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\begin{document}

\section{ Trigonometry - Basic Ideas}
  The origins of trigonometry lie in the efforts of the ancient Babylonian and Greek astronomers and astrologers to understand the motions of the sun and the visible planets.  Later trigonometry became an essential tool  for navigation and surveying.  Today trigonometry has immense applications in nearly every area of technology. For instance, trigonometry is essential to our understanding of sound, how to record it, and how to transmit it.   Compression techniques for both audio and  video signals rely on being able to represent signals in terms of trigonometric functions.  We simply could not we do much of what we do without trigonometry.   


Trigonometry can be thought of  in two ways , first as  the study of the relationships  between the lengths of the sides of  a triangle  as a function of its angles   or  second  as the  mathematics of  a circle.  Both perspectives  are useful.
We begin with a very basic question.  What is an angle?  We all have  ideas.  Given two lines that intersect,  we can speak of the angle between them as a method  of  measuring   how close the two lines are to one another.  How would we define such a  measure?   We could define it as the number one would get by measuring  the separation with a protractor. But this depends on the precision of the protractor  and although it is essential in practical situations it  is not a method that can be incorporated into mathematical  methods. 

\subsection{Angles  }

	Given an arbitary line and a point  $O$  on the line, the point $O$   divides the line into two half lines   which start at $O$.  These half lines are called {\it rays.} A second point, say $P$, on the given line  will determine one of the rays, which is denoted as  $\overrightarrow{OP}.$   An {\it angle}  consists of two rays, $\overrightarrow{OP}$  and $\overrightarrow{OQ}$ with the same starting point.  Usually these rays are derived from different lines.  The job  is to describe a method  of  measuring the separation between the two rays. 
	


	 This is what is done.  Let one of the rays say $\overrightarrow{OP} $ correspond with the positive portion of the real line and such that the point $O$  is the origin  $(0,0)$ of $\R^2$.  Next consider a circle of radius 1 centered at the origin.     Let  $P$ be chosen so that $P =(1,0) $  and let $Q$ be chosen so that  $Q =(x,y)$  is the intersection of the second ray 
$\overrightarrow{OQ}$ with the circle.  The angle can now be represented by the three points $P, O,$  and 
$Q$   as    $\angle POQ.$  

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
	   \centering
	   \includegraphics[width=3in]{angle.jpg} 
	   \caption{ $\angle POQ$  measured positively as the length of the connecting counter-clockwise arc  or negatively as the length of the  clockwise arc. }
	   \label{fig:angle}
	\end{figure}



Starting at $P$  the circle can be traversed in two ways to arrive at the point $Q$ - in a counter-clockwise fashion or in a clockwise fashion.  The corresponding arcs connecting $P$  with $Q$  can be referred to as the counter-clockwise arc and the clockwise arc.  The measurement of the angle determined by the two rays  can then also be done in two ways.  It can be measured positively by the length of the counter-clockwise arc  or it can be measured negatively as the length of the clockwise arc.  That is,  measuring counter-clockwise gives a positive number and measuring clockwise gives a negative number.  This measurement according to the length of arcs on the unit circle  is called {\it radian measure.}

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{rightangle.jpg} 
   \caption{An angle with positive counterclockwise radian measure $\frac{\pi}{2}$  and negative clockwise radian measure 
   $-\frac{3\pi}{2}$ }
   \label{fig:example}
\end{figure}

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{180deg.jpg} 
   \caption{an angle with positive radian measure $\pi$  and negative clockwise radian measure $-\pi$}
   \label{fig:example}
\end{figure}


Since the circumference of the unit circle is $2\pi,$   the circumference of a quarter circle is 
$\frac{\pi}{2}.$  Thus if the point $Q$  is located at the point $(0,1),$  angle $\angle POQ$  is said to be  a right angle, and its   Its  positive counterclockwise radian measure  is  $\frac{\pi}{2}.$   where as its negative  measure as the length of the connecting clockwise arc which  is $- \frac{3\pi}{2}$.   If $Q = (-1, 0),$  angle $\angle POQ$  is $180^o$  and has positive radian measure $\pi$,  since the connecting arc is a half circle.   In this case the  clockwise arc is also a half circle, so that the negative radian measure is $-\pi .$

