[\img{U4_7F5.png}{}{20em}{}\]
\subsection{The transformed sine and cosine functions}
The sine and cosine functions are new parent functions, and we can apply the simple transformations
to obtain many more functions. The general transformed function will have the form:
\[f(t)=A\sin(Bt+C)+k\]
or
\[f(t)=A\cos(Bt+C)+k\]
where the letters \(A,B,k\) determine amplitude, period and vertical
shift according to the formulas:
\[\mbox{Amplitude }=|A|, \hspace{3ex} \mbox{Period }=\frac{2\pi}{B},
\hspace{3ex} \mbox{Vertical shift }=k.\]
To find the beginning and the end of one cycle, solve the equations
\[Bt+C=0, \hspace{5ex} Bt+C=2\pi.\]
Note that the amplitude is really the same thing as the vertical stretch that we studied in
Module 1.
To draw the graph of one cycle of
some transformed sine or cosine function, we need to find the five key points
that are either on the midline, or a maximum, or a minimum for one cycle. The next example shows
how to do it.
\begin{example}
Let
\[f(t) = 3 \sin(2t - \pi)+2.\]
For this function, we have \(A=3\), \(B=2\), \(C=-\pi\), \(k=2\). So we find:
\[\mbox{Amplitude}= 3, \hspace{5ex} \mbox{Period}=\frac{2\pi}{2}=\pi, \hspace{5ex}
\mbox{Vertical shift}=2.\]
The beginning of a cycle is found by solving \(2t-\pi =0\):
\[\begin{array}{rcll}
2t-\pi & = & 0 & \\[2ex]
2t & = & \pi & \\[2ex]
t & = & \displaystyle \frac{\pi}{2} &\mbox{Beginning of cycle}
\end{array}
\]
The end of the cycle is found by solving \(2t-\pi =2\pi\):
\[\begin{array}{rcll}
2t-\pi & = & 2\pi & \\[2ex]
2t & = & 3\pi & \\[2ex]
t & = & \displaystyle \frac{3\pi}{2} &\mbox{End of cycle}
\end{array}
\]
To find the five key points, we follow the steps:
\begin{itemize}
\item We draw the midline as a dashed line. Since the midline for the parent function
was the \(t\)-axis \(y=0\), the midline for the shifted function is given by the vertical shift:
\(y=k\). So
\[\mbox{Midline: } y=2.\]
\[\img{U4_7F7.png}{}{20em}{}\]
Note that we did not mark the \(t\)-axis with any numbers. That's because we need to find the
right units for this graph, in the next step.
\item Determine the horizontal distance between one point and the next. Since one cycle
is divided into four parts by the five points, we need to divide the period by 4, and we
call this number the \textit{unit} for the graph:
\[\mbox{unit}=\frac{P}{4}=\frac{\pi}{4}.\]
\item We mark the \(t\)-axis with multiples of the unit, and we draw a grid.
In this example the multiples are
\[\frac{\pi}{4}, \hspace{3ex} \frac{2\pi}{4}=\frac{\pi}{2}, \hspace{3ex}\frac{3\pi}{4},
\hspace{3ex} \frac{4\pi}{4}=\pi, \hspace{3ex} \frac{5\pi}{4}, \hspace{3ex}
\frac{6\pi}{4} =\frac{3\pi}{2} \ldots\]
\[\img{U4_7F6.png}{}{20em}{}\]
\item
At this point it's useful to make a quick sketch of the parent function, to use as a guide
for the general shape of the transformed function:
\[\img{U4_7F13.png}{}{10em}{}.\]
The first and last of the five key points for any transformed sine function must be on the midline
(as in the parent \(\sin t\)). We know that the cycle begins at \(t=\pi/2\) and ends at \(t=3\pi/2\). So we can draw first
and last point:
\[\img{U4_7F8.png}{}{20em}{}\]
\item
To find the \(t\)-coordinates for the other points, we move to the right by one unit at time.
In this problem the first point is on a grid line, so to find the next point we just move
to the next grid line. To place the second point, we move up from
the midline by \(3\), because the amplitude is \(3\). So the second point is at
\((3\pi/4,5)\):
\[\img{U4_7F9.png}{}{20em}{}\]
\item
In the same way, the third point will be back on the midline at \((\pi,2)\), and
the fourth will be \(3\) below the midline at \((5\pi/4,-1)\):
\[\img{U4_7F10.png}{}{20em}{}\]
\item
We can now draw one cycle of the sine graph as a smooth curve, trying to imitate the shape of the
parent function:
\[\img{U4_7F11.png}{}{20em}{}\]
\end{itemize}
If we want to draw two cycles, we can add more points and repeat the shape of the graph:
\[\img{U4_7F12.png}{}{20em}{}\]
We can see from the graph that the range is \(R=[-1,5]\).
\end{example}
\begin{example}
Now we draw the graph of a transformed cosine function.
