\chapter{Trigonometry}
\section{The reference angle and its use}
In Unit 4.3 we defined the trig functions for acute angles using SOH-CAH-TOA, and we used those
definitions to find the exact trig values of \(30^\circ\), \(45^\circ\), and \(60^\circ\) angles
(all in Quadrant 1).
Then in Unit 4.4 we used the \(x,y,r\) definition to define the trig functions for any angles,
including quadrantal angles.
In this unit we combine the \(x, y, r\) definition and the values for the
\(30^\circ\), \(45^\circ\), and \(60^\circ\) angles to find \textbf{exact} trig values for all
the special angles that we studied in Unit 4.2 (the multiples of \(30^\circ\), \(45^\circ\),
and \(60^\circ\) in Quadrants 2, 3 and 4).
A key ingredient we will use is the reference angle, that we now define.
\subsection{Geometric picture of reference angle}
Given an angle \(\theta\) in standard position (and not a quadrantal angle),
the \textit{reference angle} \(\theta_R\) is the
acute, positive angle that the terminal side of \(\theta\) makes with the \(x\)-axis.
This means that the reference angle depends only on the terminal side of \(\theta\). The pictures
below show the reference angle for a terminal side in each of the four quadrants.
\[\begin{array}{cc}
\img{U4_5F1.png}{}{18em}{} & \hspace{3ex}\img{U4_5F2.png}{}{18em}{}
\end{array}
\]
\[\begin{array}{cc}
\img{U4_5F3.png}{}{18em}{} & \hspace{3ex} \img{U4_5F4.png}{}{18em}{}
\end{array}
\]
Note that the reference angle is always positive, by definition, so we draw it without an
arrow.
It is also important to note that different angles \(\theta\) may have the same reference angle \(\theta_R\).
The following pictures show two different angles \(\theta_1\) and \(\theta_2\) (one positive and one
negative) that have the
same reference angle, for each of the four quadrants. Note that an acute angle (a positive angle
smaller than \(90^\circ\)) is its own reference angle (\(\theta_1\) in the first picture).
\[\begin{array}{cc}
\img{U4_5F5.png}{}{18em}{} & \hspace{3ex}\img{U4_5F6.png}{}{18em}{}\\[2ex]
\img{U4_5F7.png}{}{18em}{} & \hspace{3ex} \img{U4_5F8.png}{}{18em}{}
\end{array}
\]
From a picture, it is possible to find how the reference angle is related to a given angle, as we
show in the following example.
\begin{example}
The picture shows a positive angle of \(147^\circ\) together with its reference angle \(\theta_R\).
\[
\img{U4_5F9.png}{}{18em}{}
\]
We see from the picture that \(\theta_R\) is the angle needed to complete a straight angle of
\(180^\circ\). In other words, \(147^\circ + \theta_R = 180^\circ\), and so \(\theta_R = 180-147
=33^\circ\).
\end{example}
If the angle is negative, it is useful to work with its size, or absolute value, as in the following example.
\begin{example}
The picture shows a negative angle \(\theta=-132^\circ\) and its reference angle \(\theta_R\).
\[
\img{U4_5F10.png}{}{18em}{}
\]
The size of \(\theta \) is \(|-132^\circ|=132^\circ\).
We see from the picture that if we add \(\theta_R\) to the \textbf{size} of \(\theta\) (not to \(\theta\) itself) we get \(180^\circ\),
that is, \(132^\circ + \theta_R =180^\circ\) and so \(\theta_R =180-132=48^\circ\).
\end{example}
\begin{example}
The picture shows a positive angle \(\displaystyle \theta=\frac{11\pi}{6}\)
and its reference angle \(\theta_R\).
\[
\img{U4_5F11.png}{}{18em}{}
\]
If we add \(\theta_R\) to \(\theta\) we get a \(2\pi\) angle, that is:
\[ \frac{11\pi}{6} +\theta_R =2\pi,\] and so \[\theta_R=2\pi-\frac{11\pi}{6}=\frac{12\pi}{6} -\frac{11\pi}{6}=\frac{\pi}{6}.\]
\end{example}
\subsection{The reference angle algebraically}
We can also find the reference angle from just the degree or radians measure of the angle, without
drawing any picture.
If the absolute value (or size) \(|\theta|\) is less than \(90^\circ\), then \(\theta_R=|\theta|\).