\subsubsection{Conversion between degrees and radians}

When we measure angles with degrees,  think of the circumference of the unit circle as divided into 360 equal segments.  Each of these is further divided into 60 equal subunits called minutes and each minute is further divided into 60 subunits called seconds.   This is what we understand.  Using ratios we can develop a method for converting from degree measure to  radian measure and conversely from radian measure to degree measure.  Now it is clear that the length of a circular arc divided by the circumference is the same no matter what style of measurement we decide to use.  We could measure in kilometers or in microns;  the ratio of arc length to circumference would be the same.  So, if we measure by degrees or by radians the fraction $\frac{arc \: length}{circumference}$   is the same.   Thus, if $d$  is the degree measurement of an angle and $r$  is the radian measure,  we know that 
$$\dfrac{d}{360}  = \dfrac{r}{2\pi} $$

Thus, for instance,  converting an angle of $60^o$ to radian measure, gives

$$\dfrac{60}{360} = \dfrac{r}{2\pi}.$$
Solving for $r$ we have $r = \frac{\pi}{3}.$

\subsubsection{Generalized notion of angle}

The description of an angle as two rays joined at their initial points  has useful generalizations.  What we have shown is that  an angle  $\angle POQ$  can be identified with two arcs on the unit circle, a counter-clockwise arc and a clockwise arc  and that the length of these arcs  are the radian measures of the angle. The length of the counter-clockwise arc is represented as a positive number and the length of the clockwise arc as a negative number.  Thus any angle as we have described it gives us two numbers a number $\theta$  where $0 \le \theta \le 2\pi$  representing the length of a counter-clockwise arc and the number $-(2\pi - \theta) = \theta -2\pi$   which is the  length of the corresponding clockwise arc.  Note, that we allow for the possibility of an arc zero length or of length  $2\pi.$ 

We know that an angle can be identified with two arcs on the unit circle.   In many applications it is useful to alter slightly the definition of what we mean by the word  ``angle'',  to refer simply to an arc on the circle  starting at $P = (0,1) $ and continuing to a point $Q$ - in either a counter-clockwise or a clockwise direction.   Of course any such arc determines an angle previously understood  so this is not a big change.  It does however now offer the advantage that the word angle now refers to only one entity - either a positively measured counter-clockwise arc or the negatively measured clockwise arc.   In this way an angle can now be represented uniquely by the measure of its length as an arc on the circle; that is  as a number $\theta,$    where $-2\pi \le \theta \le 2\pi.$ This is the first generalization.   

The second generalization  has arisen from the need to represent multiple revolutions about the circle which are followed by an angle as we presently know it.  For instance 2 full revolutions in the counter-clockwise direction followed by an angle of  $90^o$  is  represented by the length of  the path traversed by a point which starts a  $P$ and takes two full revolutions followed by a quarter revolution. The path length is $2\pi + 2\pi + \frac{\pi}{2} = 4\pi + \frac{\pi}{2}.$   

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{generalized_angle.jpg} 
   \caption{an angle consisting of one counter-clockwise complete revolution plus the poriton from $P$ to $Q$}
   \label{fig:generalized_angle}
\end{figure}


In this way any real number,  whether positive or negative,  can be used to represent such a path.  How is this done?   Let $x$ be a real number; divide $x$  by $2\pi$  to find how many full revolutions of the circle it represents.  The remainder after the division represents the length of the arc that is tacked onto  the sequence  of full circle revolutions.  For instance if  $x = 45$ radians,  dividing by $2\pi$  gives $$\frac{45}{2\pi} \approx 7.163 = 7 +0.163 \: \:{\rm radians.} $$ 
Multiplying both ends of the equation by $2\pi$ gives 
$$45 \approx 7 \times 2\pi + 2\pi \times 0.163  \approx  7(2\pi) + 1.024$$
Thus $45$  in radian measure represents aproximately  $7$ counter-clockwise revolutions plus a partial revolution of 1.024 radians.

\subsection{Right triangle trigonometry}

We begin the study of trigonometric  functions  in a special setting in which the angles considered are less than $90^o$  and form one of the angles of right triangle.  Later this can be easily generalized to the case of a generalized angle  whose measure is an arbitrary real number.  Given a right triangle 
$ \Delta POQ $   whose right angle is at the vertex $P$  and $\angle POQ = \theta ,$  the various trigonometric functions describe relationships between the sides of the triangle as a function of $\theta.$ The hypotenuse is the side $OQ$; the side $PQ$ which opposite the angel $\angle POQ $  is  said to be the {\it opposite}  side, and the side $OP$ is called the {\it adjacent}  side. See figure 
\ref{fig:right_triangle}.  If $\theta =0.$ we assume $Q=P$  and the triangle degenerates to the line segment $OP.$    Similarly if 
$ \theta = \frac{\pi}{2} $ we assume that $O=P$  and the triangle degenerates to the line segment 
$OQ.$   We also assume a notational device;  if $A$  and $B$  are any two points the length of the line segment $AB$  is denoted $\overline{AB}.$

