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\fancyfoot[CO,CE]{\it Trigonometry Tutorial by A. A. Tovar, Ph. D., Created Oct. 2002, Amended Jan. 2009.}
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\begin{document}
\title {Trigonometry Explained}
\author{Anthony A. Tovar, Ph. D.
\\Eastern Oregon University
\\1 University Blvd.
\\La Grande, Oregon, 97850 }
%\date{July 24, 2002}
\setcounter{page}{1}
\maketitle
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%\mainmatter
\chapter{Trigonometry}
\section{Introduction}
The goal of this tutorial is to explain how trigonometry could be developed
by simply drawing some triangles.
The reader is lead into ``inventing'' trigonometry - not just the HOWs, but
the WHYs as well. During course work, it is easy to ``miss the forest from
the trees.'' Here we look at the forest, and maybe a tree or two. This tutorial
is NOT meant to substitute for course work because learning mathematics
is like riding a bicycle - you learn it by doing it (as opposed to just
reading about it). This tutorial
is also very different than many reviews that merely state results. The
emphasis here is on the mental process of developing the mathematics.
It is hoped that the reader will gain a sense of discovery.
This tutorial is intended for anyone who knows a little algebra, and wants
to learn what trigonometry is about. This may include students who just finished
a course in algebra, students who recently completed a course in
trigonometry but didn't quite get the ``big picture,'' students
who had trigonometry years ago and want a review, and
educators (including those who homeschool).
This tutorial may also be useful to students in the middle
of a trigonometry course because
the best way to learn mathematics (and science, in general) is to
examine the trees for a little while, then the forest, then the trees again,
then the forest again... This zooming in and zooming out helps retention
immeasurably.
While going through this tutorial, the reader should have a pencil
and a calculator handy - learning is not a spectator sport!
\section{Why Trigonometry?}
As can be seen from the first three figures...
\begin{itemize}
\item irregular polygons can be made up of many triangles.
\item regular polygons can be made up of many triangles.
\item Irregular triangles can be made up of right triangles.
\end{itemize}
A right triangle is one in which two of the sides are at
$90 \degrees$ (or perpendicular) to each other, as shown in Fig.
(1.4).
%\pagebreak
\begin{figure}[ht]
\begin{center}
\includegraphics[width=5cm,height=3cm]{graphics/arb_fig.eps}
\caption{An irregularly shaped object with many sides can be divided
into triangles.}
\end{center}
\label{arb_fig}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.4cm,height=3cm]{graphics/reg_fig.eps}
\caption{A regularly shaped object with many sides can be divided
into triangles.}
\end{center}
\label{reg_fig}
\end{figure}
\begin{figure}[hb]
\begin{center}
\includegraphics[keepaspectratio=true,width=5cm]{graphics/triangle1.eps}
\caption{An arbitrary triangle can be divided into two right triangles.}
\end{center}
\label{triangle1}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[width=5cm,height=3cm]{graphics/right.eps}
\caption{A right triangle has a $90\degrees$ angle.
The longest side, $c$, is called the hypotenuse.}
\end{center}
\label{right}
\end{figure}
Basically, we cannot reduce right triangles into simpler figures.
Therefore, if we want to learn about the geometry of figures, we
need to study right triangles $first$.
Once we have learned about the relationships between the sides
and the angles of right triangles, we can apply the results
to more complicated (including multidimensional)
figures.\footnote{What about smooth figures, such
as circles? It ends up that we can break those into a large number
(infinite, actually) of right triangles and rectangles. Indeed,
Archimedes did this problem at about 260 B.C. to help him determine $\pi$.
It may or may not sound too bad, but try doing it using archaic
notation such as Roman numerals!}
This is a common technique in math and science. To understand complicated
ideas and structures, we start by investigating fundamental ``building blocks."
We can then apply knowledge gained to help us understand the more
complicated goings-on in the natural world.
\section{The Pythagorean Theorem}
The first thing we need to known about right triangles is that the
sides of a right triangle (a, b, and c) are not independent. Given
any two, one can calculate the third. This should not be a big
surprise. Draw two sides of a right triangle, and you have no choice
as to the length of the third line you draw. This basic result is known as
the \textbf{Pythagorean Theorem}:
\EQ
c^2 = a^2 + b^2
\EN
where $a$, $b$, and $c$ are as labeled on Fig. 1.4.
We could determine this empirically by merely drawing lots
of right triangles and measuring their sides.\footnote{This would
make a good laboratory exercise.}
There are many geometrical
proofs of the theorem\footnote{For example, do a World Wide Web search for
``proof Pythagorean theorem."}, but they are not of interest here.
