\chapter{Trigonometry}
\section{The trig functions for general angles}
The definitions of trig functions in the previous unit are only valid for acute angles
(angles smaller than 90 degrees), because
we used the angles of a right triangle to define them.
We now define trig functions for \textbf{all} angles. So the angle could be negative, obtuse, or
larger than \(180^\circ\).
\subsection{Definition using \(x,y,r\)}
First we draw an angle in standard position as in the picture:
\[\img{U4_4F1.png}{}{15em}{}\]
We choose any point \((x,y)\) on the terminal side
of the angle, and we call \(r\) the distance of this point from the origin.
So we have three numbers \(x,y,r\), that are related by the Pythagorean theorem:
\[x^2+y^2=r^2.\]
In fact, for the angle shown in the picture (in Quadrant 1), the three numbers are just
\(x=\mbox{adj}\), \(y=\mbox{opp}\) and \(r=\mbox{hyp}\) for the triangle with sides
\(x,y,r\). So we now
define the six trig functions of
angle \(\theta\) as follows:
\[
\begin{array}{ccc}
\displaystyle \sin\theta =\frac{y}{r} \hspace{5ex} &\displaystyle \cos\theta =\frac{x}{r} \hspace{5ex} & \displaystyle \tan\theta =\frac{y}{x}\\[2ex]
\displaystyle \csc\theta =\frac{r}{y} \hspace{5ex} &\displaystyle \sec\theta =\frac{r}{x} \hspace{5ex} &\displaystyle \cot\theta =\frac{x}{y}
\end{array}
\]
Of course if the angle is in Quadrant 1, this definition is the same as the previous one
using SOH-CAH-TOA. But our new definition can now be used in any quadrant.
The number \(r\) is always positive (being a distance). But the numbers \(x\) and \(y\)
could be negative or zero.
This means that the trig functions could be negative, depending on the sign of \(x\) and \(y\).
They could also be zero, and four of them
(\(\csc,\sec,\tan,\cot\)) could be undefined, because they have either \(x\) or \(y\) in the
denominator. The Pythagorean relationship \(x^2+y^2=r^2\) will remain valid even if \(x\) or \(y\) are negative or zero.
\begin{example}
We want to find all trig functions for the angle shown in the picture.
\[\img{U4_4F3.png}{}{15em}{}\]
The angle is in Quadrant 2, and the coordinates of the given point are \((x,y)=(-7,24)\).
First we find \(r\):
\[
\begin{array}{rcl}
(-7)^2+24^2&=& r^2\\[1ex]
49+ 576 & = & r^2\\[1ex]
625 &=& r^2 \\[1ex]
r&=& 25
\end{array}
\]
Now we use the definitions to find the six trig functions:
\[\begin{array}{ccc}
\displaystyle \sin\theta = \frac{y}{r} = \frac{24}{25} &
\displaystyle \cos\theta = \frac{x}{r} =
-\frac{7}{25} &
\displaystyle \tan\theta = \frac{y}{x} = -\frac{24}{7}\\[2ex]
\displaystyle \csc\theta = \frac{r}{y} =
\frac{25}{24} &
\displaystyle \sec\theta = \frac{r}{x} = -\frac{25}{7} &
\displaystyle \cot\theta = \frac{x}{y} =
-\frac{7}{24}
\end{array}
\]
\end{example}
Note that two of the trig functions are positive, and four are negative. This will always be the case when the angle is in Quadrants 2, 3, or 4.
\subsection{The ASTC rule}
In some problems, we need to determine the sign of each trig functions
in the various quadrants ahead of time.
To find the right sign, use the ASTC rule:
\[\img{U4_4F2.png}{}{10em}{}\]
\[
\begin{array}{ll}
\mbox{In Quadrant 1,}\hspace{-1ex}& \mathbf{A}\mbox{ll functions are positive.}\\
\mbox{In Quadrant 2, the}\hspace{-1ex}& \mathbf{S}\mbox{ine function is positive.}\\
\mbox{In Quadrant 3, the}\hspace{-1ex}& \mathbf{T}\mbox{angent function is positive.}\\
\mbox{In Quadrant 4, the}\hspace{-1ex}& \mathbf{C}\mbox{osine function is positive.}
\end{array}
\]
We also need to remember that \(\csc\), \(\sec\) and \(\cot\) have the same sign as
\(\sin\), \(\cos\) and \(\tan\), because they are the reciprocals.
