\magnification=1100
\overfullrule0pt

\input amssym.def
\input prepictex
\input pictex
\input postpictex


% ********************* Definitions ************************************

%\def\widetilde{\mathaccent"0365 }

\def\CC{{\Bbb C}}
\def\FF{{\Bbb F}}
\def\HH{{\Bbb H}}
\def\NN{{\Bbb N}}
\def\OO{{\Bbb O}}
\def\QQ{{\Bbb Q}}
\def\RR{{\Bbb R}}
\def\ZZ{{\Bbb Z}}

\def\cA{{\cal A}}

\def\cB{{\cal B}}
\def\cR{{\cal R}}
\def\cZ{{\cal Z}}
\def\cC{{\cal C}}
\def\cR{{\cal R}}

\def\fb{\frak{b}}
\def\fg{\frak{g}}
\def\fh{\frak{h}}
\def\fn{\frak{n}}

\def\Card{\hbox{Card}}
\def\End{\hbox{End}}
\def\Hom{\hbox{Hom}}
\def\Ind{\hbox{Ind}}
\def\Id{\hbox{Id}}
\def\codim{\hbox{codim}}

\def\diag{\hbox{diag}}
\def\id{\hbox{id}}
\def\im{\hbox{im}}
\def\tr{\hbox{tr}}
\def\Tr{\hbox{Tr}}

\def\fbox{}


% ********************* FONTS ************************************

\font\smallcaps=cmcsc10
\font\titlefont=cmr10 scaled \magstep1
\font\titlefontbold=cmbx10 scaled \magstep1
\font\titlesubfont=cmr10 scaled \magstep1
\font\sectionfont=cmbx10
\font\tinyrm=cmr10 at 8pt

% ******************** SECTION HEADERS ***************************

\newcount\sectno
\newcount\subsectno
\newcount\resultno

\def\section #1. #2\par{
\sectno=#1
\resultno=0
\bigskip\noindent{\sectionfont #1.  #2}~\medbreak}

\def\subsection #1\par{\bigskip\noindent{\it  #1} \medbreak}

%******************* MATHEMATICAL LABELS **************************

\def\prop{ \global\advance\resultno by 1
\medskip\noindent{\bf Proposition \the\sectno.\the\resultno. }\sl}
\def\lemma{ \global\advance\resultno by 1
\medskip\noindent{\bf Lemma \the\sectno.\the\resultno. }
\sl}
\def\fact{ \global\advance\resultno by 1
\medskip\noindent{\bf Fact \the\sectno.\the\resultno. }
\sl}

\def\remark{ \global\advance\resultno by 1
\medskip\noindent{\bf Remark \the\sectno.\the\resultno. }}
\def\example{ \global\advance\resultno by 1
\medskip\noindent{\bf Example \the\sectno.\the\resultno. }\sl}
\def\cor{ \global\advance\resultno by 1
\medskip\noindent{\bf Corollary \the\sectno.\the\resultno. }\sl}
\def\thm{ \global\advance\resultno by 1
\medskip\noindent{\bf Theorem \the\sectno.\the\resultno. }\sl}
\def\defn{ \global\advance\resultno by 1
\medskip\noindent{\it Definition \the\sectno.\the\resultno. }\slrm}
\def\endthm{\rm\medskip}
\def\thmend{\rm\medskip}
\def\endlemma{\rm\medskip}
\def\endfact{\rm\medskip}
\def\endexample{\rm\medskip}
\def\endprop{\rm\medskip}
\def\endcor{\rm\medskip}
\def\pf{\rm\smallskip\noindent{\it Proof. }}
\def\endpf{\qed\hfil\medskip}
\def\pfend{\qed\hfil\medskip}
\def\note{\smallbreak\noindent{Note:}}
\def\enddefn{\rm\medskip}

