\chapter{Trigonometry}
\section{Right triangle trigonometry}
At the heart of trigonometry lies the Pythagorean Theorem for right triangles. Instead of the
usual \(a,b,c\) labels for the sides, we will use the labels \textit{hyp} (for hypotenuse), \textit{opp}
(for opposite)
and \textit{adj} (for adjacent). The side labeled \(\mbox{hyp}\) is of course always the longest side
(opposite the \(90^\circ\)
angle). We will then select one of the two acute angles, and name \(\mbox{opp}\)
the side opposite to it, and
\(\mbox{adj}\) the side adjacent to it. The two figures below show the labeling.
\[\img{U4_3F1.png}{}{15em}{} \hspace{5ex} \img{U4_3F2.png}{}{15em}{}\]
Remember that the sum of all angles in any triangle is \(180^\circ\). So in a right triangle
(where one angle is \(90^\circ\)) the sum of the two acute angles is \(90^\circ\).
When two acute angles have a sum of \(90^\circ\) they are called \textit{complementary}. So for example
\(30^\circ\) and \(60^\circ\) are complementary, \(15^\circ \) and \(75^\circ\) are complementary,
and so on.
With our labeling of the sides, the Pythagorean Theorem says:
\[(\mbox{opp})^2 + (\mbox{adj})^2 = (\mbox{hyp})^2 .\]
\subsection{Definition of the trigonometric functions}
Using the three sides \(\mbox{opp}\), \(\mbox{adj}\) and \(\mbox{hyp}\) we now define
the six trigonometric functions of an acute angle \(\theta\):
\[
\begin{array}{ccccc}
\sin\theta =\frac{\mbox{opp}}{\mbox{hyp}} & \ & \cos\theta =\frac{\mbox{adj}}{\mbox{hyp}} & \ & \tan\theta =\frac{\mbox{opp}}{\mbox{adj}}\\
\ & \ \ \ & \ & \ \\
\csc\theta =\frac{\mbox{hyp}}{\mbox{opp}} & \ & \sec\theta =\frac{\mbox{hyp}}{\mbox{adj}} & \ & \cot\theta =\frac{\mbox{adj}}{\mbox{opp}}
\end{array}.
\]
The function names \(\sin\), \(\cos\), \(\tan\), \(\csc\), \(\sec\), \(\cot\)
are the short form of the full names of the functions: sine, cosine, tangent, cosecant, secant
and cotangent.
\
You can use SOH-CAH-TOA to help memorize the definition of the first three trig functions:
| \textbf{S}in \(=\) \textbf{O}pposite over \textbf{H}ypotenuse |
| \textbf{C}os \(=\) \textbf{A}djacent over \textbf{H}ypotenuse |
| \textbf{T}an \(=\) \textbf{O}pposite over \textbf{A}djacent |
Then remember that the last three are the reciprocals (just flip the fraction).
\subsection{Trig functions and right triangles}
\begin{example}
We need to find the six trigonometric functions of angle \(\theta\) in the following picture:
\[\img{U4_3F3.png}{}{15em}{}\]
We see from the picture that
\[\begin{array}{c}
\mbox{opp} = 5 \\
\mbox{adj} = 12 \\
\mbox{hyp} = 13
\end{array}\]
So, according to the definitions, we have
\[\begin{array}{ccc}
\displaystyle \sin\theta = \frac{\mbox{opp}}{\mbox{hyp}} = \frac{5}{13} &
\displaystyle \cos\theta = \frac{\mbox{adj}}{\mbox{hyp}} =
\frac{12}{13} & \displaystyle \tan\theta = \frac{\mbox{opp}}{\mbox{adj}} = \frac{5}{12}\\[2ex]
\displaystyle \csc\theta = \frac{\mbox{hyp}}{\mbox{opp}} =
\frac{13}{5} &
\displaystyle \sec\theta = \frac{\mbox{hyp}}{\mbox{adj}} = \frac{13}{12} &\displaystyle \cot\theta = \frac{\mbox{adj}}{\mbox{opp}} =
\frac{12}{5}
\end{array}
\]
\end{example}
In the previous example we were given all three sides of the triangle.
Sometimes we are only given two sides.
Then we need to find the third side using the Pythagorean theorem.
\begin{example}
We are given the triangle
\[\img{U4_3F4.png}{}{12em}{}\]
and we want to find the six trig functions for the angle \(\theta\).
