\chapter{Trigonometry}
\section{Special angles and arc length}
\subsection{Quadrantal angles}
In the previous section we defined the quadrantal angles to be the angles in standard position
whose terminal side is on the \(x\) axis or the \(y\) axis. If we start at \(\theta=0^\circ\) and
we complete a full positive rotation (counter-clockwise),
we find the five angles shown in the first picture
below, while going in the negative direction (clockwise)
we find the angles in the second picture (note that \(0^\circ\) and \(360^\circ\) are co-terminal):
\[
\begin{array}{cc}
\img{U4_2F1.png}{}{20em}{}& \img{U4_2F2.png}{}{20em}{}\\
\mbox{Positive quadrantal angles} &
\mbox{Negative quadrantal angles}
\end{array}
\]
Another way to describe the quadrantal angles is to say that they are
multiples of \(90^\circ=\pi/2\).
\subsection{Multiples of \(30^\circ\)}
There are three \(30^\circ=\pi/6\) angles in each quadrant, and so twelve in a full revolution.
Think of a pizza divided into \(12\) equal slices.
\begin{array}{cc}
\img{U4_2F3.png}{}{20em}{} & \img{U4_2F4.png}{}{20em}{}\\
\mbox{Positive multiples of \(30^\circ\)} &
\mbox{Negative multiples of \(30^\circ\)}
\end{array}
To find the radian measure of such an angle, instead of using the conversion formula,
we can count the number of slices and multiply by \(\pi/6\).
\begin{example}
To draw the angle \(150^\circ\) in standard position, we notice that \(150=180-30\), and so
we need to subtract one \(30^\circ\) slice from half a revolution:
\[
\img{U4_2F5.png}{}{20em}{}
\]
The angle is in Quadrant 2, and we draw it indicating the direction of rotation with an arrow:
\[
\img{U4_2F5a.png}{}{20em}{}
\]
Since we have \(5\) slices in this picture, the radian measure is
\(\displaystyle 5\cdot\frac{\pi}{6}=\frac{5\pi}{6}\).
\end{example}
\subsection{Multiples of \(45^\circ\)}
There are two \(45^\circ=\pi/4\) angles in each quadrant, and so eight in a full revolution.
Think of a pizza divided into \(8\) equal slices.
\[
\begin{array}{cc}
\img{U4_2F6.png}{}{20em}{}& \img{U4_2F7.png}{}{20em}{}\\
\mbox{Positive multiples of \(45^\circ\)} &
\mbox{Negative multiples of \(45^\circ\)}
\end{array}
\]
\begin{example}
To draw the angle \(-315^\circ\) in standard position, we notice that \(315=270+45\), and so
we need to add one \(45^\circ\) slice to a \(270^\circ\) angle, but going in the negative (clockwise)
direction. Or, we could also subtract a \(45^\circ\) slice from \(360^\circ\), because
\(315=360-45\). Either way, we get the seven \(45^\circ\) slices in the picture (shown in red because we are moving in the
negative direction):
\[\img{U4_2F8.png}{}{20em}{}
\]
The corresponding angle in Quadrant 1 is shown below.
\[
\img{U4_2F8a.png}{}{20em}{}
\]
Since we have \(7\) slices in this picture, each of size \(\pi/4\), the radian measure is
\(\displaystyle -7\cdot\frac{\pi}{4}=-\frac{7\pi}{4}\).
\end{example}
\subsection{Multiples of \(60^\circ\)}
There are three \(60^\circ=\pi/3\) angles in a straight angle, and so six in a full revolution.
Think of a pizza divided into \(6\) equal slices. This time the \(y\) axis does not contain a side
for the slices, and we draw it dashed in the picture:
\begin{array}{cc}
\img{U4_2F9.png}{}{20em}{} & \img{U4_2F10.png}{}{20em}{}\\
\mbox{Positive multiples of \(60^\circ\)} &
\mbox{Negative multiples of \(60^\circ\) }
\end{array}
\begin{example}
To draw the angle \(240^\circ\) in standard position, we notice that \(240=180+60\), and so
we need to add one \(60^\circ\) slice to a \(180^\circ\) angle, that will be in Quadrant 3:
\[
\img{U4_2F11.png}{}{12em}{}
\hspace{5ex} \img{U4_2F11a.png}{}{15em}{}
\]
Since we have \(4\) slices in this picture, each of size \(\pi/3\), the radian measure is
\(\displaystyle 4\cdot\frac{\pi}{3}=\frac{4\pi}{3}\).
