
\chapter{\index{ Trig Functions} Trig Functions and \index{ Sinusoidal Functions} Sinusoidal Functions }

We start by reviewing trig functions of an angle measured in radians.
Because this is review material, we will go through it rapidly.

\smallskip

We first draw the unit circle in the $xy$-plane.
The standard way of measuring an angle is to start with the positive $x$-axis
and rotate counterclockwise.  As we go through the angle, we sweep around
the circumference of the unit circle.  The \index{ radian measure} radian measure of the angle is
defined to be the distance around the circumference that we travel.  That
is, a full revolution of $360{}^\circ$ is equal to $2\pi$ radians, half
a revolution is $180^\circ=\pi$ radians, and so on.  Other frequently used
equivalences are: $90^\circ=\pi/2$ rad, $60^\circ=\pi/3$ rad,
$45^\circ=\pi/4$ rad, $30^\circ=\pi/6$ rad.  {\it All angles in this
course will be assumed to be measured in radians unless otherwise stated.
The designation ``radian'' will often be omitted, e.g., if we write
``an angle of $\pi/6$,'' we mean a $30^\circ$ angle.}

\smallskip

By a {\it \index{ negative angle} negative angle } we
mean a \index{ clockwise angle} clockwise angle.  For example, an angle of $-\pi/2$ (i.e.,
$-90^\circ$) brings us to the same position as an angle of $3\pi/2$ (which
is $270^\circ$).

\smallskip

In general, if we have a circle of radius $r$ and sweep out an angle
of $\theta$ radians, then the distance around the circumference through
which we have traveled is simply $\theta r $.  Of course, one full
circumference is $2\pi r$, corresponding to rotation through $2\pi$ radians.

\smallskip

Returning to the unit circle (see the drawing below), after rotating
through an angle of $\theta$ radians let $(x,y)$ be the point where we end up
on the circumference.  Then the six trig functions are defined as follows:


%--- figure IA --------------------------------------------------
%DIA Diagram of unit circle on 42 should be situated to the right of the
%    six formulas for the trig functions.
\begin{rightindent}{3.65in}
\begin{alignat*}{2}
\sin\theta &=y
&
\text{cosecant}\,\theta &=\frac{1}{\sin\theta}=\frac{1}{ y}\\
&&&\\
\cos\theta&=x
&
\text{secant}\,\theta &=\frac{1}{\cos\theta}=\frac{1}{ x} \\
&&&\\
\tan\theta&=\frac{\sin\theta}{\cos\theta}=\frac{y}{ x}
&
\cot\theta
 &=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}=\frac{x}{ y} \\
\end{alignat*}
\end{rightindent}
\hfill
\begin{psfigure}{1.8in}{2.0in}{figIA.ps}
\puttext{25,40}{\Small $(x,y)$}
\puttext{10,5}{\Small $\theta$}
\end{psfigure}
%-----------------------------------------------------
It is also customary to use the names $\csc \, \theta $ for $\text{cosecant}\,\theta$ and $\sec \,\theta $ for $\text{secant}\,\theta $.

From the definition of $\sin\theta$ we see that as $\theta$ increases from
$0$ to $2\pi$ (i.e., as we go once around the \index{ unit circle} unit circle), $\sin\theta$
goes from $0$ to $1$ to  $0$ to $-1$ and back to $0$.  Then as $\theta$ goes from
$2\pi$ to $4\pi$ (i.e., as we go once again around the circle), the same
values of $\sin\theta$ are repeated.  We say that the function $\sin\theta$
has {\bf period} $2\pi$, i.e., we have $\sin(\theta+2\pi)=\sin\theta$.
Here are some other relations for the trig functions that follow easily
from the above definitions: $\cos(\theta+2\pi)=\cos\theta$, 
$\sin(\theta+\pi)=-\sin\theta$, $\cos(\theta+\pi)=-\cos\theta$,
$\tan(\theta+\pi)=\tan\theta$, $\sin(-\theta)=-\sin\theta$, $\cos(-\theta)
=\cos\theta$, $\tan(-\theta)=-\tan\theta$, $\sin^2\theta+\cos^2\theta=1$,
and $\cos\theta=\sin(\frac{\pi}{ 2}-\theta)$.  The last of these relations
is familiar from high school trigonometry: the cosine of an angle is equal to 
the sine of the complementary angle.  