Let \(f(t)=-2\cos(\pi t+\pi/4)-1\).
For this function,
\[A=-2, \hspace{5ex} B=\pi, \hspace{5ex} C=\frac{\pi}{4}, \hspace{5ex} k=1.\]
So we find:
\[\mbox{Amplitude}=|-2|=2, \hspace{5ex} \mbox{Period}=P=\frac{2\pi}{\pi}=2,
\hspace{5ex} \mbox{Vertical shift}=-1.\]
Note that there is a reflection across the \(x\)-axis, because of the negative sign in front
of the function.
The beginning and end of a cycle are obtained by solving \(\pi t+\pi/4=0\), \(\pi t+\pi/4=2\pi\):
\[
\begin{array}{rcll}
\pi t + \displaystyle \frac{\pi}{4}& = & 0 &\\[2ex]
\cancel{\pi}t & = & \displaystyle -\frac{\cancel{\pi}}{4} & \\[2ex]
t & = & \displaystyle -\frac{1}{4} & \mbox{Beginning of cycle}
\end{array}
\]
\[
\begin{array}{rcll}
\pi t + \displaystyle \frac{\pi}{4}& = & 2\pi &\\[2ex]
\pi t & = & \displaystyle 2\pi -\frac{\pi}{4} & \\[2ex]
\cancel{\pi} t & = & \displaystyle \frac{7\cancel{\pi}}{4} & \\[2ex]
t & = & \displaystyle \frac{7}{4} & \mbox{End of cycle}
\end{array}
\]
We divide the period by \(4\) to find the unit:
\[\frac{P}{4}=\frac{2}{4}=\frac{1}{2}\]
The multiples of the unit are:
\[\frac{1}{2}, \hspace{5ex} \frac{2}{2}=1,\hspace{5ex} \frac{3}{2}, \hspace{5ex}
\frac{4}{2}=2, \hspace{5ex} \frac{5}{2}\ldots\]
and the midline is \(y=k=-1\).
We can now draw the grid lines and place the midline.
\[\img{U4_7F15.png}{}{20em}{}\]
The graph of the parent function is:
\[
\img{U4_7F14.png}{}{10em}{}
\]
so the reflected graph will look like:
\[
\img{U4_7F17.png}{}{10em}{}
\]
The beginning of a cycle is at \(t=-1/4\), and the end is at \(t=7/4\). Since
the amplitude is \(2\), and the graph is reflected across the \(x\)-axis, we need
to place first and last point \(2\) below the midline, at \((-1/4,-3)\) and \((7/4,-3)\):
\[\img{U4_7F16.png}{}{20em}{}\]
This time the first point is not on the grid, and to find the other points we repeatedly add
one unit:
\[-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}, \hspace{5ex} \frac{1}{4}+\frac{1}{2}=\frac{3}{4},
\hspace{5ex} \frac{3}{4}+\frac{1}{2}=\frac{5}{4} \]
So the other points are
\[\left(\frac{1}{4},-1\right), \hspace{5ex} \left(\frac{3}{4},1\right),\hspace{5ex}
\left(\frac{5}{4}, -1\right).\]
We can now draw one cycle:
\[\img{U4_7F18.png}{}{20em}{}\]
We see from the graph that the range is \(R=[-3,1]\).
Repeating the same pattern, we can draw more cycles:
\[\img{U4_7F19.png}{}{20em}{}\]
\end{example}
Graphs of parent \(\sin\) and \(\cos\) functions
| \begin{array}{cc}
\img{U4_7F26.png}{}{12em}{} \hspace{3ex}& \img{U4_7F27.png}{}{12em}{}
\end{array} |
Period, Amplitude, Shifts and Unit
| If \(f(t)=A\sin(Bt+C)+k \hspace{3ex} \mbox{or} \hspace{3ex}
f(t)=A\cos(Bt+C)+k\), then |
| \(\mbox{Period}=P=\dfrac{2\pi}{B}, \hspace{3ex} \mbox{Amplitude}=|A|, \hspace{3ex} \mbox{Vertical shift}=k, \hspace{3ex}
\mbox{unit}=\dfrac{1}{4}P\) |
| To find beginnng and end of a cycle, solve |
| \(Bt+C=0, \hspace{5ex} Bt+C=2\pi.\) |
Steps to plot a transformed sine or cosine function
| \begin{itemize}
\item Find Period, Amplitude, Vertical shift, beginning and end of one cycle.
\item Draw the midline as a dashed line.
\item Find the unit.
\item Mark the \(x\)-axis with multiples of the unit.
\item Draw first and last of the five key points.
\item Add the unit to the first point to find the second point.
\item Repeat for the remaining key points.
\end{itemize} |
Problems
\problem
Let \(f(t)=5\sin(3t)-2\).
\begin{enumerate}
\item
Find amplitude and period for the function.
\item Sketch one cycle.
\item
Find the range of the function.