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{right_triangle.jpg} 
   \caption{right triangle}
   \label{fig:right_triangle}
\end{figure}


\begin{definition}  Given a right triangle as above,  with the above conventions the  trigonometric functions, { \it sine, cosine, tangent, cotangent, secant, cosecant,} are defined as follows for 
$0 \le \theta \le \frac{\pi}{2} .$

	\begin{enumerate}
	\item  sine of $\theta$   :    $\sin \theta = \dfrac{  \overline {PQ}}{\overline{OQ}} = \dfrac{opposite}{hypotenuse}$
	\item  cosine of $\theta$ :  $\cos \theta = \dfrac{\overline{OP}}{\overline{OQ}} = \dfrac{adjacent}{hypotenuse}$
	\item tangent of $\theta$ :  $\tan \theta = \dfrac{\overline{PQ}}{\overline{OP}} = \dfrac{opposite}{adjacent},$   for $\theta \ne \frac{\pi}{2}.$
	\item cotangent $\theta$ :   $ \cot \theta = \dfrac{\overline{OP}}{\overline{PQ}} = \dfrac{adjacent}{opposite}, $  for $\theta \ne 0$
	\item  secant of $\theta$  :   $  \sec \theta = \dfrac{\overline{OQ}}{\overline{OP}}= \dfrac{hypotenuse}{adjacent},$  for $\theta \ne \frac{\pi}{2}.$	
	
	\item  cosecant of $\theta$  :   $  \csc \theta = \dfrac{\overline{OQ}}{\overline{PQ}}= \dfrac{hypotenuse}{opposite}, $       for $\theta \ne 0.$
	
	\end{enumerate}
\end{definition}

Note that for $\theta = \frac{\pi}{2}$  the triangle $ \Delta POQ $  collapses to the straight line segment
 $OQ$    so that $\overline {OP} =0, $  and therefore the restrictions on the definition of tangent and secant are necessary.   Similarly if $\theta = 0$,  the triangle collapses to the line segment $OP,$ so that $\overline{PQ}$  becomes zero, and the restrictions on the definitions of cotangent and cosecant are again necessary.


The trigonometric functions of angles,  $\theta$  equal to $0,  \frac{\pi}{2},   \frac{\pi}{4}, \frac{\pi}{3}$  and $\frac{\pi}{6}$  can be easily calculate  by considering the triangle $\Delta POQ$  and assigning appropriate lengths to various sides.  Specifically, we have the following.

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=2in]{60_degree.jpg} 
   \caption{If $\theta = \frac{\pi}{3},$  then $\Delta POQ$ is half an equilateral triangle}
   \label{fig:60_degree}
\end{figure}



\begin{remark}{\rm \bf Basic Results}
\begin{enumerate}\rm
\item  if $\theta =0,$  then $\overline{PQ}  =0,$  so $\sin 0 =0$  and $\tan 0 =0.$
\item  if $\theta = \frac{\pi}{2},$  then $\overline{OP} =0,$  so 
	$\cos \frac{\pi}{2} = 0$  and $\cot \frac{\pi}{2} =0.$
\item  if $\theta = \frac{\pi}{4} = 45^o, $  then $\Delta POQ$   is a right triangle  and we choose 
	$\overline{OP} =1 = \overline{PQ},$  then the hypotenuse $\overline{OQ}$  has length $\sqrt{2}.$  	And, knowing all the dimensions of the triangle, the values of all the trigonometric functions can be 	calculated.  In particular 
	\begin{enumerate}
		\item $\tan \frac{\pi}{4} = 1 = \cot \frac{\pi}{4},$   
		\item $\sin \frac{\pi}{4} =\frac{1}{\sqrt{2}} = \cos \frac{\pi}{4} $
	
	\end{enumerate}
\item  if $\theta = \frac{\pi}{3} = 60^0,$  then $\Delta POQ$  is half of an equilateral triangle,  and if we chose $\overline{OQ} $  to be of length $2,$   it follows that $\overline{OP} = 1,$   and by the pythagorean theorem  $\overline{PQ} = \sqrt{3}.$   See figure \ref{fig:60_degree}.     