Otherwise, working in degrees, we subtract (or add) \(180^\circ\) from the given angle
until we get a number (positive or negative) whose absolute value is less than \(90\). Then the absolute
value of that number (always positive) will be the reference angle.
If we are working in radians, we subtract (or add) \(\pi\), and proceed in the same way.
\begin{example}
Suppose \(\theta = 585^\circ\). Subtracting \(180\) we find
\[585-180= 405.\]
Subtract 180 again:
\[405=180 = 225.\]
Again:
\[225-180=45.\]
This last number is less than \(90\), so the reference angle is \(\theta_R = 45^\circ\).
\end{example}
\begin{example}
Suppose \(\displaystyle \theta = -\frac{37\pi}{18}\).
Add \(\pi\), to reduce the size of the number:
\[-\frac{37\pi}{18}+\pi = -\frac{37\pi}{18}+\frac{18\pi}{18}=-\frac{19\pi}{18}.\]
\(19\pi/18\) is larger than \(\pi/2\). So we reduce the size by adding \(\pi\) again:
\[-\frac{19\pi}{18}+\pi =-\frac{19\pi}{18} +\frac{18\pi}{18} = -\frac{\pi}{18}.\]
We found a (negative) number whose size is less than \(\pi/2\). So the absolute
value of that number is
the reference angle:
\[\theta_R = \frac{\pi}{18}.\]
\end{example}
\subsection{Exact trig values for multiples of \(30^\circ\), \(45^\circ\) and \(60^\circ\)}
We now use the reference angle to find the exact value of the trig functions for the multiples
of \(30^\circ\), \(45^\circ\) and \(60^\circ\) that are not in Quadrant 1. The crucial tool is the following
rule.
| Suppose \(f(\theta) \) is any of the six trig functions. Then |
| \(f(\theta)=\pm f(\theta_R)\) |
| where the sign is determined by the ASTC rule. |
Since \(\theta_R\) is always an acute angle, if it is one of
\(30^\circ\), \(45^\circ\) and \(60^\circ\) we can find \(f(\theta_R)\) by using either
the standard \(30\)-\(60\)-\(90\) triangle or the standard \(45\)-\(45\)-\(90\) triangle.
Then we need to determine the quadrant for \(\theta\) and apply the ASTC rule to decide if
\(f(\theta)\) is positive or negative.
\begin{example}
Suppose we want to find \(\sin(240^\circ)\). First we find the reference angle:
\[240-180=60\]
so \(\theta_R=60^\circ\). Draw the standard \(30\)-\(60\)-\(90\) triangle:
\[\img{U4_3F9.png}{}{10em}{}.\]
Then we find
\[\sin(\theta_R)=\sin(60^\circ)=\frac{\mbox{Opp}}{\mbox{Hyp}}=\frac{\sqrt{3}}{2}.\]
Now notice that \(180< 240 < 270\), and so \(240^\circ\) is in Q3. By the ASTC rule,
the sine function is negative in Q3. So we conclude that
\[\sin(240^\circ) =-\frac{\sqrt{3}}{2}.\]
\end{example}
\begin{example}
We want to find the exact value of \(\cos(-7\pi/4)\).
We find \(\theta_R\) by adding \(\pi\):
\[-\frac{7\pi}{4}+\pi=-\frac{7\pi}{4}+\frac{4\pi}{4}=-\frac{3\pi}{4},\]
\[-\frac{3\pi}{4}+\pi=-\frac{3\pi}{4}+\frac{4\pi}{4}=\frac{\pi}{4}.\]
So \(\theta_R= \pi/4=45^\circ\). Draw the standard \(45\)-\(45\)-\(90\) triangle:
\[\img{U4_3F23.png}{}{8em}{}.\]
Then we find
\[\cos(\theta_R)=\cos \left(\frac{\pi}{4}\right)=\frac{\mbox{Adj}}{\mbox{Hyp}}=\frac{1}{\sqrt{2}}.\]
To find the quandrant for \( -\frac{7\pi}{4}\), count seven \(45^\circ\) slices in clockwise
direction:
\[\img{U4_5F18.png}{}{15em}{}\]
and find that \(-7\pi/4\) is in Q1.
By the ASTC rule
all trig functions are positive there. So we find
\[\cos\left(-\frac{7\pi}{4}\right) = \frac{1}{\sqrt{2}}.\]
\end{example}
\subsection{Finding angles from a given trig value}
Suppose we know the value of a trig function, and we are given an interval where the angle must be.