To remember the four letters ASTC in the right order, use the sentence
\begin{center}
\textbf{A}ll \textbf{S}tudents \textbf{T}ake \textbf{C}alculus
\end{center}
A common way to use the ASTC rule is to determine the quadrant of an angle from the sign of two trig
functions.
\begin{example}
Suppose we know that \(\cos \theta > 0\) and \(\tan \theta < 0\). Each inequality gives us two possible quadrants:
\[\cos \theta > 0 \Rightarrow \mbox{Q1 or Q4}\]
because in Q1 (\textbf{A}STC) \textbf{A}ll are positive, and in Q4
(AST\textbf{C})
\textbf{C}os is positive.
\[\tan \theta < 0 \Rightarrow \mbox{Q2 or Q4}\]
because the tangent function is positive in Q1 (\textbf{A}STC) and in
Q3 (AS\textbf{T}C), and so it is negative in Q2 and Q4.
Since both conditions must be satisfied, we see that only Q4 is possible. So \(\theta\) is
in Quadrant 4.
\end{example}
Of the three numbers \(x\), \(y\) and \(r\), we know that \(r\) is always positive.
Knowing the value of a trig function means knowing one of the six ratios we can make with
\(x,y,r\). This may not be enough to know the sign of \(x\) or \(y\).
But if we also know the sign of another trig function (other than the reciprocal of the given
one, that of course will have the same sign), then using the ASTC rule
we can determine the sign of both
\(x\) and \(y\). Then the Pythagorean theorem \(x^2+y^2=r^2\) can be used to find the third number,
and determine all six trig functions.
\begin{example}
Suppose we know that \(\tan \theta =3/4\) and we would like to find all other trig functions.
Because \(\tan \theta =y/x\), we know that
\[\frac{y}{x} =\frac{3}{4}.\]
But we cannot conclude that \(y=3\) and \(x=4\). First of all, there could be a common factor
that gets simplified. So \(y=6\) and \(x=8\) would give the same ratio:
\[ \frac{y}{x}=\frac{6}{8}=\frac{3}{4}.\]
This common factor would not affect the value of the other trig functions, because it would
appear and be simplified in all other trig ratios. But there is another more important missing
piece of information: we cannot conclude that \(x\) and \(y\) are both positive, because they
could also be both negative: if \(y=-3\) and \(x=-4\), then
\[\frac{y}{x}=\frac{-3}{-4} = \frac{3}{4}.\]
If \(x\) and \(y\) are both positive, then the angle \(\theta\) will be in Q1. But if they are
both negative it will be in Q3 (use the ASTC rule).
So we need another piece of information to decide where the angle is. It is enough to know just the sign
of another trig function. If for example we know that \(\sin \theta\)
is negative, then we know that the angle cannot be in Q1, and it must be in Q3.
Then \(y=-3\), \(x=-4\), and we find \(r\) from the Pythagorean theorem:
\[
\begin{array}{rcl}
(-3)^2 + (-4)^2& =& r^2\\
9+16 &=& r^2\\
25 &=& r^2\\
r&=& 5
\end{array}
\]
Once we know \(x,y\) and \(r\), we can find all trig ratios:
\[\begin{array}{cc}
\displaystyle \sin\theta = \frac{y}{r} = \frac{-3}{5}=-\frac{3}{5} &
\displaystyle \cos\theta = \frac{x}{r} =
\frac{-4}{5} = -\frac{4}{5}\\[2ex]
\displaystyle \tan\theta = \frac{y}{x} = \frac{-3}{-4}=\frac{3}{4} &
\displaystyle \csc\theta = \frac{r}{y} =
\frac{5}{-3}=-\frac{5}{3}\\[2ex]
\displaystyle \sec\theta = \frac{r}{x} = \frac{5}{-4} =-\frac{5}{4}&
\displaystyle \cot\theta = \frac{x}{y} =
\frac{-4}{-3}=\frac{4}{3}
\end{array}.
\]
\end{example}
\subsection{The trig values of quadrantal angles}
Recall that quadrantal angles are those whose terminal side (when drawn in standard position)
is either on the \(x\)-axis or the \(y\)-axis.
If \((x,y)\) is on the \(x\)-axis, then \(y=0\), and if \((x,y)\) is on the \(y\)-axis, then \(x=0\).
We will see that for any quadrantal angle, two trig functions will be zero,
two will be undefined, and two will be either \(1\) or \(-1\).
\begin{itemize}
\item Positive \(x\)-axis
Quadrantal angle \(\theta=0\) has the terminal side on the positive
\(x\)-axis.