It can also be seen from our definitions as follows:
the $x$-coordinate of the point on the unit circle you arrive at when you
rotate through $\theta$ counterclockwise starting at the positive $x$-axis is
the same as the $y$-coordinate of the point you arrive at when you
rotate through $\theta$ clockwise starting at the positive $y$-axis.
Sometimes one also needs the following formulas for the trig functions of
a sum of two angles: $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin
\beta$, and $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$.  As
a special case we have the double angle formulas $\sin(2\alpha)=2\sin\alpha
\cos\alpha$ and $\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha$.

%%% To update!
A simple geometric explanation of this will be given in class. 

\smallskip
Also notice the sign of the trig functions in the different quadrants:
$\sin x$ is positive in the 1st and 2nd quadrants (for angles between
0 and $\pi$), $\cos x$ is positive in the 1st and 4th, and $\tan x$ is
positive in the 1st and 3rd.
 


\section{\index{ Graphs of Trig Functions} Graphs of Trig Functions }  
The most important graphs to be familiar with are the graphs of the
functions $y=\sin x$, $y=\cos x$, and $y=\tan x$.  (We're now using $x$ for
the angle in radians; $x$ no longer denotes a coordinate of a point on the
unit circle.)  For certain $x$ the trig functions have easily stated exact
values, namely, when $x$ is a multiple of $\pi/6$ or of $\pi/4$.  Here it
is useful to draw the $30^\circ-60^\circ-90^\circ$ triangle with hypotenuse
$1$ and the $45^\circ-45^\circ-90^\circ$ triangle with hypotenuse $1$.
Using these triangles, we arrive at a table of values
of the trig functions, from which we can sketch the graphs on the next page.


\begin{figure}[htb]
%--- figure IB --------------------------------------------------
%DIA Diagram with 2 triangles at top of p. 43.

\ \hrulefill \

\begin{center}
\begin{psfigure}{3in}{3.0in}{figIB.ps}
\puttext{20, 18}{$1/2$}
\puttext{0, 0}{\Small$\sqrt{3}/2$}
\puttext{-5, 20}{\Small$1$}
\puttext{-5, -45}{\Small$\sqrt{2}/2$}
\puttext{20, -30}{\Small$\sqrt{2}/2$}
\puttext{-5,-25}{\Small$1$}
\end{psfigure}
\hfill
\begin{minipage}[b]{2in}
%\small
\begin{tabular}{cccc}
$x$	      & $\sin x$	    & $\cos x$	         &  $\tan x$\\
$0$	      & $0$		    & $1$		 &  $0$\\
$\frac{\pi}{6}$ & $\frac{1}{2}$	    & $\frac{\sqrt{3}}{2}$ &  $\frac{\sqrt{3}}{3}$\\    
$\frac{\pi}{4}$ & $\frac{\sqrt{2}}{2}$  & $\frac{\sqrt{2}}{2}$ &  $1$\\                               
$\frac{\pi}{3}$ &	$\frac{\sqrt{3}}{2}$  & $\frac{1}{2}$        & $\sqrt{3}$\\                    
$\frac{\pi}{2}$ & $1$                 & $0$                & $\infty$\\                      
$\frac{2\pi}{3}$& $\frac{\sqrt{3}}{2}$  & $-\frac{1}{2}$       & $-\sqrt{3}$\\
$\frac{3\pi}{4}$& $\frac{\sqrt{2}}{2}$  & $-\frac{\sqrt{2}}{2}$& $-1$\\
$\frac{5\pi}{6}$& $\frac{1}{2}$         & $-\frac{\sqrt{3}}{2}$& $-\frac{\sqrt{3}}{3}$\\      
${\pi}$       & $0$                 & $-1$               & $0$\\                           
$\frac{7\pi}{6}$& $-\frac{1}{2}$        & $-\frac{\sqrt{3}}{2}$& $\frac{\sqrt{3}}{3}$\\             
$\frac{5\pi}{4}$& $-\frac{\sqrt{2}}{2}$ & $-\frac{\sqrt{2}}{2}$& $1$\\                           
$\frac{4\pi}{3}$& $-\frac{\sqrt{3}}{2}$ & $-\frac{1}{2}$       & $\sqrt{3}$\\                    
$\frac{3\pi}{2}$& $-1$		    & $0$                & $\infty$\\                     
$\frac{5\pi}{3}$& $-\frac{\sqrt{3}}{2}$ & $\frac{1}{2}$        & $-\sqrt{3}$\\                   
$\frac{7\pi}{4}$& $-\frac{\sqrt{2}}{2}$ & $\frac{\sqrt{2}}{2}$ & $-1$\\                          
$\frac{11\pi}{6}$&$-\frac{1}{2}$        & $\frac{\sqrt{3}}{2}$ & $-\frac{\sqrt{3}}{3}$\\        
$2\pi$         &  $0$    	    & $1$                & $0$ \\
\end{tabular}
\end{minipage}