Using the reference angle, we can
find all the angles in that interval that have the given trig function value.
\begin{example}
Suppose we know that \(\displaystyle \sin \theta=-\frac{1}{2}\) and \(\theta \) is in the interval \(0\leq \theta
\leq 360^\circ\). We will proceed in three steps.
\textbf{ Step 1.} Determine the quadrants.
Since the sine is
negative, the solutions must be in Q3 and Q4, by the ASTC rule.
\textbf{Step 2.} Find and draw the reference angle.
The value \(-1/2\) of the trig function reminds us of the \(30\)-\(60\)-\(90\) triangle.
\[\img{U4_3F9.png}{-3em}{10em}{} \hspace{3ex} \sin\theta_R =\frac{\mbox{Opp}}{\mbox{Hyp}}=\frac{1}{2}\Longrightarrow \theta_R= 30^\circ\]
After drawing it, we find that the reference angle is \(30^\circ\).
We draw a \(30^\circ\) reference angle in Q3 and Q4:
\[\img{U5_4F1.png}{}{12em}{}\]
\textbf{Step 3.} Find all angles in the given interval.
We draw all possible angles that are in the interval \([0,360^\circ]\)
and have terminal sides there:
\[\img{U5_4F2.png}{}{12em}{}\]
So there are two angles in the interval \([0,360^\circ]\) with \(\sin \theta =-1/2\):
\(\theta_1=210^\circ\) and \(\theta_2=330^\circ\).
\end{example}
\begin{example}
Suppose now the given trig value is the same as in the previous example:
\(\displaystyle \sin \theta=-\frac{1}{2}\), but \(\theta \) must be in the interval
\([-180^\circ,180^\circ]\).
Then the only possible values for \(\theta\) are the two negative angles \(\theta_1=-30^\circ\),
\(\theta_2=-150^\circ\), as shown below:
\[\img{U5_4F3.png}{}{12em}{}\]
\end{example}
In the next example, we use radians and a larger interval, and we find more possible values for \(\theta\).
\begin{example}
We are given that
\[\cos \theta =\frac{1}{\sqrt{2}}\]
and \(\theta\) is in the interval \([-2\pi,2\pi]\).
\textbf{1.} By the ASTC rule, the cosine is positive in Q1 and Q4.
\textbf{2.}
The reference angle is \(\pi/4\), and we draw it in Q1 and Q4:
\[\img{U5_4F4.png}{}{12em}{}\]
\textbf{3.}
There are two positive angles in \(0\leq x \leq 2\pi\) and two negative
angles in \(-2\pi\leq x \leq 0\) with those terminal sides, and so four total solutions in
the interval \([-2\pi,2\pi]\):
\[\theta_1=\frac{\pi}{4}, \hspace{2ex} \theta_2=\frac{7\pi}{4}, \hspace{2ex} \theta_3=-\frac{\pi}{4},
\hspace{2ex} \theta_4=-\frac{7\pi}{4}.\]
\[\img{U5_4F5.png}{}{12em}{}\]
\end{example}
\subsection{The unit circle}
The circle with center at \((0,0)\) and radius \(1\) is called the \textit{unit circle}. So if \((x,y)\) is a point on the
unit circle, the corresponding value of \(r\) will be \(r=1\), and using the
\(x,y,r\) definition of the sine and cosine functions, we find
\[
\begin{array}{cc}
\cos \theta = \dfrac{x}{1} = x & \sin \theta = \dfrac{y}{1}=y.
\end{array}
\]
So a point on the unit circle
corresponding to angle \(\theta\) will have coordinates \((x,y)=(\cos \theta, \sin \theta)\), as shown in the picture below.
\[
\img{UnitCircleEx.png}{}{18em}{}
\]
The next picture shows all the special and quadrantal angles together with the corresponding
values\((\cos \theta, \sin \theta)\) on the unit circle.
\[
\img{UnitCircle.png}{}{30em}{}
\]
Problems
\problem
Find the reference angle for each of the angles given in the pictures.