%Homemade Struts:
\newbox\strutAbox
\setbox\strutAbox=\hbox{\vrule height 12pt depth6pt width0pt}
\def\strutA{\relax\copy\strutAbox}
\newbox\strutBbox
\setbox\strutBbox=\hbox{\vrule height 10pt depth5pt width0pt}
\def\strutB{\relax\copy\strutBbox}
\newbox\strutDbox
\setbox\strutDbox=\hbox{\vrule height 11pt depth5pt width0pt}
\def\strutD{\relax\copy\strutDbox}
%high strut:
\newbox\strutHbox
\setbox\strutHbox=\hbox{\vrule height 11pt depth1pt width0pt}
\def\strutH{\relax\copy\strutHbox}
%low strut:
\newbox\strutLbox
\setbox\strutLbox=\hbox{\vrule height 1pt depth5pt width0pt}
\def\strutL{\relax\copy\strutLbox}



% hack to ignore lots of typed stuff....
\def\ignore#1{\relax}

% ******************  QED SIGNS  *********************************

\def\qed{\hbox{\hskip 1pt\vrule width4pt height 6pt depth1.5pt \hskip 1pt}}

\def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
\hbox{\vrule width.#2pt height#1pt \kern#1pt
\vrule width.2pt}
\hrule height.2pt}}}}
\def\square{\mathchoice\sqr54\sqr54\sqr{3.5}3\sqr{2.5}3}
\def\whiteslug{\bf $ \square $ \rm}  % open square


%*************** EQUATIONS WITH NUMBERS **************

\def\formula{\global\advance\resultno by 1
\eqno{(\the\sectno.\the\resultno)}}
\def\formulano{\global\advance\resultno by 1 (\the\sectno.\the\resultno)}
\def\tableno{\global\advance\resultno by 1
\the\sectno.\the\resultno. }
\def\lformula{\global\advance\resultno by 1
\leqno(\the\sectno.\the\resultno)}

%************Commutative diagrams**********************

\def\mapright#1{\smash{\mathop
        {\longrightarrow}\limits^{#1}}}

\def\mapleftright#1{\smash{\mathop
        {\longleftrightarrow}\limits^{#1}}}


\def\mapsrightto#1{\smash{\mathop
        {\longmapsto}\limits^{#1}}}

\def\mapleft#1{\smash{
   \mathop{\longleftarrow}\limits^{#1}}}

\def\mapdown#1{\Big\downarrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}

\def\lmapdown#1{{\hbox{$\scriptstyle#1$}}
\llap {$\vcenter{\hbox{\Big\downarrow}}$} }

\def\mapup#1{\Big\uparrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\mapne#1{\Big\nearrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\mapse#1{
%{$\vcenter{
\hbox{$\scriptstyle#1$}
%$}
\rlap{ $\vcenter{\hbox{$\searrow$}}$ }  }
\def\mapnw#1{\Big\nwarrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
\def\mapsw#1{
%\Big
\swarrow
   \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}


%********** DATING ******************************************
\def\monthname {\ifcase\month\or January\or February\or March\or April\or
May\or June\or
July\or August\or September\or October\or November\or December\fi}

\newcount\mins  \newcount\hours  \hours=\time \mins=\time
\def\now{\divide\hours by60 \multiply\hours by60 \advance\mins by-\hours
     \divide\hours by60         % NOTE: \divide only gives integer answers.
     \ifnum\hours>12 \advance\hours by-12
       \number\hours:\ifnum\mins<10 0\fi\number\mins\ P.M.\else
       \number\hours:\ifnum\mins<10 0\fi\number\mins\ A.M.\fi}
\def\today {\monthname\ \number\day, \number\year}


%**************** PAGE HEADERS *************************

\nopagenumbers
\def\runningtitle{\smallcaps angles}
\headline={\ifnum\pageno>1\eoheadline\else\firstheadline\fi}
\def\names{\smallcaps a.\ ram}
%\def\firstheadline{\noindent Preliminary Draft \hfill  \today}
\def\firstheadline{}
\def\eoheadline{\ifodd\pageno\oddheadline\else\evenheadline\fi}
\def\oddheadline{\tenrm\hfil\runningtitle\hfil\folio}
\def\evenheadline{\tenrm\folio\hfil{\names}\hfil}