\end{example}
The quadrantal angles and the multiples of \(30^\circ\), \(45^\circ\), \(60^\circ\) discussed
in this unit are called \textit{special angles}.
The next picture summarizes all the special angles in the interval \([0,2\pi]\).
\[
\img{U4_2F19.png}{}{30em}{}
\]
\subsection{Arc length}
An \textit{arc} is a portion of a circle. We will now discuss how to find the length of an arc, that we will denote by \(s\).
Every arc of a circle can be identified by the angle
determined by its endpoints, as in the following picture, where the arc (drawn in red) is determined
by the angle \(\theta\):
\[
\img{U4_2F15.png}{}{20em}{}
\]
Recall the formula for the circumference of a circle:
\[C=2\pi r.\]
One way to think about this formula is that \(2\pi\) is the radian measure of a full revolution
(or \(360^\circ\)) that the terminal side makes to describe a circle.
If instead of the length of the whole circle we want only a part of it, we replace
\(C\) with \(s\) and \(2\pi\) with the
radian measure of the angle \(\theta\) determined by the arc:
| Arc length formula |
|---|
| \(s=\theta r \hspace{2ex}\) |
| where \(\theta\) is in radians |
So given any two of \(s\),\(\theta\) and \(r\), we can find the third quantity using \(s=\theta r\).
But it is important to remember that if an angle is given in degrees, it must be converted to radians
before using it in the formula.
\begin{example} Suppose a circle has radius $r = 9$ in and a circular arc
is determined by angle $\theta = 60^{\circ}$. To find the length \(s\) of the circular arc,
we first need to convert
\(\theta\) to radians:
\[ 60^{\circ} \cdot \frac{\pi}{180^{\circ}} = \frac{\pi}{3}. \]
Now we can plug $r$ and $\theta$ into the arc length formula:
\[ s = r\theta = 9\left(\frac{\pi}{3}\right) = 3\pi \approx \fbox{$9.425\mbox{ in. }$} \]
\end{example}
Problems
\problem
Let \(\theta=135^\circ\).
Draw \(\theta\) in standard position.
\begin{sol}
Since \(135=90+45\), we add \(45^\circ\) to a right angle:
\[\img{U4_1F18.png}{}{20em}{}\]
\end{sol}
\mproblem
Let \(\theta=210^\circ\).
Draw \(\theta\) in standard position.
\problem
Let \(\displaystyle \theta=\frac{5\pi}{3}\).
\begin{enumerate}
\item
Find a different, positive angle \(\theta_1\) co-terminal with \(\theta\), in radians.
\item
Find a negative angle \(\theta_2\) co-terminal with \(\theta\), in radians.
\item
Convert \(\theta\), \(\theta_1\) and \(\theta_2\) to degrees.
\item
Draw \(\theta\), \(\theta_1\) and \(\theta_2\) in standard position.
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\[\theta_1=\frac{5\pi}{3}+2\pi = \frac{5\pi}{3} +\frac{6\pi}{3}=\frac{11\pi}{3}.\]
\item
\[\theta_2=\frac{5\pi}{3}-2\pi=\frac{5\pi}{3}-\frac{6\pi}{3}=-\frac{\pi}{3}.\]
\item
\[\theta=\frac{5\pi}{3}\cdot \frac{180}{\pi}=300^\circ.\]
Instead of using the conversion formula for \(\theta_1\) and \(\theta_2\), it is easier to add and
subtract \(360^\circ\):
\[\theta_1=300+360=660^\circ,\]
\[\theta_2=300-360=-60^\circ.\]
\item
\[\img{U4_1F19.png}{}{20em}{}\]
\end{enumerate}
\end{sol}
\mproblem
Let \(\displaystyle \theta=\frac{7\pi}{4}\).
\begin{enumerate}
\item
Find a different, positive angle \(\theta_1\) co-terminal with \(\theta\), in radians.