We may note that if the sides of a right triangle are all whole
numbers, then the lengths of
these sides are called {\bf Pythagorean triplets}:\\
$3^2 + 4^2 = 5^2\\
5^2 + 12^2 = 13^2\\
6^2 + 8^2 = 10^2\\
7^2 + 24^2 = 25^2\\
8^2 + 15^2 = 17^2\\
9^2 + 12^2 = 15^2\\
9^2 + 40^2 = 41^2\\
10^2 + 24^2 = 26^2\\
11^2 + 60^2 = 61^2\\
12^2 + 16^2 = 20^2\\
12^2 + 35^2 = 37^2\\
13^2 + 84^2 = 85^2 $\\
A computer program was written to generate these triplets.\footnote{Can you figure out
what the next triplet is? Answer at end of tutorial. Hint: it starts with $14^2 +$.}
Dividing both sides of the Pythagorean Theorem by $c$ yields
\EQ
\left(\frac{a}{c}\right)^2 +
\left(\frac{b}{c}\right)^2 = 1
\label{pythag_ratio}
\EN
This equation reemphasizes that
there are only two independent factors of a right triangle. It also suggests
that we look at the ratios of the sides, which we'll do in the next section.
\begin{APP}
{\bf Construction}\\
Before laying a concrete foundation, construction workers make string lines
to determine where their wooden ``forms" should go. To insure that their string
lines (and hence their forms) are at exactly 90$\degrees$, they measure
3 feet from a corner and make a mark on the string. On the crossing
string they make a mark 4 feet from the corner. If the distance
between the two marks is exactly 5 feet, then the corner is ``square."
\begin{figure}[h]
\begin{center}
\includegraphics[width=6cm,height=5cm]{graphics/construction.eps}
\caption{Pythagorean Triplets are used to determine if two strings
are at right angles to each other.}
\end{center}
\label{construction}
\end{figure}
They are really using the 3-4-5 Pythagorean triplet. They also
make extensive use of the 6-8-10 triplet. They tend to not
use the 5-12-13 triplet (nor do they tend to know about that one)
because you want a triangle that has angles closer to $45 \degrees$.
\end{APP}
\section{Trigonometry: the Main Idea}
Now, instead of looking at arbitrary right triangles, let's examine
right triangles that share a common angle. These right
triangles are called \textbf{similar triangles}:
\begin{figure}[h]
\begin{center}
\includegraphics[width=7cm,height=4cm]{graphics/similar.eps}
\caption{Similar triangles.}
\end{center}
\label{similar_triangles}
\end{figure}
Again, we can draw pairs of similar triangles, and start
measuring their sides. If we do this we would immediately
see an obvious pattern.\footnote{This too would make a good laboratory exercise.}
This pattern is the \textbf{law of similar
triangles}:
\begin{quote}
{\bf
For a right triangle with a given angle, $A$,
the ratio of any two sides is constant
independent of the size of the triangle.}
\end{quote}
For example, in Fig. 1.6
\begin{equation}
\frac{a_1}{b_1} =
\frac{a_2}{b_2}
\end{equation}
This ratio is the same for all right triangles which have the
same angle, $A$, associated with them. Of course this ratio
is different for different values of $A$. Thus, there is
a function that specifies this ratio for different angles -
it is called the tangent function, or $tan(A)$.
We can similarly define the sine and cosine functions:
\EQ
sin(A)
= \frac{a}{c}
= \frac{opposite}{hypotenuse}
\label{sine_def}
\EN
\EQ
cos(A)
= \frac{b}{c}
= \frac{adjacent}{hypotenuse}
\label{cosine_def}
\EN
\EQ
tan(A)
= \frac{a}{b}
= \frac{opposite}{adjacent}
= \frac{sin(A)}{cos(A)}
\EN
where, as before, $a$, $b$, and $c$ are defined as in Fig. (1.4).
A memory device sometimes used to for these relationships involves
the (fictional) American Indian chief {\bf SOH CAH TOA}:
{\bf S}ine is {\bf O}pposite over {\bf H}ypotenuse.
{\bf C}osine is {\bf A}djacent over {\bf H}ypotenuse.
{\bf T}angent is {\bf O}pposite over {\bf A}djacent.\\
With three sides, there are six possible ratios
(OH, OA, AH, AO, HA, HO) of the sides of a right
triangle. So, we further define the
cosecant, secant, and cotangent functions as
\EQ
csc(A)
= \frac{1}{sin(A)}
\EN
\EQ
sec(A)
= \frac{1}{cos(A)}
\EN
\EQ
cot(A)
= \frac{1}{tan(A)}
\EN
Now, the ratio of any two sides has a definition.
Inclines are often encountered in everyday life.
The slope of an incline is, of course, rise/run. However,
this is precisely how the tangent function is defined.
The percent grade of an incline is just the slope measured
in percent, so that
\EQ
slope = percent \;grade = tan(\theta )
\EN
For example, a $20 \%$ grade has a slope of 0.2 and an angle of
elevation of $tan^{-1}(0.2) = 11.3 \degrees$.
\begin{APP}
{\bf The Height of a Pyramid}\\
Thales of Miletus (624-547 B.C.), often regarded as the first physicist,
was asked to measure the height of a pyramid.
Since there is no vertical surface on a pyramid, he couldn't simply
measure it. Furthermore, trigonometry hadn't been invented yet.