\end{center}


%--- figure IC --------------------------------------------------
%DIA Three graphs of trig functions in middle of p. 46.
%


\begin{minipage}[b]{2.5in}
\begin{psfigure}{2.5in}{1.5in}{figICa.ps}
\puttext{0,-5}{\Small$\pi$}
\puttext{60,-5}{\Small$2\pi$}
\puttext{0,-50}{Graph of $y = \sin(x)$}
\end{psfigure}
\begin{psfigure}{2.5in}{1.5in}{figICb.ps}
\puttext{0,-5}{\Small$\pi$}
\puttext{60,-5}{\Small$2\pi$}
\puttext{0,-50}{Graph of $y = \cos(x)$}
\end{psfigure}
\end{minipage}
\hfill
\begin{psfigure}{2.5in}{3.0in}{figICc.ps}
\puttext{15,-3}{\Small$\frac{\pi}{2}$}
\puttext{-12,-3}{\Small$-\frac{\pi}{2}$}
\puttext{26,-3}{\Small$-\pi$}
\puttext{-26,-3}{\Small$\pi$}
\puttext{0,-55}{Graph of $y = \tan(x)$}
\end{psfigure}


\bigskip

\ \hrulefill \

\end{figure}


%-----------------------------------------------------




\section{Sinusoidal Functions}  
A \index{ sinusoidal function} sinusoidal function is a function made up from the function $y=\sin x$
(or $y=\sin t$, if time is playing the role of the $x$-variable) by
inserting constants in various places.  The reason for inserting these
constants is that the pure sine function $y=\sin x$ almost never occurs in
the real world, but curves with a sine-like appearance (this is what the
word ``sinusoidal'' means) arise frequently.

\smallskip

At this point you should review the material at the end of the section on
functions: horizontal and vertical shifts, horizontal and vertical
expansion, and the last batch of homework problems in that section.



%Example 1
\begin{example}
Write the cosine function as a shift of the sine function.

\smallskip

The graphs show that the cosine function is obtained by shifting the sine
function $\pi/2$ to the left.  Thus, $\cos x=\sin(x-(-\frac{\pi}{ 2}))=\sin
(x+\frac{\pi}{ 2})$.  This relation can also be obtained using the identities
listed above: $\cos x=\cos(-x)=\sin(\frac{\pi}{ 2}-(-x))=\sin(x+\frac{\pi}{
2})$.

\end{example}

\medskip

We now give the general form for a sinusoidal function $y=f(x)$.  We
start with $y=\sin x$.