\end{enumerate}

\end{remark}


Following is an example of how these ideas may be immediately applied.
\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{CNtower.jpg} 
   \caption{Find the height of the CN tower }
   \label{fig:CNtower}
\end{figure}


\begin{example}  {\rm A George Brown College student with surveying equipment has determined that  the shadow  of the CN tower downtown Toronto  is $3,200$ feet  long  at a time that the elevation of the sun was  $29.5^o.$  How high is the tower?}
\end{example}


{\bf Solution:}    
Let $h$  stand for the height of the tower. Then we must have that $\tan29.5  = \dfrac{h}{3210}.$  Solving for $h$  gives,
$$h = 3210\times \tan 29.5 =3210\times 0.56577 = 1816.1.$$
Not a bad result.  The true height is 1,815 feet  four inches.
\endproof

In the next section we show how to extend the definitions of the trigonometric functions  to the case in which the angle $\theta$  can be any real number.

\subsection{ Trigonometric functions of arbitrary angles}

For the definitions of trigonometric functions given in the previous section,  the various ratios of one side to another  of the triangle $\Delta POQ $  were independent of the size of the triangle.  Thus had we wished we could have chosen a triangle of some standard size.  This is what will do here.  
\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{trig_def.jpg} 
   \caption{Trigonometry definitions with standard triangle}
   \label{fig:trig_def}
\end{figure}

For the following definitions assume that the triangle is such that the hypotenuse $OQ$  has length 1.  See figure \ref{fig:trig_def}.  Note the similarity of \ref{fig:trig_def}  to that of figure \ref{fig:angle},  the difference being that the point $P$ has now been moved along the x axis so that there is a right angle at $P$.

 Since the hypotenuse $OQ$ has length 1,  the definitions of the trigonometric functions become simpler.  The sine of the angle $\angle POQ$  is simply the y coordinate of the point $Q$, the cosine  is the x coordinate,  the tangent is the ratio $\dfrac{y}{x},$  and so on.

This definition allows a simple extension that lets us define trigonometric functions of an arbitrary angle. As we have seen, an arbitrary angle $\theta$, which may include several complete rotations in either in the counter-clockwise  or the clockwise directions,  is determined  point $Q=(x,y)$  anywhere on the unit circle. We then define the trigonometric functions of such an angle in the same way.  That is,  

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{trig_def2.jpg} 
   \caption{Trigonometric definitions for an arbitrary angle}
   \label{fig:trig_def2}
\end{figure}


\begin{definition}   In relation to figure \ref{fig:trig_def2},  the trigonometric functions of an arbitrary angle 
$\theta$  are defined as follows.
\begin{enumerate}
	\item $\sin \theta =y$
	\item $\cos \theta =x$
	\item $\tan \theta = \dfrac{y}{x}$ for $x \ne 0$
	\item $\cot \theta = \dfrac{x}{y}$  for $y \ne 0$
	\item $\sec \theta = \dfrac{1}{x}$   for $x \ne 0$
	\item $\csc \theta = \dfrac{1}{y} $ for $y \ne 0$
	\end{enumerate}
	\label{trig_definitions}

\end{definition}

We remark  that if $Q$  lies in the first quadrant then it's coordinates range from $0$  to $1,$  whereas if $Q$  lies in the second quadrant,  the $x$ coordinate ranges from $0$  to $-1$  while the $y$ coordinate is still positive and ranges from $0$  to $1$.  In the third quadrant,  both coordinates range from $0$  to $-1$, and in the fourth quadrant $x$  ranges from $0$ to $1$  and $y$  ranges  from $0$  to $-1$.

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{sine.jpg} 
   \caption{graph of $\sin \theta$  for  $0 \le \theta \le 2\pi$ }
   \label{fig:sine}
\end{figure}

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{minus_sine.jpg} 
   \caption{Showing $\sin(-\theta) = -y = -\sin \theta $}
   \label{fig:minus_sine}
\end{figure}