%**************** TITLE *************************
\vphantom{$ $}  %My kludge to get the first page to move down a bit
\vskip.75truein
\centerline{\titlefont Numbers}
\bigskip
\centerline{\rm Arun Ram}
%${}^\ast$ 
\centerline{Department of Mathematics}
\centerline{University of Wisconsin, Madison}
\centerline{Madison, WI 53706 USA}
\centerline{{\tt ram@math.wisc.edu}}
\medskip
\centerline{Version: \today}

%\footnote{}{\tinyrm 
%${}^\ast$ 
%Research partially supported by the National Security Agency
%and by EPSRC Grant GR K99015 at the Newton Institute for
%Mathematical Sciences.}
%\footnote{}{\tinyrm
%\noindent {Keywords:} exponential functions, trigonometric functions}

\bigskip

%**************** ABSTRACT *************************
%\noindent{\bf Abstract.}



\bigskip\noindent
$$PICTURE$$
$\pi$ is the distance half way around a circle of radius $1$.
\smallskip\noindent
Measure angles according to the distance traveled on a circle of
radius $1$.  
$$PICTURE$$
The angle $\theta$ is measured by traveling a distance $\theta$ on
a circle of radius $1$.
\medskip\noindent
Stretch both $x$ and $y$ to get a circle of radius $r$.
$$PICTURE$$
The distance $\theta$ stretches to $r\theta$.
Hence, the 
$$(\hbox{arc length along an angle $\theta$ on a circle of radius $r$})
=r\theta.$$
\smallskip\noindent
The distance $2\pi$ around a circle of radius $1$ stretches to
$2\pi r$ around a circle of radius $r$.  
So the circumference of a circle is $2\pi r$ is the circle is radius $r$.

\bigskip
To find the area of a circle first approximate with a 
polygon inscribed in the circle.
$$PICTURE$$
the eight triangles form an octagon $P_8$ in the circle.
The area of the octagon is almost the same as the area of the
circle.
\smallskip\noindent
Unwrap the octagon.
$$PICTURE$$
The area of the octagon is the area of the $8$ triangles.  The
area of each triangle is ${1\over2}bh$.  So the area of the
octagon is ${1\over2}Bh$.

Take the limit as the number of triangles in the interior polygon
gets larger and larger (the polygon gets closer and closer to being
the circle).  Then
$$\eqalign{
\hbox{Area of the circle}
&= \lim_{n\to\infty} \Big(\hbox{area of an $n$-sided polygon $P_n$}\Big) \cr
&= \lim_{n\to \infty} \Big({1\over2}Bh\Big) \cr
&PICTURE\hbox{total base}\quad\hbox{height of triangle} \cr
&= {1\over2}(2\pi r)(r) \cr
&PICTURE\hbox{length of an unwrapped circle}
\quad\hbox{radius of the circle} \cr
&= \pi r^2. \cr}
$$
So the area of a circle is $\pi r^2$ if the circle is radius $r$, and the
$$\eqalign{
(\hbox{area of an arc of angle $\theta$ for a circle of radius $r$})
&= {\theta\over 2\pi}\cdot\left(\hbox{area of the whole circle}\right) \cr
&= {\theta\over2\pi}\cdot \pi r^2 = {\theta r^2\over 2}. \cr
}$$

\bigskip\noindent
{\bf Trigonometric functions}
\bigskip

\smallskip
\itemitem{} $\sin\theta$ is the $y$-coordinate of a point
at distance $\theta$ on a circle of radius $1$,
\smallskip
\itemitem{} $\cos\theta$ is the $x$-coordinate of a point
at distance $\theta$ on a circle of radius $1$,
\smallskip
\itemitem{} $\displaystyle{
\tan \theta = {\sin\theta\over \cos\theta} }$,
\smallskip
\itemitem{} $\displaystyle{
\cot \theta = {\cos\theta\over \sin\theta} }$,
\smallskip
\itemitem{} $\displaystyle{
\sec \theta = {1\over \cos\theta} }$,
\smallskip
\itemitem{} $\displaystyle{
\csc \theta = {1\over \sin\theta} }$,