The hypothenuse is missing. The Pythagorean theorem gives us:
\[9^2+6^2=\mbox{hyp}^2.\]
Solving, we find:
\[\begin{array}{rcl}
81+36&=&\mbox{hyp}^2\\[2ex]
117&=& \mbox{hyp}^2\\[2ex]
\mbox{hyp}&=&\sqrt{117}
\end{array}
\]
Now we use the definitions to find the six trig functions:
\[\begin{array}{ccc}
\displaystyle \sin\theta = \frac{\mbox{opp}}{\mbox{hyp}} = \frac{9}{\sqrt{117}} &
\displaystyle \cos\theta = \frac{\mbox{adj}}{\mbox{hyp}} =
\frac{6}{\sqrt{117}} &
\displaystyle \tan\theta = \frac{\mbox{opp}}{\mbox{adj}} = \frac{9}{6}=\frac{3}{2}\\[2ex]
\displaystyle \csc\theta = \frac{\mbox{hyp}}{\mbox{opp}} =
\frac{\sqrt{117}}{9} & \displaystyle \sec\theta = \frac{\mbox{hyp}}{\mbox{adj}} = \frac{\sqrt{117}}{6} & \displaystyle \cot\theta = \frac{\mbox{adj}}{\mbox{opp}} =
\frac{6}{9}=\frac{2}{3}
\end{array}
\]
\end{example}
\subsection{Trig functions for the \(30^\circ\) and \(60^\circ\) angles}
We now find the values of the trig functions for a right triangle that has an acute angle of
\(30^\circ\). So the other acute angle will be \(60^\circ\). Such a triangle is
called a \textit{30-60-90 triangle}. To find the trig values of this triangle,
draw it as in the picture below:
\[\img{U4_3F9.png}{-3em}{12em}{} \hspace{5ex} 1\lt \sqrt{3}\approx 1.7 \lt 2\]
To remember where to put the three numbers \(1,\sqrt{3}\) and \(2\), remember that
\(\sqrt{3}\) is about \(1.7\), so it must be the side of intermediate length. Of course 1 is the
short side, and 2 is the long side.
We can use this triangle and the definitions of trig functions to find all trig values
for \(30^\circ\)
(or \(\frac{\pi}{6}\))
and \(60^\circ\) (or \(\frac{\pi}{3}\)). But before we do that, let's check that the three numbers
\(1,2,\sqrt{3}\) are the right ones to use for the \(30\)-\(60\)-\(90\) triangle.
First we check that if we use the given numbers \(1\) and \(2\)
for two of the sides, the angles
are really \(30^\circ\) and \(60^\circ\). To do this, we double the triangle as in the picture below:
\[\img{U4_3F10.png}{}{12em}{}\]
We see from the picture that the doubled triangle is equilateral,
because all sides have length 2. But that means that
all its angles are \(60^\circ \), and the smaller angle in the \(30\)-\(60\)-\(90\) triangle
is half of that, or \(30^\circ \).
So we see that the numbers \(1\) and \(2\) for the short side and the hypotenuse really give a
\(30\)-\(60\)-\(90\) triangle.
To check that \(\sqrt{3}\) is right for the base of our triangle,
we use the Pythagorean theorem:
\[b^2+1^2=2^2\]
so
\[b^2=3\]
and \[b=\sqrt{3}.\]
To find the trig values for the \(\displaystyle 30^\circ=\frac{\pi}{6}\) angle,
draw the \(30\)-\(60\)-\(90\) triangle with the sides marked as follows:
\[\img{U4_3F11.png}{}{12em}{}\]
Using the definitions, we find:
\[\begin{array}{ccc}
\displaystyle \sin 30^\circ=\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} &
\displaystyle \cos 30^\circ =\cos \left(\frac{\pi}{6}\right) =
\frac{\sqrt{3}}{2} & \displaystyle \tan30^\circ =\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\\[3ex]
\displaystyle \csc 30^\circ =\csc\left(\frac{\pi}{6}\right)=2&
\displaystyle \sec 30^\circ =\sec \left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} & \displaystyle \cot 30^\circ = \cot \left(\frac{\pi}{6}\right) =\sqrt{3}
\end{array}
\]
To find the trig values for the \(\displaystyle 60^\circ=\frac{\pi}{3}\) angle, draw the triangle:
\[\img{U4_3F12.png}{}{12em}{}\]
Using the definitions, we find:
\[\begin{array}{ccc}
\displaystyle \sin 60^\circ=\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} &
\displaystyle \cos 60^\circ =\cos \left(\frac{\pi}{3}\right) =
\frac{1}{2} & \displaystyle \tan 60^\circ =\tan\left(\frac{\pi}{3}\right) =\sqrt{3}\\[3ex]
\displaystyle \csc 60^\circ =\csc\left(\frac{\pi}{3}\right)=
\frac{2}{\sqrt{3}}&
\displaystyle \sec 60^\circ =\sec \left(\frac{\pi}{3}\right) =2 & \displaystyle \cot 60^\circ = \cot \left(\frac{\pi}{3}\right) =
\frac{1}{\sqrt{3}}
\end{array}
\]
Using the trig functions, we can now find all sides of a \(30\)-\(60\)-\(90\) triangle as soon as we
know just one side.
\begin{example}
Find all sides of the triangle in the picture:
\[\img{U4_3F13.png}{}{12em}{}\]
First we label the sides:
\[\img{U4_3F15.png}{}{15em}{}\]
Now use one of the first three trig functions (\(\sin, \cos\) or \(\tan\)). To decide which
one, note that we already know the hypotenuse, that appears in the definition of \(\sin\) and \(\cos\).