\subsection{Graphs of the trigonometric functions}
The graphs of the trigonometric functions are an important tool in understanding what goes on.  The graphs can be visualized by slowly considering what happens to the x  and y  coordinates of the point $Q =(x,y) $  as it moves around the circle.  
In the case of the sine function, the y  coordinate of $Q$  begins at $y = 0$  when  the corresponding angle $\theta = 0$  and $Q$ corresponds to the point
$P = (1,0).$   As $\theta$  increases,  the y  coordinate increases  until it reaches a value of $y=1$ and the point $Q$  corresponds to the point $(0,1)$  on the  y  axis.  Then as $\theta$ moves from $90^o$ or 
$\frac{\pi}{2}$  radians  to $180^o$ or $\pi$ radians, the y coordinate decreases to a value of $0$.  Finally, as $\theta $  moves from $\pi$  to $360^o $  or $\ 2\pi$ radians,  the y coordinate behaves in the same fashion as before,  except that now the values are negative.  The graph for $0 \le \theta \le 2\pi$  can be seen in figure \ref{fig:sine}.

\begin{figure}[h] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{sine2pi.jpg} 
   \caption{Graph of $\sin \theta$  for $-2\pi \le \theta \le 2\pi$}
   \label{fig:sine2pi}
\end{figure}

Now what about negative values of $\theta$?   Suppose the angle $\theta$  is measured positively in the counter-clockwise direction and is determined by the point $Q = (x,y).$    Then $-\theta$  is measured in the clockwise direction.  It is determined by the point $\widetilde{Q} = (x, -y)$; see fiigure 
\ref{fig:minus_sine}.  
This allows us to expand the graph to include negative values of $\theta; $    see figure \ref{fig:sine2pi}




Further for angles greater than $2\pi$  or less that $-2\pi$  the values of the trigonometric functions of such angles are also determined by the coordinates of a point  on the unit circle.  For instance  if $\theta  = n \times 2\pi  +  \omega,$   then 
$\theta$  consists of $n$ complete revolutions followed by an angle $\omega$   in the range 
$0 \le \omega \le 2\pi.$   Definition \ref{trig_definitions}  tells us that $\sin \theta$    is   defined  as the y coordinate of the point $Q$  on the unit circle  which had been determined by the angle $\omega. $   

Saying this in a slightly different way,  the definitions of the trigonometric functions of an arbitrary angle first expressed in Definition  \ref{trig_definitions}  may be rephrased as follows .

 \begin{definition} Given an angle $\omega,$ where $0 \le \omega \ 2\pi,$ a positive integer $n$  and a trigonometric function $f$
	\begin{enumerate}
	\item if $ \theta = 2n\pi   + \omega,$   then $f( \theta ) = f (\omega)$
	 \item if $\theta =- 2n\pi  - \omega,$   then $f(\theta ) = f( - \omega). $
	 \end{enumerate}
\end{definition}

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=4in]{sine_graph.jpg} 
   \caption{Graph of the sine function on $[-6\pi, 6\pi]$}
   \label{fig:sine_graph}
\end{figure}


In other words, in each interval 
$[2\pi, 4\pi], [4\pi, 6\pi], \cdots, [2n\pi,  2(n+1)\pi], \cdots,$  
the 6 trigonometric functions replicate  their behavior on the interval $[0,2\pi]$.  Likewise  on each of the intervals 
$[-4\pi, -2\pi], [-6\pi, -4\pi], \cdots, [-2(n+1\pi, -n\pi] \cdots,$   
they replicate the behavior on $[-2\pi,0]$.   In particular the graph of the sine function on the interval 
$[-6\pi, 6\pi]$ is as seen in figure   \ref{fig:sine_graph}.

\subsubsection{ The graph of the cosine function}




As with the sine function  to visualize the graph of the cosine function we need to examine the values of the x coordinate of the point $Q =(x,y) $  
as it moves about the unit circle.  Let  $\theta$ let be the counter-clockwise angle made by the ray
$\overrightarrow{OQ}$ 
with the positive x axis.  If $\theta =0,$  the point $Q$ corresponds to the point $(1,0),$ so that 
$\cos 0 = 1.$ 
As $Q$  moves  counter clockwise, the  x coordinate decreases until it becomes zero when 
$\theta = \frac{\pi}{2} = 90^o.$  Thus, 
$\cos \dfrac{\pi}{2}   =0$.  Then as 
$\theta$  goes from 
$\frac{\pi}{2}$   to $\pi ,$ 
 the x coordinate goes from 
 $0$  to $-1,  $  so that $\cos \pi =-1$.  Similar analysis shows that as $\theta$  goes from 
 $\pi$  to $\dfrac{3\pi}{2},\:\: \cos \theta$  goes from $-1$  to $0$,  and as $\theta$ goes then from 
 $\frac{3\pi}{2}$  to $2\pi,$  $\cos \theta $  goes form $0$ to plus $1.$  
 
  \begin{figure}[h] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{Maplecos.jpg} 
   \caption{Graph of the cosine on $[-2\pi. 2\pi]$}
   \label{fig:cosine}
\end{figure}
 
