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\noindent
Class Notes: March 19th, 2009

\bigskip
\section*{Part I:  Why is there no text for this class? (Just in case you were wondering.)}


Textbooks that are usually used for a course like this generally fall
into two categories: 
\begin{enumerate}
\item Many are too low level (ie. they tell you what you need to know
  but not why it works or why you need to use it) 
\item Many focus more on how to teach, giving you really broad,
  general issues that one encounters in teaching.  
\end{enumerate}
Neither of these options fit into the philosophy of the course.  Our
focus shouldn't be in amassing algorithms but rather making educated
choices about when to use what algorithms and why we would use
them. In addition, a major goal of this course is to try to make
connections between what you learn as an undergraduate and how it
relates to the secondary curriculum.


\section*{Part II: A quick reminder about probability}

Let's start off with some easy probability (ie: discrete probability).

\medskip
\noindent{\bf Example 1}:  
Draw a card out of a standard deck.  What is the probability that it is red? 

The probability is $\dfrac{\text{the number of red cards}}{\text{the
    total number of cards}} = \dfrac{26}{52} = \dfrac{1}{2}$ 

\vspace{.2in}
\noindent
{\bf Example 2}:  Let's try the same problem but now with a deck of
cards including Jokers (of which there are 2).  Now our total number
of cards is 54.  


The probability is $\dfrac{\text{the number of red cards}}{\text{the
    total number of cards}} = \dfrac{26}{54} = \dfrac{13}{27}$

\vspace{.2in}
Now let's talk about probability with independent events (or trial).
They key is that if our events/trials are independent then we multiply
the individual probabilities. 

\medskip
\noindent
{\bf Example 3}:  If I flip 3 coins, what is the probability that
they're all heads? 


The probability that any one coin will come up heads is $\frac{1}{2}$.
So the probability that all the coins will come up heads is:
$$\left( \dfrac{1}{2}\right)\left( \dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)  = \dfrac{1}{8} $$
 
 
Note that we could list all possible solutions to this problem, see
there are eight possible and that only one gives all heads, but
listing is only feasible for small numbers.  It wouldn't always work
out so easily. 
 
\bigskip
What if our events are not independent?
 
\medskip
\noindent
{\bf Example 4}: What is the probability that we draw 2 red cards from
a standard deck? 
 
 
Again, since the probability of one event is $\frac{\text{the
    number of red cards}}{\text{the total number of cards}}$ 
and since we multiply probabilities of independent events, 
we just need account for the fact that there will be one less red card
in our deck when we calculate the probability of pulling the second
red card out.  After taking this into account, we can treat this as
two independent events, viewing the second draw as just choosing a
card from a smaller deck.  The resulting probability is
$$\left( \dfrac{\text{number of red cards}}{\text{total
    number of cards}}\right) \left( \dfrac{\text{number of red cards}
  - 1}{\text{total number of cards} - 1}\right) = \left(
\dfrac{26}{52}\right) \left( \dfrac{25}{51}\right) $$
 

\vspace{0.2in}
What if the colors we wanted were different?
 
\medskip\noindent
{\bf Example 5}:  What is the probability that we draw one red card
and one black card from a standard deck? 
  
Here we have to work a little harder, since what happens in the first
event determines what we want the second event to be.  Let's look at
it two different ways.
 
 
We can calculate $P\left(\text{one red, one black}\right) + P\left(
\text{one black, one red}\right)$.  This is 
$$\left( \dfrac{26}{52}\right) \left( \dfrac{26}{51}\right) + \left(
\dfrac{26}{52}\right) \left( \dfrac{26}{51}\right) = 2\left(
\dfrac{26}{52}\right) \left( \dfrac{26}{51}\right) = \dfrac{26}{51}.$$

\medskip
Alternatively, we can calculate the probability of not pulling out one
red and one black.  In other words, we would calculate $1 - \left[
  P\left(\text{two blacks}\right) + P\left( \text{two
    reds}\right)\right]$ as
$$1 - \left[ \left( \dfrac{26}{52}\right) \left( \dfrac{25}{51}\right)
  + \left( \dfrac{26}{52}\right) \left( \dfrac{25}{51}\right)\right] =
1 - \dfrac{25}{51} = \dfrac{26}{51}.$$