\medskip\noindent
Since the equation of a circle of radius $1$ is $x^2+y^2=1$ this
forces
$$\sin^2\theta + \cos^2\theta = 1.$$
The pictures
$$PICTURE\qquad\hbox{and}\qquad PICTURE$$
show that
$$\sin(-\theta) = -\sin\theta
\qquad\hbox{and}\qquad
\cos(-\theta)=\cos\theta.$$
Also
$$PICTURE\qquad\hbox{and}\qquad PICTURE$$
show that 
$$
\eqalign{
\sin 0 &= 0 \cr \cos 0 &= 1\cr}
\qquad\hbox{and}\qquad
\eqalign{
\sin {\pi\over2} &= 1, \cr
\cos {\pi\over2} &= 0. \cr}
$$
Draw the graphs
$$PICTURE
\qquad\hbox{and}\qquad
PICTURE,$$
by seeing how the $x$ and $y$ coordinates change as you
walk around the circle.

There are five trig identities to remember:
$$\matrix{
\sin(x+y) = \sin x\cos y + \cos x\sin y, \cr
\cos(x+y) = \cos x\cos y - \sin x\sin y, \cr
\sin^2 x + \cos^2 x = 1, \cr
\sin(-x) = -\sin x \quad\hbox{and}\quad \cos(-x) = \cos x, \cr
}$$
As well as the two triangles
$$PICTURE \qquad\hbox{and}\qquad PICTURE.$$
From these triangles,
$$\matrix{
\sin{\pi\over6}={1\over2}\hfill &\qquad 
&\cos{\pi\over6}={\sqrt{3}\over2}\hfill \cr
\sin{\pi\over3}={\sqrt 3\over2}\hfill 
&&\cos{\pi\over3}={1\over2}\hfill \cr
\sin {\pi\over 4}={1\over\sqrt 2}\hfill
&&\cos {\pi\over 4}={1\over \sqrt 2}={\sqrt 2\over 2} \hfill \cr
}$$

Since the only trig identities I remember are identities for
sines and consines I usually
verify trig identities by first writing them
completely in terms of sines and cosines.

\medskip\noindent 
{\bf Example}.  Verify
$\displaystyle{
{\sec B\over \cos B}-{\tan B\over \cot B}=1.}$
$$\eqalign{
{\sec B\over\cos B}-{\tan B\over \cot B}
&=
{\left(\displaystyle{1\over \cos B}\right)\over \cos B}
-{\left(\displaystyle{\sin B\over \cos B}\right)\over 
\left(\displaystyle{\cos B\over \sin B}\right)}
\cr
&={1\over \cos^2B} -{\sin^2B\over\cos^2B}
={1-\sin^2B\over \cos^2B}={\cos^2B\over\cos^2B}=1.}$$

\smallskip\noindent 
{\bf Example}.  
Verify $\displaystyle{
\cot \alpha-\cot \beta 
={\sin (\beta -\alpha )\over \sin\alpha\sin\beta}}$.
\smallskip
$$\eqalign{
\hbox{Left Hand Side}
&= \cot\alpha -\cot\beta
={\cos\alpha\over\sin\alpha}-{\cos\beta\over\sin\beta}\cr
&\cr
&={\cos\alpha\sin\beta -\cos\beta \sin\alpha
\over\sin\alpha\sin\beta}\cr
&\cr
\hbox{Right Hand Side}
&= {\sin (\beta -\alpha)\over\sin\alpha\sin\beta}
={\sin\beta\cos (-\alpha )+\cos\beta\sin (-\alpha )
\over\sin\alpha\sin\beta}\cr
&\cr
&={\sin\beta\cos\alpha +\cos\beta (-\sin\alpha)
\over\sin\alpha\sin\beta}
={\sin\beta\cos\alpha -\cos\beta \sin\alpha
\over\sin\alpha\sin\beta}. \cr
}$$
So
$$
\hbox{Left Hand Side} = \hbox{Right Hand Side}.
$$