However, he did have a notion of similar triangles, which is all
he needed.
\begin{figure}[h]
\begin{center}
\includegraphics[width=10cm,height=4cm]{graphics/pyramid.eps}
\caption{Thales measured the height of a pyramid using similar triangles.}
\end{center}
\label{pyramid}
\end{figure}
He basically stuck a stick in the ground and measured its height
and the length of the shadow it gave off. He then measured the
length of the shadow of the pyramid. Using the law of similar
triangles,
\EQ
\frac{Height\; of\; Stick}{Length\; of\; Stick\; Shadow} =
\frac{Height\; of\; Pyramid}{Length\; of\; Pyramid\; Shadow}
\EN
From this expression it is easy to determine the height of the pyramid.
\end{APP}
\section{Degrees and Radians}
One can measure angles in degrees or radians. Degrees exist in
everyday usage, but not all are familiar with radians.
Since radian measure is based on the circle,
we first review the the circle's properties.
The distance from the center of a circle to any point on the
circle is called the {\bf radius}, $r$. Twice this distance
is the diameter.
The area of a circle is given by the formula
\EQ
A = \pi r^2
\EN
where
$\pi$ is a transcendental number (it's an infinite nonrepeating
decimal which is also not the square root of an integer). It's
value is approximately
$\pi = 3.14159265...$\footnote{A common (but bad) fractional approximation
is $22/7$ which is only accurate to 2 decimal places. The
approximation $355/113$ is accurate to 6 decimal places.}
The perimeter of a circle is called the
{\bf circumference}. It is a little more than six times the
radius:
\EQ
C = 2 \pi r
\EN
These formulas could be determined by just drawing lots
of circles.\footnote{Or use a cone, a string, a calipers, and
a measuring tape.} These formulas are some of the most
beautiful theorems in classical geometry.
We can use the circumference formula to determine the
length of an arc. For example, a
$45 \degrees$ circular arc has one-eighth the length of
a whole circle, or $2\pi r/8 = \pi r/4$.
\begin{figure}[h]
\begin{center}
\includegraphics[keepaspectratio,width=4cm]{graphics/45_arc.eps}
\caption{A $45 \degrees$ arc.}
\end{center}
\label{45_arc}
\end{figure}
If the circle is chosen
to have a radius of one, then the arc length is just
$\pi/4$. This is used as an alternate way of specifying
angles. {\bf An angle in radians is represented by the length of a
unit-radius arc}. For example, instead of $45 \degrees$, we use
$\pi/4$ radians. No wonder we use degrees instead of radians for
everyday usage! However,
for mathematical calculations, it is far easier to
use radian measure\footnote{In advanced mathematics it only
makes sense to use radian measure. For example, it can be shown
that $sin(\theta ) \approx \theta$ for small angles $\theta$, but
only if we measure $\theta$ in radians}. One can relate the two:
\EQ
\frac{Angle\; in\; radians}{Angle\; in\; degrees} =
\frac{\pi}{180}
\EN
This conversion factor can easily be remembered since a full circle is
$2\pi$ radians and is also $360 \degrees$. (Thus,
$\pi$ radians is $180 \degrees$.)
For the above example,
\EQ
\frac{Angle\; in\; radians}{45} =
\frac{\pi}{180}
\EN
Cross multiplying yields
\EQ
Angle\; in\; radians =
\frac{45 \pi}{180} = \frac{\pi}{4}
\EN
Thus, $45 \degrees$ is $\pi/4$ radians, as obtained above.
\section{The Unit Circle}
Since right triangles have only two sides which are independent, we
can arbitrarily choose the length of one of the sides.\footnote{As long as
we don't choose zero, of course!} Let's choose the longest side to
be of length 1. Notice that I didn't say one what - could be one meter,
one mile, or one inch. Besides saying the length of the hypotenuse
is 1, let's also specify that one of the corners of our triangle
lies on some $xy$ coordinate system
(in the figure below, the solid dots represents the origin).
Furthermore, the corner must one of the two corners that touches
the hypotenuse. The angles are always measured from the $x$-axis.
Let's look at some ``triangles'' as shown in Fig. 1.9.
\begin{figure}[h]
\begin{center}
\includegraphics[keepaspectratio=true,width=10cm]{graphics/3_triangles.eps}
\caption{Six Triangles with a common origin.}
\end{center}
\label{3_triangles}
\end{figure}
The first triangle in the figure is at $0 \degrees$. It's not
much of a triangle. Since the longest side is 1, the base ($x$-coordinate)
is also 1. The height ($y$-coordinate) of the triangle is 0
($1^2 + 0^2 = 1^2$).
The second triangle has a $30 \degrees$ angle. It is not
obvious what the length of the sides are. This triangle
is discussed later\footnote{For those who can't wait, the short side
is $1/2$ and the longer side is $\sqrt{3}/2$.}.
The third triangle is at $45 \degrees$. It has
the same base as height. If we call this value $x$, then
from the Pythagorean Theorem
\begin{equation}
x^2 +x^2 = 1^2 \;\; \Rightarrow \;\;
2(x^2) = 1 \;\; \Rightarrow \;\;
x^2 = 1/2 \;\; \Rightarrow \;\;
x = 1/\sqrt{2} = \sqrt{2}/2
\end{equation}
The $60 \degrees$ triangle is the same as the $30 \degrees$ triangle with
the base and height exchanged.