\bigskip
Note:  One thing that is often confusing is that sometimes we discuss
probability, and sometimes we discuss ``the odds'' of something
happening.  These describe the same concept, but in different ways,
and so this confuses many students (and teachers!).  If we say that the
odds of an event occurring are 3 to 5, what we mean is that there are
3 ways it can happen and 5 ways it can fail.  Thus, there are a total
of 3 + 5 = 8 ways the event could turn out.  So the probability of the
event happening would be $\frac{3}{8}$.  It is important to keep such
potential points of confusion in mind so that we can point them out
and avoid them. 

\medskip
If we have a large number of trials then we can say there are 
$\infty$-many events.  This leads to probability distributions.  We
can think of these distributions as a function over the sample space so
that the total area under the curve is equal to 1.  This represents
all possible outcomes.  Then calculating the probability just
corresponds to finding the area under the curve corresponding to the
events we are concerned with.


\includegraphics[scale=.8]{Normalcurve.pdf}

In the high-school curriculum, all distributions are normal.  The
midpoint is the mean (usually denoted by $\mu$) and 
$68.2\%$ of the data will fall within $\pm 1$ standard deviation
($\sigma$) from the mean.   


All problems on the Regents exam essentially come down to a linear
change of variables to adjust what the mean and standard deviation are
in each situation, and then picking off what you are interested in
from the chart.


If a random sample (size $n$) of values $x_i$ is taken from the
collection is large enough then
$$\dfrac{\sum x_i}{\text{n}} \longrightarrow \mu\ 
\text{as $n$ gets large.}$$

And the standard deviation, denoted $\sigma$, is equal to the square
root of the variance: 
$$\sigma = \sqrt{\dfrac{\sum\limits_{i = 1}^{n}(x_{i} - \mu)^{2}}{n}}$$


\section*{Part III:  Enough of that, let's talk about Trig!}

One of the major themes of geometry in secondary school is the idea of
congruent triangles. 

\medskip
\noindent
\textit{Definition}:  Two triangles are congruent of corresponding sides math up
in terms of length and corresponding angles match up in terms of angle
measure.  
\medskip

We can think of two triangles as congruent if we can get from one to
the other using rigid motions in the plane, that is we can
flip/translate/rotate our way from one triangle to the other. 

We have the usual ways to determine if two triangles are congruent:

\vspace{-.8\baselineskip}
\begin{itemize}\addtolength{\itemsep}{-0.5\baselineskip}
\item Side Angle Side (SAS) 
\item Side Side Side (SSS)
\item Angle Side Angle (ASA)
\item Angle Angle Side (AAS)
\end{itemize}


If we have Side Side Angle (SSA), this does not guarantee us congruent
triangles.  It will match one half of cases (because there we could
construct two possible triangles that have the same two sides and
angle). 


If we have Angle Angle Angle (AAA), this leads to the idea of similar
triangles (ie: triangles in which the edge lengths are in proportion).
Note, we really only need two angle measures to get AAA because a
triangle's interior angles must sum to 180$^{\circ}$ in Euclidean geometry. 


If we have two similar triangles, $\triangle \text{ABC} \sim \triangle
\text{DEF}$, then we can think of putting the bigger one inside of the
smaller one and we have the following relations: 

\noindent
\begin{minipage}{2.9in}
\includegraphics[scale=.7]{Similartriangles.pdf}
\end{minipage}
\begin{minipage}{3.5in}
$$\dfrac{|\text{AE}|}{|\text{AB}|} = \dfrac{|\text{AF}|}{|\text{AC}|}
= \dfrac{|\text{EF}|}{|\text{BC}|}$$ 

Furthermore, we'll have EF parallel to BC (this follows from the
alternate interior angle theorem).
\end{minipage}

It seems that all that really matters is the measure of the angle at
A/D, so we might as well restrict our attention to similar right
triangles, which will be determined completely by one of the non-right
angles. 


\centerline{\includegraphics[scale=.65]{Similartrianglesright.pdf}}


We can find invariants related to similarity by looking at ratios
corresponding to the angle $\theta$ and we can give these ratios
names.