\smallskip\noindent 
{\bf Example}.  Verify 
$\displaystyle{{\tan A-\sin A\over \sec A}
={\sin^3A\over 1+\cos A}}.$
\medskip\indent
$\displaystyle{
{\tan A-\sin A\over \sec A} =^?
{\sin^3A\over 1+\cos A} }$
\medskip\noindent
So \qquad$(1+\cos A)(\tan A-\sin A)
=^? \sin^3 A \sec A.$
\medskip\noindent
So \qquad
$\tan A + \cos A\tan A - \sin A -\sin A\cos A
=^? \sin^3 A\sec A.$
\medskip\noindent
So \qquad
$\displaystyle{
{\sin A\over \cos A} + \cos A\left({\sin A\over \cos A}\right) - \sin A
-\sin A\cos A
=^? \sin ^3 A \left({1\over \cos A}\right). }$
\medskip\noindent
So \qquad
$\displaystyle{
{\sin A\over \cos A} + \sin A - \sin A
-\sin A\cos A
=^? \sin ^3 A \left({1\over \cos A}\right). }$
\medskip\noindent
So \qquad
$\displaystyle{
{\sin A-\sin A \cos^2 A\over \cos A}
=^? {\sin ^3 A\over\cos A}
}$
\medskip\noindent
So \qquad
$\displaystyle{
\sin A-\sin A \cos^2A =^? {\sin^3 A\over\cos A}. }$
\medskip\noindent
So \qquad
$1-\cos^2A =^? \sin^2 A.$
\medskip\noindent
YES, because $\sin^2A + \cos^2 A =1.$

\bigskip
and so
$$
\matrix{
{\rm adam}(t) ~~~=
~~\hbox{$x$-coordinate of the point on a circle of radius 1}\hfill 
\cr
\phantom{{\rm adam}(t) ~~~=~~}
~~\hbox{which is distance $d$ from the point (1,0),}\hfill 
&\qquad\qquad\hbox{and} \cr
\cr
{\rm eve}(t) ~~~=
~~\hbox{$y$-coordinate of the point on a circle of radius 1} \hfill\cr
\phantom{{\rm adam}(t) ~~~=~~}
~~\hbox{which is distance $d$ from the point (1,0).} \hfill\cr
}$$
$$
\beginpicture
\setcoordinatesystem units <2cm,2cm>         % sets scale
\setplotarea x from -1.7 to 1.7, y from -1.7 to 1.7    % sets plot size up
\plot -1.5 0  1.5 0 /
\plot 0 -1.5  0 1.5 /
\plot 0 0  0.5 .866025 /
\plot 0.5 0  0.5 .866025 /
\put{$({\rm adam}(0),{\rm eve}(0))$}[bl] at 1.1 0.1      %
\put{$({\rm adam}(d),{\rm eve}(d))$}[bl] at .6 0.9      %
\put{$\bullet$} at .5 .866025 %
\put{$\bullet$} at 1 0      %
\put{$y$} at -.2 1.4      %
\put{$x$} at 1.4 -.2      %
%Circle
\ellipticalarc axes ratio 1:1 360 degrees from 1 0 center at 0 0
\endpicture
$$
The triangle in this picture is
$$
\beginpicture
\setcoordinatesystem units <2cm,2cm>         % sets scale
\setplotarea x from -0.3 to 1.2, y from -0.3 to 1.2    % sets plot size up
\plot 0 0  0.5 0 /
\plot 0 0  0.5 .866025 /
\plot 0.5 0  0.5 .866025 /
\put{$1$}[r] at 0.1 0.4      %
\put{${\rm adam}(d)$}[t] at 0.25 -0.1      %
\put{${\rm eve}(d)$}[l] at .6 0.4      %
\endpicture
\qquad\qquad\qquad\qquad
\beginpicture
\setcoordinatesystem units <2cm,2cm>         % sets scale
\setplotarea x from -0.3 to 1.2, y from -0.3 to 1.2    % sets plot size up
\plot 0 0  0.5 0 /
\plot 0 0  0.5 .866025 /
\plot 0.5 0  0.5 .866025 /
\put{hypotenuse}[r] at 0.1 0.4      %
\put{adjacent}[t] at 0.25 -0.1      %
\put{opposite}[l] at .6 0.4      %
\endpicture
$$
and so
$$
{\rm adam}(d) = {\hbox{opposite}\over \hbox{hypotenuse}}
\qquad\hbox{and}\qquad
{\rm eve}(d) = {\hbox{adjacent}\over \hbox{hypotenuse}}
$$
for a right triangle with angle $d$.