 Finally repeating the analysis for negative angles 
 $-2\pi  \le \theta \le 0,$  we arrive at the graph in figure \ref{fig:cosine}.  As with the sine function , the behavior of the cosine function  on intervals  $[2n\pi, 2(n+1)\pi]$ and  $[-2(n+1)\pi,  -2n\pi]$  replicates the behavior on  $[0, 2\pi]$  and $[-2\pi, 0].$  Figure \ref{fig:cos}  shows the graph of the cosine on the interval  
$[-6\pi, 6\pi]$ 

\begin{figure}[p] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=4in]{big_cos.jpg} 
   \caption{Graph of the cosine on the interval$[-6\pi, 6\pi]$ }
   \label{fig:cos}
\end{figure}

\begin{figure}[p] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{sin_cos.jpg} 
   \caption{ Graph of the cosine in red is a translate of the graph of the sine in green  and vice-versa}
   \label{fig:sin_cos}
\end{figure}	
\subsubsection{Beginning Identities}



In figure  \ref{fig:sin_cos}  we show the a portion of graph of the cosine function superimposed on a portion of the graph of the sine function.  Notice that the red colored graph of the  cosine   is simply a translation to the left by $\frac{\pi}{2}$ of the  green colored graph of   the  sine function.  Similarly, the sine function is a translation to the right by $\frac{\pi}{2}$   of the cosine function. We can express these observations analytically  as follows.

\begin{result} {\bf Translation Identities}
	\begin{enumerate}
	\item  $ \cos \theta =\sin(\theta + \frac{\pi}{2}) $
	\item $\sin \theta = \cos(\theta - \frac{\pi}{2}) $
	\end{enumerate}
\end{result}


Examining the graph of the sine function as in figure \ref{fig:sine_graph},  we see that the reflection about the y axis of   the portion of the sine graph for positive $\theta$ gives the negative of the graph of the sine for negative $\theta.$  In other words,  to obtain the portion graph of the sine for negative $\theta,$   take the portion  for positive $\theta,$   flip it about the y axis and then reflect the result about the x axis.

 As for the cosine function, examining figure \ref{fig:cosine}  we see that the reflection about the y axis  of the the portion of the graph for $\theta >0$  corresponds precisely with the graph of the cosine for negative $\theta$.  Said in another way, to obtain the portion of the graph of the cosine for negative $\theta,$  take the portion for positive $\theta,$ and simply flip it about the y axis.
 
 
     These ideas can be expressed as follows.

\begin{result}:  {\bf Reflection Identities}
	\begin{enumerate}
	\item For any $\theta, \: \:\sin(-\theta) = - \sin \theta$
	\item For any angle $\theta, \: \:   \cos(-\theta) = \cos \theta$
	\end{enumerate}
\end{result}	
	
	Reflecting back on the definition of the sine and cosine functions,  an angle $\theta$ is determined by a  point $Q = (x,y)$  on the unit circle  so that  $\sin \theta = y$   and $\cos \theta = x$.  However lets consider the triangle $\Delta POQ$  of figure \ref{fig:angle} where $0 \le \theta \le \frac{\pi}{2}.  $ Since this is a right triangle we then know that $x^2 + y^2 = 1$  by the Pythagorean theorem.   For other values of $\theta $  a similar triangle can be constructed .  The result is that  $\sin^2 \theta + \cos^2 \theta =1 $ 	
\begin{result}:  {\bf Pythagorean Identity}
For any angle $\theta,  \: \: \sin^2 \theta + \cos^2 \theta =1$	
	
\end{result}	

Again focusing on the definitions in terms of the coordinates of the point $Q = (x,y)$   determined by an angle $\theta,$ we know that$ \tan \theta = \dfrac{y}{x}   = \dfrac {\sin \theta}{\cos \theta}$   and 
 $\cot \theta = \dfrac{x}{y}   = \dfrac {\\cos \theta}{\sin \theta}$.  Similarly 
 $\sec \theta = \dfrac{1}{x} = \dfrac{1}{\cos \theta}$   and $\csc \theta = \dfrac{1}{y}  = \dfrac{1}{\sin \theta}.  $ This is summarized in the following statement.