\[
\begin{array}{ccc}
\img{U4_5F12.png}{}{10em}{} \hspace{5ex} &
\img{U4_5F13.png}{}{10em}{} \hspace{5ex} &
\img{U4_5F14.png}{}{10em}{} \\
\mbox{a.} \hspace{5ex} & \mbox{b.} \hspace{5ex} & \mbox{c.}
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
We see from the picture that if we add \(\theta_R\) to \(180^\circ\), we get \(213^\circ\). So we need
to solve
\[\theta_R + 180=213\]
and we find
\[\theta_R=213-180=33^\circ.\]
\item
The given angle is negative, and larger (in size) than \(180^\circ = \pi\) by the size of
\(\theta_R\). So if we reduce the size \(6\pi/5\) by \(\theta_R\), we will get \(\pi\):
\[\frac{6\pi}{5} -\theta_R =\pi \]
So
\[\theta_R = \frac{6\pi}{5} - \pi = \frac{6\pi}{5} - \frac{5\pi}{5} = \frac{\pi}{5}.\]
\item
If we add \(\theta_R\) to \(4\pi/7\), we get \(\pi\):
\[\frac{4\pi}{7} + \theta_R = \pi\]
\[\theta_R= \pi - \frac{4\pi}{7}=\frac{7\pi}{7}-\frac{4\pi}{7}=\frac{3\pi}{7}.\]
\end{enumerate}
\end{sol}
\mproblem
Find the reference angle for each of the angles given in the pictures.
\[
\begin{array}{ccc}
\img{U4_5F15.png}{}{10em}{} \hspace{5ex} &
\img{U4_5F16.png}{}{10em}{} \hspace{5ex} &
\img{U4_5F17.png}{}{10em}{} \\
\mbox{a.} \hspace{5ex} & \mbox{b.} \hspace{5ex} & \mbox{c.}
\end{array}
\]
\problem
Find the reference angle for the angles:
\[
\begin{array}{ccc}
\mbox{a.}\hspace{1ex} -670^\circ \hspace{5ex} & \mbox{b.}\hspace{1ex}
\displaystyle \frac{12\pi}{5}\hspace{5ex} &
\mbox{c.}\hspace{1ex} 355^\circ
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
We add \(180\) till we get the size of the number less than \(90\):
\[-670+180=-490\]
\[-490+180=-310\]
\[-310+180=-130\]
\[-130+180=50.\]
So \(\theta_R = 50^\circ\).
\item
Reduce the size by subtracting \(\pi\):
\[\frac{12\pi}{5}-\pi=\frac{12\pi}{5} -\frac{5\pi}{5}=\frac{7\pi}{5}.\]
\[\frac{7\pi}{5} -\pi =\frac{7\pi}{5}-\frac{5\pi}{5} =\frac{2\pi}{5}.\]
So \[\theta_R=\frac{2\pi}{5}.\]
\item
\(355-180=175\), and
\(175-180=-5\).
So \(\theta_R = 5^\circ\).
\end{enumerate}
\end{sol}
\mproblem
Find the reference angle for the angles:
\[
\begin{array}{ccc}
\mbox{a.}\hspace{1ex} \displaystyle -\frac{14\pi}{3} \hspace{5ex} & \mbox{b.}\hspace{1ex}
\displaystyle 285^\circ\hspace{5ex} &
\mbox{c.}\hspace{1ex} -440^\circ
\end{array}
\]
\problem
Find the exact value of all six trig functions for \(\theta=-4\pi/3\).
\begin{sol}
Find the reference angle by adding \(\pi\):
\[-\frac{4\pi}{3}+\pi=-\frac{4\pi}{3}+\frac{3\pi}{3}=-\frac{\pi}{3}.\]
So \(\theta_R = \pi/3=60^\circ\), and the trig functions for \(\theta_R\)
are found with the standard
standard \(30\)-\(60\)-\(90\) triangle:
\[\img{U4_3F9.png}{}{10em}{}\]
\[\begin{array}{ccc}
\displaystyle \sin 60^\circ = \frac{\mbox{Opp}}{\mbox{Hyp}} = \frac{\sqrt{3}}{2} &
\displaystyle \cos 60^\circ = \frac{\mbox{Adj}}{\mbox{Hyp}} =
\frac{1}{2} &
\displaystyle \tan 60^\circ = \frac{\mbox{Opp}}{\mbox{Adj}} = \frac{\sqrt{3}}{1}=\sqrt{3}\\[3ex]
\displaystyle \csc 60^\circ = \frac{\mbox{Hyp}}{\mbox{Opp}} =
\frac{2}{\sqrt{3}} &
\displaystyle \sec 60^\circ = \frac{\mbox{Hyp}}{\mbox{Adj}} = \frac{2}{1}=2 &
\displaystyle \cot 60^\circ = \frac{\mbox{Adj}}{\mbox{Opp}} =
\frac{1}{\sqrt{3}}
\end{array}
\]
To find the quadrant for \(\theta\), count four \(60^\circ\) slices clockwise:
\[\img{U4_5F19.png}{}{15em}{}.\]
So \(\theta\) is in Quadrant 2. By the ASTC rule, \(\sin\) and \(\csc\) are positive, and
all others negative.