The fifth triangle has zero base, but a height of 1. It has a $90 \degrees$ angle.
The last triangle in the figure has an angle of $135 \degrees$. The
base is $-\sqrt{2}/2$. (Yes, a negative value!)\footnote{Negative
angles may take a little bit of getting used to, but
wait, it gets weirder!} The height of the triangle is a positive $\sqrt{2}/2$.
A couple of quick notes before we look at the trig functions.
Right triangles with angles between $0 \degrees$ and $90 \degrees$ are
said to be in the ``first quadrant."
Triangles with angles between $90 \degrees$ and $180 \degrees$ are
said to be in the ``second quadrant."
Triangles with angles between $180 \degrees$ and $270 \degrees$ are
said to be in the ``third quadrant."
Triangles with angles between $270 \degrees$ and $360 \degrees$ are
said to be in the ``fourth quadrant."
A triangle with an angle of $405 \degrees$ is the same as a triangle
with an angle of $45 \degrees$, so that we can add or subtract
$360 \degrees$'s as often as we wish.\footnote{I told you it gets weirder. You can
tell a friend you climbed stairs oriented at $405 \degrees$! It makes more sense if
you think about a car in a spin-out. A $45 \degrees$ spin-out is very different from a
$405 \degrees$ spin-out! But, if we're only interested in the final orientation of the car,
then the results are the same.}
For example, a triangle with
an angle of $-30 \degrees$ is the same as a triangle with an angle
of $330 \degrees$, and thus is in quadrant IV.
Again, get use to the idea of negative sides and angles, which we must have
when using an $xy$ coordinate system.
Now, let's look at the $sin$ and $cos$ trig functions. All of the other
trig functions can be obtained from those two. The ``adjacent" side
is always on the x-axis, and the ``opposite" side is always on the y-axis.
Since the hypotenuse is always 1,
\begin{equation}
sin(\theta ) = opposite = height = y-axis
\end{equation}
and
\begin{equation}
cos(\theta ) = adjacent = base = x-axis
\end{equation}
where, by convention, we use the Greek letter $\theta$ (called ``theta")
to represent the angle.
Looking at the triangles in Fig. (1.9), the
reader should be able to confirm that:
\begin{center}
\begin{tabular}{||l|c|c||}
\hline \hline
$\theta$ & $sin(\theta )$ & $cos(\theta )$\\
\hline \hline
$0\degrees$ & $0$ & $1$ \\
\hline
$45\degrees$ & $\sqrt{2}/2$ & $\sqrt{2}/2$ \\
\hline
$90\degrees$ & $1$ & $0$ \\
\hline
$135\degrees$ & $\sqrt{2}/2$ & $-\sqrt{2}/2$ \\
\hline
$180\degrees$ & & \\
\hline
$225\degrees$ & & \\
\hline
$270\degrees$ & & \\
\hline
$315\degrees$ & & \\
\hline
$360\degrees$ & & \\
\hline \hline
\end{tabular}
\end{center}
\noindent
({\bf Yes, actually take a minute or two and confirm these NOW! Then Fill in the
rest of the table, we'll need the answers in a bit.})
\vspace{0.5cm}
When we examined the previous figure, we said that all of the solid dots
were at the origin. With this, we can represent a right triangle as a single
point in the $xy$ plane. The point used is the other end of the hypotenuse.
Thus, for example, the $0 \degrees$ ``triangle'' would be represented by
the point at x=1, y=0, or (1,0). Similarly, the $45 \degrees$ triangle
would be represented by a point at ($\sqrt{2}/2$, $\sqrt{2}/2$). Since
we decided to always make the hypotenuse of length one, all possible
triangles must be represented as some point on a circle. This is called
the ``unit circle.''
\begin{figure}[h]
\begin{center}
\includegraphics[width=7cm,height=7cm]{graphics/unit_circle.eps}
\caption{The Unit Circle.}
\end{center}
\label{unit_circle}
\end{figure}
Again, keep in mind that the $x$-axis can be thought of
as the ``cosine axis" and the $y$-axis can be thought of as
the ``sine axis." Thus, from the figure, $cos( 30 \degrees ) = \sqrt{3}/2$,
and $sin( 30 \degrees ) = 1/2$.
%\pagebreak[4]
In a course in trigonometry, we are expected to have the unit circle with
the values for sine and cosine of the various angles memorized! It's not
really very hard. The sine and cosine values are all either
$0$, $1$, $1/2$, $\sqrt{3}/2$, or $\sqrt{2}/2$. Then you have to determine
the sign. Here's how to proceed.
Mentally, or actually draw a circle, and a line from the center
to the point corresponding to the angle given. If the angle is $0 \degrees$,
$90 \degrees$, $180 \degrees$, or $270 \degrees$, it's a $0$ or a $1$. Then
determine the sign. If it's not one of these the answers not a $0$ or a $1$.
These four angles are symmetric. Four other symmetric angles are
$45 \degrees$, $135 \degrees$, $225 \degrees$, and $315 \degrees$.
