\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 $$ \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 cylinder \plot -3 .5 3 .5 / \ellipticalarc axes ratio 3:1 90 degrees from -4.5 1 center at -3 1 \ellipticalarc axes ratio 3:1 90 degrees from 3 .5 center at 3 1 \plot -3 2 3 2 / \ellipticalarc axes ratio 3:1 180 degrees from -3 3 center at -3 2.5 \plot -3 3 3 3 / %top edge of cylinder \ellipticalarc axes ratio 3:1 180 degrees from 3 2 center at 3 2.5 \plot -4.5 2.5 -4.5 1 / \plot 4.5 2.5 4.5 1 / % 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 % 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 / %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 7:3.5 -90 degrees from -3.8 1 center at -3.8 1.35 \ellipticalarc axes ratio 13:3.5 90 degrees from 3.2 1 center at 3.2 1.35 \ellipticalarc axes ratio 7:5 -90 degrees from -3.8 1.5 center at -3.8 2 \ellipticalarc axes ratio 13:5 90 degrees from 3.2 1.5 center at 3.2 2 \setlinear \setdashes \plot -3.8 2.5 3.2 2.5 / \plot -3.8 1.7 3.2 1.7 / \ellipticalarc axes ratio 7:3.5 90 degrees from -3.8 1.7 center at -3.8 1.35 \ellipticalarc axes ratio 13:3.5 -90 degrees from 3.2 1.7 center at 3.2 1.35 \ellipticalarc axes ratio 7:5 90 degrees from -3.8 2.5 center at -3.8 2 \ellipticalarc axes ratio 13:5 -90 degrees from 3.2 2.5 center at 3.2 2 \endpicture $$ $$ \beginpicture \setcoordinatesystem units <1cm,1cm> % sets scale \setplotarea x from -5 to 5, y from -2 to 2 % sets plot size up \put{$\bullet$} at -3 2 % \put{$\bullet$} at -2 2 % \put{$\bullet$} at -1 2 % \put{$\bullet$} at 0 2 % Top dots \put{$\bullet$} at 1 2 % \put{$\bullet$} at 2 2 % \put{$\bullet$} at 3 2 % \put{$\bullet$} at -3 0.5 % \put{$\bullet$} at -2 0.5 % \put{$\bullet$} at -1 0.5 % \put{$\bullet$} at 0 0.5 % Bottom dots \put{$\bullet$} at 1 0.5 % \put{$\bullet$} at 2 0.5 % \put{$\bullet$} at 3 0.5 % %Flagpole \plot -4.5 2.5 -4.5 1.62 / \plot -4.5 1.38 -4.5 1.12 / \plot -4.5 .88 -4.5 0 / \plot -4.25 2.5 -4.25 1.62 / \plot -4.25 1.38 -4.25 1.12 / \plot -4.25 .88 -4.25 0 / \ellipticalarc axes ratio 1:1 360 degrees from -4.5 2.5 center at -4.375 2.5 \put{$*$} at -4.375 2.5 \ellipticalarc axes ratio 1:1 180 degrees from -4.5 0 center at -4.375 0 % Vertical edges %\plot -3 2 -3 .5 / \plot -2 2 -2 .5 / \plot -1 2 -1 .5 / %\plot 0 2 0 .5 / %\plot 1 2 1 .5 / \plot 2 2 2 .5 / %\plot 3 2 3 .5 / \setquadratic %0 crossing \plot 3 .5 2.98 .7 2.8 .8 / %bottom corner \plot 3 2 2.98 1.8 2.8 1.7 / %top corner \setlinear %top edge \plot 2.8 1.7 2.15 1.7 / \plot 1.85 1.7 1.15 1.7 / \plot 0.85 1.7 0.15 1.7 / \plot -0.15 1.7 -0.85 1.7 / \plot -1.15 1.7 -1.85 1.7 / \plot -2.15 1.7 -2.85 1.7 / \plot -3.15 1.7 -4.1 1.7 / \ellipticalarc axes ratio 2:1 180 degrees from -4.65 1.7 center at -4.65 1.6 \plot -4.65 1.5 -3.8 1.5 / %bottom edge \plot 2.8 .8 2.15 .8 / \plot 1.85 .8 1.15 .8 / \plot 0.85 .8 0.15 .8 / \plot -0.15 .8 -0.85 .8 / \plot -1.15 .8 -1.85 .8 / \plot -2.15 .8 -2.85 .8 / \plot -3.15 .8 -4.1 .8 / \ellipticalarc axes ratio 2:1 180 degrees from -4.65 1 center at -4.65 .9 \plot -4.65 1 -3.8 1 / \setquadratic \plot -3.8 1.5 -3.7 1.45 -3.5 1.25 -3.05 .8 -3 .5 / \plot -3.4 1.35 -3.05 1.7 -3 2 / \plot -3.8 1 -3.7 1.05 -3.6 1.15 / % single crossing \plot 0 .5 .05 .8 .4 1.15 / \plot .6 1.35 .95 1.7 1 2 / \plot 0 2 .05 1.7 .5 1.25 .95 .8 1 .5 / \endpicture $$ $$ \beginpicture \setcoordinatesystem units <1cm,1cm> % sets scale \setplotarea x from -5 to 5, y from -2 to 2 % sets plot size up \put{$\bullet$} at -3 2 % \put{$\bullet$} at -2 2 % \put{$\bullet$} at -1 2 % \put{$\bullet$} at 0 2 % Top dots \put{$\bullet$} at 1 2 % \put{$\bullet$} at 2 2 % \put{$\bullet$} at 3 2 % \put{$\bullet$} at -3 .5 % \put{$\bullet$} at -2 .5 % \put{$\bullet$} at -1 .5 % \put{$\bullet$} at 0 .5 % Bottom dots \put{$\bullet$} at 1 .5 % \put{$\bullet$} at 2 .5 % \put{$\bullet$} at 3 .5 % %Border of annulus \plot -3 .5 3 .5 / \plot -3 -.5 3 -.5 / \ellipticalarc axes ratio 2:1 180 degrees from -3 .5 center at -3 0 \ellipticalarc axes ratio 2:1 180 degrees from 3 -.5 center at 3 0 \plot -3 2 3 2 / \plot -3 -2 3 -2 / \ellipticalarc axes ratio 5:4 180 degrees from -3 2 center at -3 0 \ellipticalarc axes ratio 5:4 180 degrees from 3 -2 center at 3 0 % Vertical edges %\plot -3 2 -3 .5 / \plot -2 2 -2 .5 / \plot -1 2 -1 .5 / %\plot 0 2 0 .5 / %\plot 1 2 1 .5 / \plot 2 2 2 .5 / %\plot 3 2 3 .5 / \setquadratic %0 crossing \plot 3 .5 3.02 .8 3.2 1 / %bottom corner \plot 3 2 3.05 1.7 3.2 1.5 / %top corner \plot -3.8 1.5 -3.7 1.45 -3.5 1.25 -3.05 .8 -3 .5 / \plot -3.4 1.35 -3.05 1.7 -3 2 / \plot -3.8 1 -3.7 1.05 -3.6 1.15 / \ellipticalarc axes ratio 1.8:2 180 degrees from -3.8 1 center at -3.8 0 \ellipticalarc axes ratio 2.8:2 180 degrees from 3.2 -1 center at 3.2 0 \ellipticalarc axes ratio 2.8:3 180 degrees from -3.8 1.5 center at -3.8 0 \ellipticalarc axes ratio 3.8:3 180 degrees from 3.2 -1.5 center at 3.2 0 \setlinear \plot -3.8 -1 3.2 -1 / \plot -3.8 -1.5 3.2 -1.5 / \setquadratic % single crossing \plot 0 .5 .05 .8 .4 1.15 / \plot .6 1.35 .95 1.7 1 2 / \plot 0 2 .05 1.7 .5 1.25 .95 .8 1 .5 / \endpicture $$ \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}