 
\vfill\eject
\end




% Last Edit 7 February 1997

%            This document is written in Plain TeX
%            The macros: prepictex.tex, pictex.tex, and postpictex.tex are also
%             required for the full compilation of the document.

\magnification=1000
\overfullrule0pt
\nopagenumbers

\input prePicTeX
\input PicTeX
\input postPicTeX
\input amssym.def

\vphantom{$ $}  %My kludge to get the first page to move down a bit
\vskip2truein


$$
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$$
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\ellipticalarc axes ratio 2:1 180 degrees from -4.65 1.7  center 
at -4.65 1.6 
\plot -4.65 1.5  -3.8 1.5 /
%bottom edge
\plot 2.8 .8  2.15 .8 /
\plot 1.85 .8  1.15 .8 /
\plot 0.85 .8  0.15 .8 /
\plot -0.15 .8 -0.85 .8 /
\plot -1.15 .8 -1.85 .8 /
\plot -2.15 .8 -2.85 .8 /
\plot -3.15 .8  -4.1 .8 /
\ellipticalarc axes ratio 2:1 180 degrees from -4.65 1  center 
at -4.65 .9 
\plot -4.65 1  -3.8 1 /
\setquadratic
\plot  -3.8 1.5  -3.7 1.45 -3.5 1.25  -3.05 .8  -3 .5 /
\plot  -3.4 1.35  -3.05 1.7  -3 2 /
\plot  -3.8 1  -3.7 1.05  -3.6 1.15 /
% single crossing
\plot  0 .5  .05 .8  .4 1.15 /
\plot  .6 1.35  .95 1.7  1 2 /
\plot 0 2  .05 1.7  .5 1.25  .95 .8  1 .5 /
\endpicture
$$