\item
Find a negative angle \(\theta_2\) co-terminal with \(\theta\), in radians.
\item
Convert \(\theta\), \(\theta_1\) and \(\theta_2\) to degrees.
\item
Draw \(\theta\), \(\theta_1\) and \(\theta_2\) in standard position.
\end{enumerate}
\problem
Draw the angles in standard position:
\begin{enumerate}
\item
\(\theta_1=225^\circ\)
\item
\(\displaystyle \theta_2=-\frac{4\pi}{3}\)
\item
\(\displaystyle \theta_3=-\frac{3\pi}{2}\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\(225=180+45\), so we need to add a \(45^\circ\) slice to a straight angle:
\[
\img{U4_2F12.png}{}{15em}{}
\]
\item
\(\displaystyle \frac{4\pi}{3}=\pi +\frac{\pi}{3}\), so we need to add a
\(\displaystyle \frac{\pi}{3}\) angle (or \(60^\circ\)) to a straight angle, but going
in the clockwise direction, because the given angle is negative:
\[
\img{U4_2F13.png}{}{15em}{}
\]
\item
\(\displaystyle \frac{3\pi}{2}= 3\cdot \frac{\pi}{2}\), so it's three right angles, in the
clockwise direction:
\[
\img{U4_2F14.png}{}{15em}{}
\]
\end{enumerate}
\end{sol}
\mproblem
Draw the angles in standard position:
\begin{enumerate}
\item
\(\displaystyle \theta_1=\frac{3\pi}{4}\)
\item
\(\theta_2=300^\circ\)
\item
\(\displaystyle \theta_3=-\pi\)
\end{enumerate}
\problem \
\begin{enumerate}
\item
Find the length of the arc determined by an angle of \(200^\circ\) in a circle of radius \(4 \) in.
Find your answer as an exact value, without decimal approximations, then use your calculator
to find an approximation with two decimal places.
\item
Find the radius of the circle for which an arc of length \(12\) cm
determines an angle of \(40^\circ\). Approximate your answer to two decimal places.
\item Find the degree measure of the angle determined by an arc of length \(35\) cm in a circle of radius \(29 \) cm, approximated to two decimal places.
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
First convert \(200^\circ\) to radians:
\[200\cdot \frac{\pi}{180}=\frac{10\pi}{9}.\]
Then use \(s=\theta r\):
\[\begin{array}{rcll}
s&=& \displaystyle \frac{10\pi}{9} \cdot 4 \\[2ex]
&=& \displaystyle \frac{40\pi}{9} \mbox{ in} & \mbox{(exact value)}\\[2ex]
& = & 13.96 \mbox{ in} & \mbox{(decimal approximation)}
\end{array}. \]
\item
\(\displaystyle 40^\circ = 40\cdot \frac{\pi}{180}=\frac{2\pi}{9}\), so using \(s=\theta r\) with \(s=12\) we find
\[
\begin{array}{rcl}
12& = & \displaystyle \frac{2\pi}{9} r\\[2ex]
r & = & \displaystyle 12\cdot \frac{9}{2\pi}=\frac{54}{\pi}\approx 17.19 \mbox{in}
\end{array}\]
\item
Using \(s = \theta r\), we find
\[\theta = \frac{s}{r} = \frac{35}{29}.\]
But we must remember that the formula used the radian measure of \(\theta\).
So we need to convert this answer to degrees:
\[\frac{35}{29}\cdot \frac{180}{\pi}\approx 69.15^\circ.\]
\end{enumerate}
\end{sol}
\mproblem \
\begin{enumerate}
\item
Find the length of the arc determined by an angle of \(175^\circ\) in a circle of radius \(13 \) in.
Approximate your answer to two decimal places.
\item
Find the radius of the circle for which an arc of length \(7\) cm
determines an angle of \(170^\circ\). Find your answer as an exact value,
without decimal approximations, then use your calculator
to find an approximation with two decimal places.
\item Find the degree measure of the angle determined by an arc of length \(27\) cm
in a circle of radius \(43 \) cm, approximated to two decimal places.
\end{enumerate}