If the angle is any of these, the sine and the cosine are both $\sqrt{2}/2$.
Then check for sign. Again, remember the $x$-axis is the cosine axis and the
$y$-axis is the sine axis. You can visualize these in your head (we'll do
an example in a moment) without the need for writing
anything down. What about the others? Easy as well. There are only
two possibilities - the short side is $1/2$ and the long side is $\sqrt{3}/2$.
Let's do an example.
\begin{EX}
Determine $sin(240 \degrees )$.
\end{EX}
\begin{SOL}
First, draw a circle and a line corresponding to $240 \degrees$:
\includegraphics[width=4cm,height=4cm]{graphics/angle_example.eps}
Since the sine axis is the y-axis, $sin(240 \degrees )$ is the length of the
dotted line in the figure. It's not one of those evenly space angles, so
the answer is either $1/2$ or $\sqrt{3}/2$. It's the long one, so it
must be $\sqrt{3}/2$. Checking for sign, we can seen that the y-coordinate
value of the point is negative, so $sin(240 \degrees ) = -\sqrt{3}/2$.
\noindent
Checking with the calculator (in degrees mode, of course),
$sin(240 \degrees ) = -.866025$.
However, $-\sqrt{3}/2 = -.866025$ in agreement with our calculation.
\end{SOL}
Let's go back and figure out the lengths of the sides of a
$30 \degrees$/$60 \degrees$ triangle. We start by drawing an equilateral
triangle (all sides are have length $1$). Since all the angles are the
same and the sum of the angles must be $180 \degrees$, they must be
$60\degrees$ each.
\begin{center}
\includegraphics[width=4cm,height=4cm]{graphics/30_60_triangle.eps}
\end{center}
It can easily be seen that by cutting one of the
$60\degrees$ angles in two, 2 of the sides of the resulting right
triangle are $1$ and $1/2$. The length of the third side, from the
Pythagorean theorem is
\begin{equation}
b = \sqrt{c^2 - a^2} = \sqrt{3/4} = \sqrt{3}/2
\end{equation}
\section{Visualizing the Trigonometric Functions}
It is often useful to have a graph of the trigonometric functions in your
mind so that you can do simple sense checks for your calculations.
Using the table of the sine and cosine function that you (dutifully)
filled in, we can make an initial plot of the sine function:
\begin{center}
\includegraphics[angle=90,width=9cm,height=5cm]{graphics/sine2.eps}
\end{center}
Again, to fill in the data points all we need to is to obtain
a protractor to help you draw triangles with various known
angles. Measure each triangle's base height and hypotenuse.
The sine is just the height divided by the hypotenuse. Record
the points in a table, fill in the graph. If you were to do
this, your plot would look something like
%\begin{figure}[ht]
\begin{center}
\includegraphics[width=7cm,height=3cm]{graphics/sin.eps}
\end{center}
The function has a maximum value of 1,
a minimum value of -1, and an average value of zero. It is
periodic with period $2\pi$.
The next plot is of cos(x). It has the same maximum, minimum,
average, and period as sin(x). Indeed, it is merely a sine wave shifted
a quarter of a cycle.
\begin{center}
\includegraphics[width=7cm,height=3cm]{graphics/cos.eps}
\end{center}
The following is a plot of tan(x). It has a period of $\pi$
and has a vertical asymptote at $\pi/2$ and every $\pi$ thereafter.
\begin{center}
\includegraphics[width=7cm,height=3cm]{graphics/tan.eps}
\end{center}
\section{Some Applications}
Trigonometry underlies just about everything remotely connected
to ``high-tech.'' Trigonometry is used in everything
from building bridges to finding stars to surveying. We could come
up with literally thousands of examples, and it would still fall
woefully short of the power of trigonometry.
We'll start with an example from optics.
\begin{APP}
Different materials have different optical properties. One parameter
that demonstrates this is refractive index. For example, air has
a refractive index of 1.0, water is 1.33, and most glass is about 1.5.
When light strikes a boundary between two different media, some of the
light is reflected off and some is transmitted through (or, ``refracted'').
The light that is reflected bounces off at the same angle it hits with
much like a billiard ball hitting a pool table cushion (this is the
``law of reflection''). The light that
is refracted does not go straight through - it goes at an angle given
by Snell's Law:
\EQ
n_1 sin(\theta_1 ) =
n_2 sin(\theta_2 )
\EN
If sunlight hits a lake at $37 \degrees$ (from the normal), what angle does it
travel through the water?
In this case, Snell's Law is
\EQ
n_{water} sin(\theta_{water} ) =
n_{air} sin(\theta_{air} )
\EN
or
\EQ
1.33 sin(\theta_{water} ) =
sin(37 \degrees )
\EN
Dividing both sides by 1.33 and taking the inverse sine yields
$\theta_{water} = 26.9 \degrees$ (from the normal).
While this is a trivial example, many scientists have spend their
entire lifetimes designing telescopes and optical fiber systems basically
using only these two laws.
Their field is known as ``Geometric Optics.''
\end{APP}
\begin{APP}
You see an airplane traveling in the sky at $18 \degrees$ from the horizontal.