$$
\beginpicture
\setcoordinatesystem units <1cm,1cm>         % sets scale
\setplotarea x from -5 to 5, y from -2 to 2    % sets plot size up
\put{$\bullet$} at -3 2      %
\put{$\bullet$} at -2 2      %
\put{$\bullet$} at -1 2      %
\put{$\bullet$} at  0 2      %   Top dots
\put{$\bullet$} at  1 2      %
\put{$\bullet$} at  2 2      %
\put{$\bullet$} at  3 2      %   
\put{$\bullet$} at -3 .5          %
\put{$\bullet$} at -2 .5          %
\put{$\bullet$} at -1 .5          %
\put{$\bullet$} at  0 .5          %  Bottom dots
\put{$\bullet$} at  1 .5          %
\put{$\bullet$} at  2 .5          %
\put{$\bullet$} at  3 .5          %
%Border of annulus
\plot -3 .5 3 .5 / 
\plot -3 -.5  3 -.5 / 
\ellipticalarc axes ratio 2:1 180 degrees from -3 .5 center at -3 0
\ellipticalarc axes ratio 2:1 180 degrees from 3 -.5 center at 3 0
\plot -3 2  3 2 /
\plot -3 -2  3 -2 /
\ellipticalarc axes ratio 5:4 180 degrees from -3 2 center at -3 0
\ellipticalarc axes ratio 5:4 180 degrees from 3 -2 center at 3 0
% Vertical edges
%\plot -3 2  -3 .5 /
\plot -2 2  -2 .5 /
\plot -1 2  -1 .5 /
%\plot  0 2   0 .5 /
%\plot  1 2   1 .5 /
\plot  2 2   2 .5 /
%\plot  3 2   3 .5 /
\setquadratic
%0 crossing
\plot  3 .5  3.02 .8  3.2 1   /  %bottom corner
\plot  3 2  3.05 1.7  3.2 1.5 /  %top corner
\plot  -3.8 1.5  -3.7 1.45 -3.5 1.25  -3.05 .8  -3 .5 /
\plot  -3.4 1.35  -3.05 1.7  -3 2 /
\plot  -3.8 1  -3.7 1.05  -3.6 1.15 /
\ellipticalarc axes ratio 1.8:2 180 degrees from -3.8 1 center at -3.8 0
\ellipticalarc axes ratio 2.8:2 180 degrees from 3.2 -1 center at 3.2 0
\ellipticalarc axes ratio 2.8:3 180 degrees from -3.8 1.5 center at -3.8 0
\ellipticalarc axes ratio 3.8:3 180 degrees from 3.2 -1.5 center at 3.2 0
\setlinear
\plot -3.8 -1  3.2 -1 /
\plot -3.8 -1.5  3.2 -1.5 /
\setquadratic
% single crossing
\plot  0 .5  .05 .8  .4 1.15 /
\plot  .6 1.35  .95 1.7  1 2 /
\plot 0 2  .05 1.7  .5 1.25  .95 .8  1 .5 /
\endpicture
$$
\bigskip
\centerline{\bf Figure 1.}



\vfill\eject
\end


\n {\bf Example}.  Verify $\sin 3x=3\sin x\cos^2x-\sin^3x$ and $\cos
3x=\cos^3x-3\sin^2x\cos x$
\begin{eqnarray*}
e^{i3x}&=& \cos 3x+i\sin 3x\\
&=& (e^{ix})^3=(\cos x+i\sin x)^3\\
&=& (\cos^2x+2i\sin x\cos x+i^2\sin^2x)(\cos x+i\sin x)\\
&&\\
&=&\cos^3x+i\sin x\cos^2x+2i\sin x\cos^2x-2\sin^2x\cos x\\
&&\qquad\qquad -\sin^2x\cos x-i\sin^3x\\
&&\\
&=& \cos^3x+3i\sin x\cos^2x-3\sin^2x\cos x-i\sin^3x\\
&&\\
&=&\cos^3x-3\sin^2x\cos x+i(3\sin x\cos^2x-\sin^3x).
\end{eqnarray*}

\n So
\begin{eqnarray*}
\cos 3x+i\sin 3x&=& \cos^3x-3\sin^2x\cos x\\
&&\\
&&\quad +i(3\sin x\cos^2x-\sin^3x).
\end{eqnarray*}

\n So
$$
\cos 3x=\cos^3x-3\sin^2x\cos x
$$
and
$$
\sin 3x=3\sin x\cos^2x-\sin^3x.
$$
\smallskip

\centerline{$\mbox{\fib}\;\;\bullet\;\;\mbox{\fib}$}
\smallskip

\n A function $f(x)$ is continuous at $x=a$ if $f(x)$ doesn't jump to
$f(a)$.

%%INSERT FIGURE
\bigskip


\begin{tabular}{c}
Continuous\\
at $x=a$\end{tabular} \hspace*{1.0in}\begin{tabular}{c}
Not continuous\\
at $x=a$\end{tabular} 

\medskip

\n In other words:
\begin{enumerate}
\item[{}] A function $f(x)$ is continuous at $x=a$

\item[{}] if $f(x)$ gets closer and closer to $f(a)$

\item[{}] as $x$ gets closer and closer to $a$.
\end{enumerate}

\n In other words;
\begin{enumerate}

\item[{}] A function $f(x)$ is continuous at $x=a$ if
$$
\lim_{x\to a} f(x)=f(a).
$$
\end{enumerate}