\smallskip

{\bf Step 1.~Expansion.}  Vertically, we expand by
a factor of $A$, which is called the {\it \index{ amplitude} amplitude }.  We do this by
replacing $\sin x$ by $A\sin x$.  This makes the function go to a maximum
of $A$ and a minimum of $-A$ (rather than a maximum of $1$ and a minimum
of $-1$ as was the case for $\sin x$).  Next, we expand 
horizontally in such a way that the function repeats its pattern from
$x$ to $x+B$ (rather than from $x$ to $x+2\pi$ as with $\sin x$).  $B$ is
called the {\it \index{ period } period}.  That is, we want to expand horizontally by a factor
of $B/2\pi$.  We do this by replacing $x$ by $x/(B/2\pi)=\frac{2\pi}{ B}x$
inside the sine.  To summarize, the result of the vertical and horizontal
expansion is to replace $\sin x$ by the function $A\sin\left(\frac{2\pi}{ B}x
\right)$.  The graph of this new function is shown below.  Notice that
as $x$ goes from 0 to $B$, the value of the function goes from 0 to $A$
to 0 to $-A$ and back to 0.

%--- figure ID --------------------------------------------------
\begin{center}
\begin{psfigure}{5in}{2.5in}{figID.ps}
\puttext{-57,24}{\Small$A$}
\puttext{-57,-24}{\Small$A$}
\puttext{2,-5}{\Small$B$}
\puttext{55,-5}{\Small$2B$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------


{\bf Step 2.~Shifting.}  We now want to move the whole \index{ sine wave} sine wave upward
a certain amount $D$ (called the {\it \index{ vertical shift} vertical shift }) and over to the
right a certain amount $C$ (called the {\it \index{ phase shift} phase shift }).  We do this
by adding $D$ outside the function, and replacing $x$ by $x-C$ inside
the function.  The result is:
\[
       y=A\sin\left(\frac{2\pi}{ B}(x-C)\right)\>+D.
\]
In the graph of this function below, we indicate the meaning of the four
constants $A$, $B$, $C$, and $D$.

\medskip

%--- figure IE --------------------------------------------------
\begin{center}
\begin{psfigure}{\textwidth}{2.5in}{figIE.ps}
\puttext{50,13}{\Small$A$}
\puttext{7,40}{\Small$B$}
\puttext{-28,27}{\Small$C$}
\puttext{-50,-17}{\Small$D$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

%Example 2
\begin{example}
Suppose that the \index{ mean daily temperature} mean daily temperature (i.e., the temperature in
${}^\circ$F averaged over the 24-hour period) on the $t$-th day of the year
in Seattle is tabulated over a period of many years, and the results for
each day are averaged.  This is what we call the ``\index{ normal mean
temperature} normal mean
temperature '' on the $t$-th day.  For simplicity, suppose that the year
consists of $12$ months of $30$ days each.  Take $t=1$ to be January 1, and so
on, until $t=360$ for December $30$.  Further assume: (1) the normal mean
temperature function $y(t)$ is sinusoidal; (2) the coldest mean temperature
is $37{}^\circ$ on January 30; and (3) the warmest mean temperature is
$65{}^\circ$ on July 30.  Find a formula for $y(t)$.
%%% To update! For local temperatures.

\smallskip
We graph $y(t)$ by first plotting the two points $t=30$, $y=37$ and $t=210$
(i.e., July 30), $y=65$.  We then draw a sine wave in such a way that its
minimum is at the first point, its maximum is at the second, and its period
is a year, i.e., $360$ days.  We also draw a horizontal dotted line half-way
between the coldest and warmest temperature, i.e., at $y=51$.


%--- figure IF --------------------------------------------------
\begin{center}
\begin{psfigure}{\textwidth}{2in}{figIF.ps}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

\medskip

Now we read the four constants $A$, $B$, $C$, $D$ off the graph.  The
\index{ amplitude} amplitude $A$ is $14$, the \index{ period} period $B$ is $360$, and the \index{ vertical shift} vertical shift (the
distance up to the dotted line) is 51.  To find $C$ we want to know when
the curve crosses the dotted line on its way from a minimum to a maximum,
i.e., when the curve looks like $\sin x$ looks when it crosses the origin.
This can be found by taking the $t$-value half-way between the coldest day
(which is $t=30$) and the warmest day (which is $t=210$).  This $t$-value
is $(30+210)/2=120$ (i.e., April 30).  Thus, the \index{ phase shift} phase shift is $C=120$,
and our equation is 
\[
     y(t)=14\sin\left(\frac{2\pi}{ 360}(t-120)\right)+51.
\]

\end{example}

%