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
For this function, we have
\[A=5,\hspace{5ex} B=3, \hspace{5ex} C=0, \hspace{5ex} k=-2.\]
So the amplitude is \(5\), and the period is \(P=2\pi/3\).
\item
A cycle begins at \(t=0\) (by solving \(3t=0\)) and ends at \(t=2\pi/3\) (by solving \(3t=2\pi\).
The vertical shift
is \(-2\), so the midline is \(y=-2\).
The unit is
\[\mbox{unit }=\frac{1}{4}P=\frac{1}{4}\frac{2\pi}{3}=\frac{\pi}{6},\]
and the multiples are
\[\frac{\pi}{6}, \hspace{5ex} \frac{2\pi}{6}=\frac{\pi}{3},
\hspace{5ex} \frac{3\pi}{6} = \frac{\pi}{2}, \hspace{5ex}
\frac{4\pi}{6}=\frac{2\pi}{3}\ldots\]
Moving up by \(5\) (the amplitude) from the midline we get to \(-2+5=3\), and moving down we get to
\(-2-5=-7\).
So the \(y\)-coordinates of the five key points will be at \(-2\), \(3\), \(-7\), and the points are:
\[(0,-2), \hspace{5ex} \left(\frac{\pi}{6},3\right),\hspace{5ex} \left(\frac{\pi}{3},-2\right),
\hspace{5ex} \left(\frac{\pi}{2},-7\right), \hspace{5ex} \left(\frac{2\pi}{3},-2\right).\]
\[\img{U4_7F20.png}{}{12em}{}\]
\item
We see from the graph that the range is \(R=[-7,3]\).
\end{enumerate}
\end{sol}
\mproblem
Let \(f(x)=4\sin(2x)+3\).
\begin{enumerate}
\item
Find amplitude and period for the function.
\item Sketch one cycle.
\item
Find the range of the function.
\end{enumerate}
\problem
Let \(\displaystyle f(x)=3\cos\left(\pi x-\frac{\pi}{3}\right)-1\).
\begin{enumerate}
\item
Find amplitude and period for the function.
\item Sketch one cycle.
\item
Find the range of the function.
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
For this function, we have
\[A=3,\hspace{5ex} B=\pi, \hspace{5ex} C=-\frac{\pi}{3}, \hspace{5ex} k=-1.\]
So the amplitude is \(3\), and the period is \(P=2\pi/\pi=2\).
\item
The horizontal shift is found by solving \(\pi x- \pi/3=0\):
\[
\begin{array}{rcll}
\pi x -\displaystyle \frac{\pi}{3}& = & 0 &\\[2ex]
\cancel{\pi}x & = & \displaystyle \frac{\cancel{\pi}}{3} & \\[2ex]
x & = & \displaystyle \frac{1}{3} & \mbox{Horizontal shift}
\end{array}
\]
and the vertical shift
is \(-1\), so the midline is \(y=-1\).
The unit is
\[\mbox{unit }=\frac{1}{4}P=\frac{2}{4}=\frac{1}{2},\]
and the multiples are
\[\frac{1}{2}, \hspace{5ex} \frac{2}{2}=1,
\hspace{5ex} \frac{3}{2}, \hspace{5ex} \frac{4}{2}=2\ldots\]
To find the \(x\)-coordinates of the key points, we begin at the horizontal shift and keep adding
the unit:
\[\frac{1}{3}, \hspace{5ex} \frac{1}{3}+\frac{1}{2}=\frac{5}{6}, \hspace{5ex}
\frac{5}{6}+\frac{1}{2}=\frac{4}{3}, \hspace{5ex} \frac{4}{3}+\frac{1}{2}=\frac{11}{6},
\hspace{5ex}\frac{11}{6}+\frac{1}{2}=\frac{7}{3} \ldots\]
Moving up by \(3\) (the amplitude) from the midline we get to \(-1+3=2\), and moving down we get to
\(-1-3=-4\).
So the \(y\)-coordinates of the five key points will be at \(-1\), \(2\), \(-4\).
The first point is at the horizontal shift, and it's a maximum (as for the parent function).
The five key points are:
\[\left(\frac{1}{3}, 2\right) \hspace{5ex} \left(\frac{5}{6},-1\right),
\hspace{5ex} \left(\frac{4}{3},-4\right), \hspace{5ex} \left(\frac{11}{6},-1\right),
\hspace{5ex} \left(\frac{7}{3},2\right).\]
\[\img{U4_7F22.png}{}{12em}{}\]
\item
We see from the graph that the range is \(R=[-4,2]\).
\end{enumerate}
\end{sol} \mproblem
Let \(\displaystyle f(x)=-2\cos\left(3x+\frac{\pi}{4}\right)+3\).
\begin{enumerate}
\item
Find amplitude and period for the function.
\item Sketch one cycle.
\item
Find the range of the function.
\end{enumerate}