\begin{result}:   {\bf Basic Identities}
		\begin{enumerate}
		\item $ \tan \theta =  \dfrac {\sin \theta}{\cos \theta}$  
		\item $\cot \theta   = \dfrac {\cos \theta}{\sin \theta}$
		\item $\tan \theta =  \dfrac{1}{\cot \theta}$
		\item $ \sec \theta  =  \dfrac{1}{\cos \theta}$
		\item $ \csc \theta = \dfrac{1}{ \sin \theta}$
		\end{enumerate}
\end{result}

	
\subsubsection{Graph of tangent, cotangent, secant, and cosecant}

\begin{figure}[p] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=3in]{tan1.jpg} 
   \caption{Graph of the tangent on the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$}
   \label{fig:tan1}
\end{figure}

\begin{figure}[p] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=4in]{tan2.jpg} 
   \caption{Graph of the tangent on an extended domain}
   \label{fig:tan2}
\end{figure}

The graphs of the remaining trigonometric functions can be derived from those of sine and cosine.  For instance, for the tangent function,  since  $\tan \theta  = \dfrac{ \sin \theta}{\cos \theta},$   we know:
	\begin{itemize}
	\item  if  $  \theta =0, $ then $\sin \theta =0$  and therefore $\tan \theta  = 0;$ 
  	\item  if $ 0 < \theta < \frac{\pi}{2},$  then both $\sin \theta $  and z$\cos \theta$  are positive, but as 
		$\theta $  gets close to $\frac{\pi}{2}, \: \: \cos \theta$  gets close to zero  whereas  $\sin \theta $  		gets close to $1$.  The result is that $\tan \theta $  approaches  $+ \infty$;
	\item  if  $-\frac{ \pi}{2} < \theta < 0$  a similar analysis tells us that $\tan \theta $  is negative and  		approaches $-\infty$  as  $\theta$  approaches $-\frac{\pi}{2}.$
	\end{itemize}	
The graph of the tangent function over the interval for $- \frac{\pi}{2}  \theta < \frac{\pi}{2}$  is shown in diagram \ref{fig:tan1}. Note that the graph becomes asymptotic to the vertical lines through $-\frac{\pi}{2}$  and $\frac{\pi}{2}$.

A similar analysis will show that the graph of the tangent will be the same on all adjoining intervals of length $\pi$ - namely the intervals $(\frac{\pi}{2},  \frac{3\pi}{2}), \: \: (\frac{3\pi}{2},  \frac{5\pi}{2}) \cdots$     and the intervals  
$(-\frac{3\pi}{2}, -\frac{\pi}{2}), \: \: (-\frac{5\pi}{2}, -\frac{3\pi}{2}), \cdots.$  The graph of the tangent over an extended domain is shown in diagram \ref{fig:tan2}.   Observe that  the $\cos \theta = 0$  if $\theta $  is a  multiple of  $\frac{\pi}{2}$ ;  that is  $\cos \frac{n\pi}{2} =0$  for all integers $n,$ whether they be positive or negative.  Consequently at all such points  $\theta, \: \:  \tan \theta $  is not defined  and the graph of the tangent is asymptotic to the vertical line through $\theta   =  \frac{n\pi}{2}$.

The graphs  of the cotangent, the secant, and the cosecant  can be derived similarly from those of the sine and the cosine.  The analysis is omitted and left as an exercise for the student . The graphs of the cotangent, secant and cosecant are shown in figures \ref{fig:cot}, \ref{fig:sec}, and \ref{fig:csc} respctively.  Observe that just as the graph of the cosine is a translation of the graph of the sine, so the graph of secant is a translation of the graph of cosecant.   Also observe that the distinct curves described by both the secant and the cosecant are tangent to the horizontal lines through $(0,1)$  and $(0, -1)$,  namely the lines $y = 1$  and $ y= -1$.

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=4in]{cot.jpg} 
   \caption{Graph of the cotangent}
   \label{fig:cot}
\end{figure}

\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=4in]{sec.jpg} 
   \caption{Graph of the secant}
   \label{fig:sec}
\end{figure}	
	
	
\begin{figure}[htbp] %  figure placement: here, top, bottom, or page
   \centering
   \includegraphics[width=4in]{csc.jpg} 
   \caption{example caption}
   \label{fig:Graph of the cosecant}
\end{figure}
	
\end{document}






