So we can now find all trig functions for \(\theta\):
\[\begin{array}{ccc}
\displaystyle \sin\left(-\frac{4\pi}{3}\right) = \frac{\sqrt{3}}{2} &
\displaystyle \cos\left(-\frac{4\pi}{3}\right) =
-\frac{1}{2} &
\displaystyle \tan\left(-\frac{4\pi}{3}\right) = -\sqrt{3}\\[2ex]
\displaystyle \csc\left(-\frac{4\pi}{3}\right) =
\frac{2}{\sqrt{3}} &
\displaystyle \sec\left(-\frac{4\pi}{3}\right) = -2 &
\displaystyle \cot\left(-\frac{4\pi}{3}\right) =
-\frac{1}{\sqrt{3}}
\end{array}
\]
\end{sol}
\mproblem
Find the exact value of all six trig functions for \(\theta=-5\pi/6\).
\problem
Find the exact value of all six trig functions for \(\theta=225^\circ\).
\begin{sol}
Find the reference angle by subtracting \(180\):
\[225-180=45.\]
So \(\theta_R=45^\circ\). Draw the standard \(45\)-\(45\)-\(90\) triangle and find the trig
functions for \(\theta_R\):
\[\img{U4_3F23.png}{}{8em}{},\]
\[\begin{array}{ccc}
\displaystyle \sin 45^\circ = \frac{\mbox{Opp}}{\mbox{Hyp}} = \frac{1}{\sqrt{2}} &
\displaystyle \cos 45^\circ = \frac{\mbox{Adj}}{\mbox{Hyp}} =
\frac{1}{\sqrt{2}} &
\displaystyle \tan 45^\circ = \frac{\mbox{Opp}}{\mbox{Adj}} = \frac{1}{1}=1\\[2ex]
\displaystyle \csc 45^\circ = \frac{\mbox{Hyp}}{\mbox{Opp}} =
\frac{\sqrt{2}}{1}=\sqrt{2} &
\displaystyle \sec 45^\circ = \frac{\mbox{Hyp}}{\mbox{Adj}} = \frac{\sqrt{2}}{1}=\sqrt{2} &
\displaystyle \cot 45^\circ = \frac{\mbox{Adj}}{\mbox{Opp}} =
\frac{1}{1}=1
\end{array}
\]
Since \(225=180+45\), \(\theta\) is in Quadrant 3:
\[\img{U4_5F20.png}{}{18em}{}.\]
By the ASTC rule, \(\tan\) and \(\cot\) are positive, and all others negative.
\[\begin{array}{ccc}
\displaystyle \sin 225^\circ = -\frac{1}{\sqrt{2}} &
\displaystyle \cos 225^\circ =
-\frac{1}{\sqrt{2}} &
\displaystyle \tan 225^\circ = 1\\[2ex]
\displaystyle \csc 225^\circ =
-\sqrt{2} &
\displaystyle \sec 225^\circ = -\sqrt{2} &
\displaystyle \cot 225^\circ = 1
\end{array}
\]
\end{sol}
\mproblem
Find the exact value of all six trig functions for \(\theta=135^\circ\).
\problem
Suppose that \(\displaystyle \sin \theta =\frac{\sqrt{3}}{2}\) and \(\theta\) is in the interval \([0,2\pi]\). Find all the
possible values for \(\theta\)
\begin{sol}
\textbf{1.} The sine is positive in Q1 and Q2.
\textbf{2.}
The reference angle is \(\pi/3\).
Draw it in Q1 and Q2:
\[\img{U5_4F9.png}{}{15em}{}\]
\textbf{3.}
There are two solutions in the interval \([0,2\pi]\): \(\theta_1= \pi/3\) and \(\theta_2=2\pi/3\).
\[\img{U5_4F10.png}{}{15em}{}\]
\end{sol}
\mproblem
Suppose that \(\displaystyle \sqrt{3}\tan \theta=-\frac{1}{\sqrt{3}}\) and \(\theta \) is in the interval \([0,2\pi]\).
Find all the possible values of \(\theta\).