One minute later it is at $30 \degrees$ from the horizontal. Assuming that
the airplane is traveling at about 550 miles/hour, what is the cruising altitude
of the airplane?\\
We begin by drawing a picture:
\begin{figure}[h]
\begin{center}
\includegraphics[width=9cm,height=3cm]{graphics/airplane.eps}
\caption{Determining the cruising altitude of an airplane.}
\end{center}
\label{airplane}
\end{figure}
Since the plane is traveling at 550 miles/hour for 1/60 of an hour,
we know that the distance it travels is $550/60 = 9.167$ miles.
With respect to the figure, this is $d_1-d_2$. We also know $A_2$ and $A_1$.
The obvious solution is to look at the triangle contained by the
three large dots. However, we only know one side and one angle
of this triangle, so the ``obvious solution'' doesn't work! Looking at the
two right triangles in the figure, it follows that
\EQ
\frac{h}{d_1} = tan(\theta_1) \Rightarrow d_1 = \frac{h}{tan (\theta_1)}
\EN
\EQ
\frac{h}{d_2} = tan(\theta_2) \Rightarrow d_2 = \frac{h}{tan (\theta_2)}
\EN
Again, we don't know either $d_1$ or $d_2$, but $d_1-d_2$ is known.
Solving these for $d_1-d_2$ yields
\EQ
d_1 - d_2 = \frac{h}{tan(\theta_1 )} - \frac{h}{tan(\theta_2 )}
= h[ cot(\theta_1 ) - cot(\theta_2 ) ]
\EN
and $h$ is the only unknown. Solving for $h$ yields
\EQ
h = \frac{d_1 - d_2}{cot(\theta_1) - cot(\theta_2)}
\EN
Plugging the given data into our derived formula yields
\EQ
h = \frac{9.167}{cot(18 \degrees ) - cot(30 \degrees )}
\EN
Many calculators do not have the cotangent function, so we must
rewrite the solution in terms of the tangent function:
\EQ
h = \frac{9.167}{1/tan(18 \degrees ) - 1/tan(30 \degrees )}
\EN
Of course we must be sure our calculator is in ``degrees mode'' and
not ``radian mode.'' If the calculator gives no indication, it is
in degrees mode. The result of the calculation is 7.29 miles.
\end{APP}
\section{Trigonometric Identities}
There are a number of mathematical identities relating the trig functions
to each other. In this section we will examine some of these.
A restatement of the Pythagorean Theorem in terms of trig
functions is
\EQ
sin^2 (\theta ) + cos^2 (\theta ) =1
\EN
which is obtained by applying the definitions of sine Eq. (\ref{sine_def})
and cosine Eq. (\ref{cosine_def}) into Eq. (\ref{pythag_ratio}).
As discussed, the trig functions are the same for $45 \degrees$
as $405 \degrees$. We can add or subtract any (integer) number of
$360 \degrees$'s (or $2\pi$'s). Put another way,
the sine and cosine functions are periodic with period
$2\pi$, so that
\EQ
sin(\theta \pm 2\pi ) = sin(\theta)
\EN
\EQ
cos(\theta \pm 2\pi ) = cos(\theta)
\EN
\EQ
tan(\theta \pm \pi ) = tan(\theta)
\EN
Similarly, it can be seen from the graphs that
\EQ
sin(\theta \pm \pi ) = -sin(\theta)
\EN
\EQ
cos(\theta \pm \pi ) = -cos(\theta)
\EN
\EQ
tan(\theta \pm \pi/2 ) = -cot(\theta)
\EN
Another thing to notice from the plots of the trigonometric functions is
that the sine function is merely a cosine function phase shifted
(and vice-versa) so that
\EQ
sin(\theta \pm \pi/2 ) = \pm cos(\theta )
\EN
\EQ
cos(\theta \pm \pi/2 ) = \mp sin(\theta )
\EN
Looking at the graphs of the trigonometric functions, it can be seen that
sine is an odd function (anti-symmetric about the y-axis)
and cosine is an even function (symmetric about the y-axis) so
that
\EQ
sin(-\theta) = - sin(\theta)
\EN
\EQ
cos(-\theta) = cos(\theta)
\EN
\EQ
tan(-\theta) = -tan(\theta)
\EN
The previous identities merely involve understanding
how sine and cosine work. Therefore, most of them do not require
memorization as long as you remember what the plots of sine and cosine
look like. Some less obvious, but extremely
powerful identities are the identities for a sum of angles:
\EQ
cos(A \pm B) = cos(A)cos(B) \mp sin(A) sin(B)
\EN
\EQ
sin(A \pm B) = sin(A)cos(B) \pm cos(A) sin(B)
\EN
\EQ
tan(A \pm B) = \frac{tan(A) \pm tan(B)}{1 \mp tan(A) tan(B)}
\EN
You should note that the previous identities (except the first one) can
be obtained from these last three ``sum'' identities.
There are many other trigonometric identities, but these are the
basic ones.
We now have a good understanding of right triangles, and we've
looked at some applications in physics. As an example
of geometry, we apply these results to a higher-order figure -
the non-right triangle.