\medskip
\begin{minipage}{2.6in}
\centerline{{\includegraphics[width=2.5in]{Trigfuntions.pdf}}}
\end{minipage}
\qquad
\begin{minipage}{3in}
\[\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\quad  \cos\theta =
\dfrac{\text{adj}}{\text{hyp}}\quad  \tan\theta =
\dfrac{\text{opp}}{\text{adj}}\]   
\[\csc\theta = \dfrac{1}{\sin\theta}\quad  \sec\theta =
\dfrac{1}{\cos\theta}\quad  \cot\theta = \dfrac{1}{\tan\theta}\] 
\end{minipage}

\bigskip\noindent
\begin{minipage}{3.5in}
Since these are ratios, we may as well choose the triangle with
hypotenuse equal to 1.  This will give $x=\cos\theta$ and 
$y=\sin\theta$.

Then, the well-known Pythagorean Theorem, tells us that
$x^{2} + y^{2} = 1$ and so 
$$\cos^{2}\theta + \sin^{2}\theta = 1.$$
Many other standard trig identities are just the Pythagorean Theorem
in disguise.
\end{minipage}
\begin{minipage}{3in}
\centerline{\includegraphics[width=2.5in]{TrigID.pdf}}
\end{minipage}


\vspace{.5in}

\noindent\begin{minipage}{2in}
\includegraphics[width=1.5in]{Lawofcosines.pdf}
\end{minipage}
\begin{minipage}{4in}
\par We can also derive a generalization of the Pythagorean
Theorem called the {\bf Law of Cosines}. 


It says that in any triangle with side lengths $a$, $b$ and $c$, we
have $$c^{2} = a^{2} + b^{2} - 2ab\cos\theta$$ where $\theta$ is the
angle opposite the side of length $c$.
\end{minipage}

\bigskip\bigskip
How do we prove this?  We drop an altitude down and then use the
Pythagorean Theorem on the two resulting right triangles. 

\centerline{\includegraphics[height=1.5in]{Lawofcosinesproof.pdf}}

For the triangle on the left, we get $x^{2} + y^{2} = b^{2}$ and for
the triangle on the right we'll have $(a - x)^{2} + y^{2} = c^{2}$.

Solving the equation for the triangle on the left for $y^{2}$ will
give us $y^{2} = b^{2} - x^{2}$. Now substitute this into the
equation for the triangle on the right, yielding 
  $$(a - x)^{2} + b^{2} - x^{2} = c^{2}.$$
Expanding the equation gives $a^{2} + 2ax + x^{2} + b^{2} - x^{2} =
c^{2}$ and canceling the $x^{2}$ terms gives us
  $$a^{2} + 2ax + b^{2} = c^{2}.$$

Now notice that $\cos\theta = {x}/{b}$; solving for $x$ gives
$b\cos\theta = x$.  Making the substitution for $x$ in the above
equation gives us the law of
cosines:  
  $$c^{2} = a^{2} + b^{2} - 2ab\cos\theta.$$

\bigskip\bigskip
Also we can obtain the {\bf Law of Sines}.  This says that in any
triangle, if the angle of measure $A$ is opposite the side of length
$a$,  the angle of measure $B$ is opposite side $b$, and the angle of
measure $C$ is opposite side $c$, we have
\[ \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}. \]


\bigskip
How do we prove it?  Again, drop an altitude down from an angle and
convert the problem into one involving right triangles.
We'll drop the perpendicular from angle A.


\centerline{\includegraphics[scale=.8]{Lawofsines.pdf}}


Notice that we now have $\sin B = \dfrac{y}{c}$ and $\sin C =
\dfrac{y}{b}$ so, solving both equations for $y$ gives 
$y = b\sin C = c\sin B$.
Since $b\sin C = c\sin B $, we can divide  both sides by $b$ and $c$,
to get
 $$\dfrac{\sin B}{b} = \dfrac{\sin C}{c}.$$

If we use the same process, dropping an altitude from angle C, we will
get
  $$\dfrac{\sin B}{b} = \dfrac{\sin A}{a}.$$
Combining these two gives the result.
\end{document}
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