So we use one of them (it does not matter which of the two). Using the sine function, we write:
\[\sin30^\circ =\frac{\mbox{opp}}{\mbox{hyp}}.\]
Then we substitute the known values: \(\displaystyle \sin30^\circ =\frac{1}{2}\), \(\mbox{hyp}=12\):
\[\frac{1}{2}=\frac{a}{12}\]
Cross multiplying we find
\[12=2a\]
and so \[a=6.\]
To find \(b\), we could use the Pythagorean theorem. Or we can use the cosine function:
\[\cos30^\circ = \frac{\mbox{adj}}{\mbox{hyp}}\]
so
\[\begin{array}{rcl}
\displaystyle \frac{\sqrt{3}}{2}&=&\displaystyle \frac{b}{12}\\[2ex]
12\sqrt{3}&=&2b\\[2ex]
b&=&6\sqrt{3}
\end{array}
\]
\end{example}
\subsection{Trig functions for the \(45^\circ\) angle}
If an acute angle in a right triangle has measure \(45^\circ\), then the other acute angle is also
a \(45^\circ\) angle and the triangle is isosceles. Such a triangle is called a \textit{45-45-90 triangle}.
We set the equal sides equal to \(1\):
\[\img{U4_3F24.png}{}{10em}{}\]
Using the Pythagorean theorem we find
\[1^2+1^2=c^2\]
and solving we get
\[c=\sqrt{2}.\]
\[\img{U4_3F23.png}{}{10em}{}\]
Since the triangle is isosceles, choosing either of the two acute angles we have
\(\mbox{opp}=\mbox{adj}=1\) and \(\mbox{hyp}=\sqrt{2}\), so the trig functions for the
\(\displaystyle 45^\circ =\frac{\pi}{4}\) angle are:
\[\begin{array}{ccc}
\displaystyle \sin 45^\circ=\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} &
\displaystyle \cos 45^\circ =\cos \left(\frac{\pi}{4}\right) =
\frac{1}{\sqrt{2}} & \displaystyle \tan45^\circ =\tan\left(\frac{\pi}{4}\right) = 1\\[3ex]
\displaystyle \csc 45^\circ =\csc\left(\frac{\pi}{4}\right)=
\sqrt{2}&
\displaystyle \sec 45^\circ =\sec \left(\frac{\pi}{4}\right) = \sqrt{2} & \displaystyle \cot 45^\circ = \cot \left(\frac{\pi}{4}\right) = 1
\end{array}
\]
Problems
\problem
Find the six trig functions for the angle \(\theta\) in the picture:
\[\mbox{a. }\hspace{3ex} \img{U4_3F5.png}{}{12em}{} \hspace{2ex} \mbox{b. } \hspace{3ex} \img{U4_3F6.png}{}{15em}{}\]
\begin{sol}
\begin{enumerate}
\item \
\[\begin{array}{ccc}
\displaystyle \sin\theta = \frac{\mbox{opp}}{\mbox{hyp}} = \frac{4}{5} &
\displaystyle \cos\theta = \frac{\mbox{adj}}{\mbox{hyp}} =
\frac{3}{5} &
\displaystyle \tan\theta = \frac{\mbox{opp}}{\mbox{adj}} = \frac{4}{3}\\[2ex]
\displaystyle \csc\theta = \frac{\mbox{hyp}}{\mbox{opp}} =
\frac{5}{4} &
\displaystyle \sec\theta = \frac{\mbox{hyp}}{\mbox{adj}} = \frac{5}{3} &
\displaystyle \cot\theta = \frac{\mbox{adj}}{\mbox{opp}} =
\frac{3}{4}
\end{array}
\]
\item \
The missing side is \(\mbox{adj}\).
Using the Pythagorean Theorem, we find
\[
\begin{array}{rcl}
\mbox{adj}^2+ 4^2&=& 7^2 \\[2ex]
\mbox{adj}^2 +16 & = & 49\\[2ex]
\mbox{adj}^2 & = & 49-16 \\[2ex]
\mbox{adj}^2 & = & 33\\[2ex]
\mbox{adj} &=& \sqrt{33}
\end{array}.