\section{Non-right Triangles}
We could build an arbitrary non-right triangle out of two
triangles and proceed to determine relationships between
the non-right triangles angles and sides. If we were to do
this we would obtain the ``law of sines'' and the ``law of cosines:''
\begin{center}
\begin{tabular}{|l|c|}
\hline
law of sines &
$\frac{a}{sin(A)} =
\frac{b}{sin(B)} =
\frac{c}{sin(C)} $\\
\hline
law of cosines &
$c^2 = a^2 +b^2 -2ab \cos (C)$\\
\hline
\end{tabular}
\end{center}
\noindent
where the length of the side $a$ is on the opposite side of the angle $A$.
Same applies to the length $b$ and the angle $B$, and the length $c$
and the angle $C$.
Furthermore, we could obtain the formula for the area of a non-right
triangle:
\EQ
Area = \sqrt{s(s-a)(s-b)(s-c)}
\EN
where
\EQ
s = \frac{a+b+c}{2}
\EN
This area formula was first obtained by Hero of Alexandria (65-125 A.D.),
and is sometimes called ``Heron's Formula.''\footnote{He is known both
as Hero and Heron.}
Disregarding area and perimeter for a moment, a triangle has six parameters -
three sides and three angles. The basic result of the sine and cosine
laws is that given any three (independent) pieces, we can determine the
other three.\footnote{We can't find the lengths of the sides if only
the three angles are given (because we could double the size of the
triangle and the angles would be the same). The only other thing we
can tell is if the person telling us is completely full of baloney.
They are if the angles don't add up to $180 \degrees$!}
In some cases, if two angles and a single side is given, there are
two possible triangles, and these laws can be used to find the
parameters of both.
\begin{EX}
Determine everything about the following triangle:
\begin{figure}[h]
\begin{center}
\includegraphics[width=7cm,height=3cm]{graphics/nonright.eps}
\caption{SAS - side/angle/side are given.}
\end{center}
\label{nonright}
\end{figure}
\end{EX}
\begin{SOL}
Since we do not have an angle with a corresponding opposite side
given, we cannot use the law of sines. From the law of cosines
it follows that the length of the hypotenuse can be obtained from
\EQ
c^2 = 1^2 +1^2 -2 \cos (147 \degrees ) = 3.67734 \;ft^2
\EN
Square rooting both sides yields
\EQ
c = 1.9176 \;ft
\EN
After making calculations
of this sort, one MUST examine the answer for reasonableness (a reality check).
In this case, we know that the answer less than 2 (if given angle was $180\degrees$
instead of $147\degrees$) and greater than $\sqrt{2}\approx 1.414$ (if given angle
was $90\degrees$). The answer of $1.9176$ is indeed between these two values.
From symmetry, both the other angles are the same. Since the sum of the
angles of a triangle must add up to $180 \degrees$, the unknown angles are each
\EQ
\frac{180\degrees - 147\degrees}{2} = 16.5 \degrees
\EN
Without the insight of symmetry, we could use the law of sines:
\EQ
\frac{sin(\theta)}{1} =
\frac{sin(147\degrees)}{1.9176} = 0.284021
\EN
Taking the inverse sine of both sides yields
\EQ
\theta = sin^{-1}(0.284021) = 16.5\degrees
\EN
which is the same as calculated above.
Finally, the area of the triangle can be found by first finding $s$:
\EQ
s = \frac{1+1+1.9176}{2} = 1.9588
\EN
Then, from Heron's Formula, the area is\footnote{Following the same
logic in the previous reality check, you should be able to show that
the area must be between 0 $ft^2$ and 0.5 $ft^2$.}
\EQ
A = \sqrt{(1.9588)(0.9588)(0.9588)(1.9588-1.9176)} = 0.2724\;ft^2
\EN
\includegraphics[width=7cm,height=3cm]{graphics/nonright2.eps}
\end{SOL}
\section{Closing Notes}
\subsection{Summary}
Trigonometry is computational geometry. Since complicated figures can
be broken into right triangles, results learned about right
triangles can be applied to these figures. This is why one would want
to study trigonometry (we want to study complicated figures).
However, this is only half of the reason. Trigonometry
can be applied to a variety of physical situations where there are no
figures or triangles present. Recall the problem of finding the height
of a pyramid. We created imaginary triangles that only existed in our
minds to solve the problem. So, if we want to study the physical world,
we are motivated to study trigonometry.
We started by examining the relationships between the lengths of the sides and
angles of right triangles. How did we do it? We used two steps. First,
we noticed that when we draw two sides of a triangle, we have no choice as
to the length of the third side. Therefore, there must be some relationship
between the sides of a right triangle. So, now that we know that such
a relationship exists, we merely need to draw lots of triangles, make a
table of the lengths of sides, and notice a pattern. Now, this may
take a little time (somewhere between half an hour to ten hours, but
NOT a lifetime or anything), but it is certainly a reasonable way to proceed. When
we are finished we will have ``discovered'' the Pythagorean Theorem.