Choose the point \((1,0)\).
\[\img{U4_4F4.png}{}{20em}{}\]
Then \(x=1\), \(y=0\), \(r=1\), and the trig functions are:
\[\begin{array}{ccc}
\displaystyle \sin 0 = \frac{0}{1}=0 &
\displaystyle \cos 0 =\frac{1}{1} =1 &
\displaystyle \tan 0 = \frac{0}{1} = 0\\[2ex]
\displaystyle \csc 0 =\frac{1}{0} \mbox{ DNE} &
\displaystyle \sec 0 = \frac{1}{1} = 1 &
\displaystyle \cot 0 =\frac{1}{0} \mbox{ DNE}
\end{array}
\]
\item Positive \(y\)-axis
Quadrantal angle \(\displaystyle \theta = 90^\circ = \frac{\pi}{2}\) has its terminal side
on the positive \(y\)-axis. Choose the point \((0,1)\).
\[\img{U4_4F5.png}{}{20em}{}\]
Then \(x=0\), \(y=1\), \(r=1\), and the trig functions are:
\[\begin{array}{cc}
\displaystyle \sin 90^\circ =\sin \left(\frac{\pi}{2}\right) = \frac{1}{1}=1 &
\displaystyle \cos 90^\circ =\cos\left(\frac{\pi}{2}\right)=\frac{0}{1} =0\\[2ex]
\displaystyle \tan 90^\circ =\tan\left(\frac{\pi}{2}\right) = \frac{1}{0} \mbox{ DNE} &
\displaystyle \csc 90^\circ = \csc \left(\frac{\pi}{2}\right) =\frac{1}{1} =1\\[2ex]
\displaystyle \sec 90^\circ = \sec \left(\frac{\pi}{2}\right) = \frac{1}{0} \mbox{ DNE} &
\displaystyle \cot 90^\circ =\cot \left(\frac{\pi}{2}\right) =\frac{0}{1} =0
\end{array}
\]
\item Negative \(x\)-axis
Quadrantal angle \(\displaystyle \theta = 180^\circ = \pi\) has its terminal side
on the negative \(x\)-axis. Choose the point \((-1,0)\).
\[\img{U4_4F6.png}{}{20em}{}\]
Then \(x=-1\), \(y=0\), \(r=1\), and the trig functions are:
\[\begin{array}{cc}
\displaystyle \sin 180^\circ =\sin \pi = \frac{0}{1}=0 &
\displaystyle \cos 180^\circ =\cos \pi=\frac{-1}{1} =-1 \\[2ex]
\displaystyle \tan 180^\circ =\tan \pi = \frac{0}{-1} =0 &
\displaystyle \csc 180^\circ = \csc \pi =\frac{1}{0} \mbox{ DNE}\\[2ex]
\displaystyle \sec 180^\circ = \sec \pi = \frac{1}{-1} =-1 &
\displaystyle \cot 180^\circ =\cot \pi =\frac{-1}{0} \mbox{ DNE}
\end{array}
\]
\item Negative \(y\)-axis
Quadrantal angle \(\displaystyle \theta = 270^\circ = \frac{3\pi}{2}\) has its terminal side
on the negative \(y\)-axis. Choose the point \((0,-1)\).
\[\img{U4_4F7.png}{}{20em}{}\]
Then \(x=0\), \(y=-1\), \(r=1\), and the trig functions are:
\[\begin{array}{cc}
\displaystyle \sin 270^\circ =\sin \left(\frac{3\pi}{2}\right) = \frac{-1}{1}=-1 &
\displaystyle \cos 180^\circ =\cos \left(\frac{3\pi}{2}\right)=\frac{0}{-1} =0\\[2ex]
\displaystyle \tan 180^\circ =\tan \left(\frac{3\pi}{2}\right) = \frac{-1}{0} \mbox{ DNE} &
\displaystyle \csc 180^\circ = \csc \left(\frac{3\pi}{2}\right) =\frac{1}{-1} =-1\\[2ex]
\displaystyle \sec 180^\circ = \sec \left(\frac{3\pi}{2}\right) = \frac{-1}{0} \mbox{ DNE} &
\displaystyle \cot 180^\circ =\cot \left(\frac{3\pi}{2}\right) =\frac{0}{-1} =0
\end{array}
\]
\end{itemize}
Of course there are many more quadrantal angles.
But they will all be co-terminal with one of the four angles discussed above.
A good strategy is to always draw a picture to find the values for \(x,y\) and \(r\).
Problems
\problem
Find the value of all trig functions for the angle shown in the picture.
\[\img{U4_4F8.png}{}{15em}{}\]
\begin{sol}
From the picture, we see that \(x=-12\), \(r=13\). Using the Pythagorean theorem,
\[
\begin{array}{rcl}
x^2 + y^2& =& r^2\\
(-12)^2+y^2 &=& 13^2\\
144+y^2 &=& 169\\
y^2&=& 25\\
y&=& \pm \sqrt{25}\\
y &=& \pm 5
\end{array}.
\]
Because \(\theta\) is in Q3, \(y\) must be negative. So \(y=-5\).