\]
Then we use the definitions:
\[\begin{array}{ccc}
\displaystyle \sin\theta = \frac{\mbox{opp}}{\mbox{hyp}} = \frac{4}{7} &
\displaystyle \cos\theta = \frac{\mbox{adj}}{\mbox{hyp}} =
\frac{\sqrt{33}}{7} & \displaystyle \tan\theta = \frac{\mbox{opp}}{\mbox{adj}} = \frac{4}{\sqrt{33}}\\[2ex]
\displaystyle \csc\theta = \frac{\mbox{hyp}}{\mbox{opp}} =
\frac{7}{4}&
\displaystyle \sec\theta = \frac{\mbox{hyp}}{\mbox{adj}} = \frac{7}{\sqrt{33}} & \displaystyle \cot\theta = \frac{\mbox{adj}}{\mbox{opp}} =
\frac{7}{4}
\end{array}
\]
\end{enumerate}
\end{sol}
\mproblem
Find the six trig functions for the angle \(\theta\) in the picture:
\[\mbox{a. }\hspace{3ex} \img{U4_3F7.png}{}{8em}{} \hspace{2ex} \mbox{b. } \hspace{3ex} \img{U4_3F8.png}{}{10em}{}\]
\problem
Find all sides of the triangles in the picture:
\[\mbox{a. }\hspace{3ex} \img{U4_3F14.png}{}{12em}{} \hspace{2ex} \mbox{b. } \hspace{3ex} \img{U4_3F16.png}{}{13em}{}\]
\begin{sol}
\begin{enumerate}
\item
Label the sides of the triangle:
\[\img{U4_3F17.png}{}{15em}{}\]
The given side is opposite the given angle. We use the sine function to find \(c\):
\[
\begin{array}{rcl}
\sin60^\circ &=& \displaystyle \frac{\mbox{opp}}{\mbox{hyp}}\\[1ex]
\displaystyle \frac{\sqrt{3}}{2}&=& \displaystyle \frac{7}{c}\\[1ex]
c\sqrt{3}&=& 14\\[1ex]
c&=& \displaystyle \frac{14}{\sqrt{3}}
\end{array}
\]
Now use the tangent function to find \(a\):
\[
\begin{array}{rcl}
\tan60^\circ &=& \displaystyle \frac{\mbox{opp}}{\mbox{adj}}\\[1ex]
\displaystyle \sqrt{3}&=& \displaystyle \frac{7}{a}\\[1ex]
a\sqrt{3}&=& 7\\[1ex]
a&=& \displaystyle \frac{7}{\sqrt{3}}
\end{array}
\]
\item \
Label the sides of the triangle:
\[\img{U4_3F18.png}{}{15em}{}\]
The given side is adjacent the given angle. We use the cosine function to find \(c\):
\[
\begin{array}{rcl}
\displaystyle \cos\left(\frac{\pi}{6}\right) &=& \displaystyle \frac{\mbox{adj}}{\mbox{hyp}}\\[1ex]
\displaystyle \frac{\sqrt{3}}{2}&=& \displaystyle \frac{9}{c}\\[1ex]
c\sqrt{3}&=& 18\\[1ex]
c&=& \displaystyle \frac{18}{\sqrt{3}}
\end{array}
\]
Now use the tangent function to find \(a\):
\[
\begin{array}{rcl}
\displaystyle \tan\left(\frac{\pi}{6}\right) &=& \displaystyle \frac{\mbox{opp}}{\mbox{adj}}\\[1ex]
\displaystyle \frac{1}{\sqrt{3}}&=& \displaystyle \frac{a}{9}\\[1ex]
9&=& a\sqrt{3}\\[1ex]
a&=& \displaystyle \frac{9}{\sqrt{3}}
\end{array}
\]
\end{enumerate}
\end{sol}
\mproblem
Find all sides of the triangles in the picture:
\[\mbox{a. }\hspace{3ex} \img{U4_3F19.png}{}{12em}{} \hspace{5ex} \mbox{b. } \hspace{3ex} \img{U4_3F20.png}{}{12em}{}\]
\problem
Find all sides of the triangle in the picture:
\[\img{U4_3F25.png}{}{8em}{}\]
\begin{sol}
Using the sine function,
\[\sin45^\circ =\frac{\mbox{opp}}{\mbox{hyp}}=\frac{1}{\sqrt{2}}\]
So substituting the values from the picture
\[
\begin{array}{rcl}
\displaystyle \frac{b}{7}&=& \displaystyle \frac{1}{\sqrt{2}}\\[2ex]
b\sqrt{2} &=& 7\\[2ex]
b&=& \displaystyle \frac{7}{\sqrt{2}}
\end{array}
\]
\end{sol}
\mproblem
Find all sides of the triangle in the picture:
\[\img{U4_3F26.png}{}{8em}{}\]