Now, we want to determine a relationship between the angles of a right
triangle. What do we do? Same thing - draw some triangles. It should
be clear very quickly that the interior angles of a right triangle (or
for any triangle for that matter) add up to $180 \degrees$\footnote{The
proof is not very hard. I thought it up in about 10 seconds. Just
bisect an isosceles triangle.}.
Thus, given the lengths of any two sides of a right triangle, we can
determine the length of the remaining side. Given one of the angles
of a right triangle we can determine the other (the third angle is always
$90 \degrees$). What if we are given one angle and one side?
I hope you agree that the sensible thing to do is to look at many (different
sized) triangles with the same angles - these are similar triangles. Drawing
a few and making a table with the lengths of the sides, it immediately
can be seen that the ratio of any two sides is constant - this is the
law of similar triangles.
Next, we give these ratios names. For example,
the ratio of the opposite side to the adjacent side is the tangent function.
Again, we draw triangles with different angles and make a table of different
angles vs. the tangent function. We do the same for the other trigonometric
functions. We then make many copies of this so it doesn't get lost. We
keep these tables with us whenever we want to calculate anything. This
keeping of ``sacred'' tables is a time honored tradition in mathematics that
still exists today. For example, you may have a copy of these trigonometric
tables with you now. They are stored in your calculator.
A right triangle has three sides and two independent angles, or five
pieces of information. Using the Pythagorean theorem, the sum of angles
being $180 \degrees$ rule, and the trigonometric functions, we can determine
all five given any two.
We can apply this ability to an absolutely insane number of problems
in geometry and physics. As examples, we looked at non-right triangles (geometry)
and the calculation of the height of a pyramid (physics).
\subsection{What more could we do?}
We've used {\bf inverse trig functions} in the examples, but haven't talked
about them much. One thing to keep in mind is you have to be careful
sometimes since their graphs are not single-valued. What's the inverse
cosine of 3? Well, friend there's an infinite number of answers, which
one do you want? If this bothers you, don't worry too much, just use
the answer your calculator tells you. If we were to continue studying
trig, the very next thing to do would
be to study the inverse trig functions.
{\bf Vectors} go with trigonometry like peas and carrots, though
sometimes vectors can be used instead of trigonometry. For example,
it's relatively easy to derive the law of sines and law of cosines
using vectors instead of drawing triangles. As another example, consider
this problem: Joe walks
2 miles north by northeast (picks up some lunch), then 1 mile southwest
(delivers some mail), and finally 1.5 miles south. Where should you
walk if you want to meet up with Joe? You can do it by drawing lots
of triangles and applying trig laws, but it's much easier using
vectors. Therefore, another important area we could explore is vector
algebra (and later vector calculus!).
We've already hinted at the use of polar coordinates, but
other {\bf coordinate systems} are also used extensively in
mathematics and physics. Conversion between these coordinate systems
involves heavy use of trig identities. As an example, if we wanted
to specify where a particular star was right now, we could specify an azimuth
angle (facing north, how many degrees do we have to turn to face the
star), an angle of elevation (what angle do we tilt our heads at
to see the star), and the distance to the star. This alternate method
of specifying the location of the star is called spherical coordinates.
One encounters coordinate systems in physics early and often,
we could also study this.
Most of the trig identities can be derived using the Euler relation -
a theorem that uses {\bf complex numbers} to relate exponentials to
the trig functions.
Complex numbers are often left out of many curricula, but you can't
do much AC circuit analysis without it.
The trigonometric functions can all be represented by a
{\bf ``Taylor Series''} - an infinite series of polynomials
(For example, $sin(x) = x - x^3/3! + x^5/5! + ...$).
One of the things you should have learned is that all problems
cannot be solved in closed form. In physics, we don't care, we
still need to know the answer. Taylor series help us out.
An arbitrary periodic function can be represented as an infinite
sum of sine and cosine function as a {\bf ``Fourier Series''}.
If you want to look at data, you should want to look at Fourier
Series. My personal opinion is that every biologist (who collects data)
should be exposed to it.
\subsection{We've conquered trigonometry, now what?}
If trigonometry is generally regarded as daunting, calculus is thought of as
unlearnable and uninteresting. Actually, calculus is simple and profound.
Basically, calculus comes in two flavors, differential calculus and
integral calculus. Differential calculus comes from studying slopes of
curves and integral calculus comes from examining areas under curves.
Think about what the speedometer on your car is telling you. Draw some
velocity vs time graphs. Based on your graphs think about how to
draw corresponding position vs. time graphs and acceleration vs. time
graphs. If you take this seriously at all, you'll invent calculus
{\it whether you want to or not}. The same types of steps we've used
here to reason out how trigonometry can be used for calculus as well,
but that's another story...
\vspace{1 cm}
\noindent
Here is the answer to the Pythagorean Triplet problem.\footnote{The
next triplet is $14^2 + 48^2 = 50^2$. This follows since
if $a^2 + b^2 = c^2$, then (multiply both sides by 4 yields)
$(2a)^2 + (2b)^2 = (2c)^2$. The $14$ triplet is just double the $7$ triplet.}
\end{document}