Now we can find all six trig functions:
\[\begin{array}{ccc}
\displaystyle \sin\theta = \frac{y}{r} = \frac{-5}{13}=-\frac{5}{13} &
\displaystyle \cos\theta = \frac{-12}{13} =
-\frac{12}{13} &
\displaystyle \tan\theta = \frac{y}{x} = \frac{-5}{-12}=\frac{5}{12}\\[2ex]
\displaystyle \csc\theta = \frac{r}{y} =
\frac{13}{-5}=-\frac{13}{5} &
\displaystyle \sec\theta = \frac{r}{x} = \frac{13}{-12}=-\frac{13}{12} &
\displaystyle \cot\theta = \frac{-12}{-5} =
\frac{12}{5}
\end{array}
\]
\end{sol}
\mproblem
Find the value of all trig functions for the angle shown in the picture.
\[\img{U4_4F9.png}{}{15em}{}\]
\problem
Suppose that \(\tan \theta < 0\), and \(\sec \theta > 0\). Find the quadrant for \(\theta\).
\begin{sol}
\[\tan \theta < 0 \Rightarrow \mbox{Q2 or Q4}\]
\[\sec \theta > 0 \Rightarrow \mbox{Q1 or Q4}\]
So \(\theta\) is in Q4.
\end{sol}
\mproblem
Suppose that \(\sin \theta > 0\), and \(\cot \theta < 0\). Find the quadrant for \(\theta\).
\problem
Find the six trig functions for \(\theta = -270^\circ\).
\begin{sol}
Drawing a picture, we find that \(\theta\) is co-terminal with \(90^\circ\):
\[\img{U4_4F10.png}{}{15em}{}\]
So the trig values are:
\[\begin{array}{cc}
\displaystyle \sin(-270^\circ) =\sin \left(\frac{\pi}{2}\right) = \frac{1}{1}=1 &
\displaystyle \cos (-270^\circ) =\cos\left(\frac{\pi}{2}\right)=\frac{0}{1} =0 \\[2ex]
\displaystyle \tan (-270^\circ) =\tan\left(\frac{\pi}{2}\right) = \frac{1}{0} \mbox{ DNE}&
\displaystyle \csc (-270^\circ) = \csc \left(\frac{\pi}{2}\right) =\frac{1}{1} =1\\[2ex]
\displaystyle \sec (-270^\circ) = \sec \left(\frac{\pi}{2}\right) = \frac{1}{0} \mbox{ DNE} &
\displaystyle \cot (-270^\circ) =\cot \left(\frac{\pi}{2}\right) =\frac{0}{1} =0
\end{array}
\]
\end{sol}
\mproblem
Find the six trig functions for \(\theta =3\pi\).
\problem
Suppose that \(\displaystyle \cos\theta =-\frac{8}{17}\) and \(\tan \theta < 0\).
Find the values of all trig functions for angle \(\theta\).
\begin{sol}
Since \(\displaystyle \cos \theta =\frac{x}{r}\), we know that
\[\frac{x}{r}=-\frac{8}{17},\]
and since \(r\) must be positive, we must have
\[\frac{x}{r}=-\frac{8}{17} = \frac{-8}{17}.\]
that is, \(x=-8\), \(r=17\).
Using the Pythagorean theorem,
\[
\begin{array}{rcl}
x^2 + y^2& =& r^2\\
(-8)^2+y^2 &=& 17^2\\
64+y^2 &=& 289\\
y^2&=& 225\\
y&=& \pm \sqrt{225}\\
y &=& \pm 15
\end{array}.
\]
We now need to decide if \(y=15\) or \(y=-15\). The given value of \(\cos \theta\) is negative,
so we know from the ASTC rule that \(\theta\) must be in Q2 or in Q3. But we are also given
that \(\tan \theta \) is negative, and so \(\theta\) must be in Q2. This means that \(y\) is
positive, and so \(y=15\). Now we find all the trig values:
\[\begin{array}{ccc}
\displaystyle \sin\theta = \frac{y}{r} = \frac{15}{17} &
\displaystyle \cos\theta = \frac{x}{r} = -\frac{8}{15} &
\displaystyle \tan\theta = \frac{y}{x} = \frac{15}{-8}=-\frac{15}{8}\\[2ex]
\displaystyle \csc\theta = \frac{r}{y} =
\frac{17}{15}&
\displaystyle \sec\theta = \frac{r}{x} = -\frac{15}{8}&
\displaystyle \cot\theta = \frac{x}{y} =
\frac{-8}{15}=-\frac{8}{15}
\end{array}.
\]
\end{sol}
\mproblem
Suppose that \(\displaystyle \sin\theta =-\frac{7}{10}\) and \(\cos \theta > 0\).
Find the values of all trig functions for angle \(\theta\).