\documentclass[11pt]{amsart}

% The next four lines override the settings in the amsart style file.
\voffset=-0.8in    % Moves text up
\hoffset=-1.2in    % Moves text left
\textwidth=6.8in   % Sets new page text width
\textheight=9.3in  % Sets new page text height

\parindent=0pt
\parskip=8pt

\usepackage{amssymb}
\usepackage{graphicx} % For including EPS files

\input{isomath}
\input{mathenv}

%% ================================================================
\begin{document}

\thispagestyle{empty} % Removes page numbering for this page only
\centerline{\textbf{\large Exam \#2 Solutions $\cdot$ Monday November 20, 2006}}
\vspace*{.2in}
MATH 111 $\cdot$ Section 7 $\cdot$ Fall 2006  \hspace*{0.5in} Name \hrulefill

\emph{
The solutions I've given here include my explanations and thus are wordier than
I would expect your responses to be.  Only problem 13 is a writing problem; for
the rest, computational answers suffice.
}

%% ================================================================
%Graphs:

\textbf{Problem 1.}
Given the function $y=\frac{1}{3}\sin(4x+1)$, find the average value and period.

\textbf{Solution:}
The average value is what's added to the $\sin$ term, which is zero.
For the period, solve the inequality
\begin{eqnarray*}
	0 &\le 4x+1 &\le 2 \pi \\
	-1 &\le 4x &\le 2 \pi - 1\\
	-\frac{1}{4} &\le x &\le \frac{2 \pi}{4} - \frac{1}{4}.
\end{eqnarray*}
The period is the right-hand side minus the left-hand side, which is $\pi/2$.

%\textbf{Solution (ii):}  Find the phase shift by solving
%\begin{eqnarray*}
%	4x+1 &=& 0 \\
%	x &=& -\frac{1}{4}.
%\end{eqnarray*}
%Then solve
%\begin{eqnarray*}
%	4x+1&=& 2 \pi\\
%	4x&=& 2 \pi - 1\\
%	x &=& \frac{2 \pi}{4} - \frac{1}{4}.
%\end{eqnarray*}
%Then subtract the phase shift from this to obtain $\pi/2$.
%The average value is what's added to the $\sin$ term, which is zero.

%\textbf{Solution (iii):}  Graph the function in your calculator
%and observe that the period looks like $\pi/2$, and the average value
%looks like $0$.

\textbf{Problem 2.}
Given the following graph, find the maximum value, minimum value, and
amplitude.

\includegraphics{e2f1.eps}

\textbf{Solution:}  Read off the values max = 5, min = -1, and amplitude
= 3.  Or, amplitude is $(\max-\min)/2 = (5+1)/2=3$.

\textbf{Problem 3.}
Write an equation for a sine function which passes through all the points
in the following data table:
$$
\begin{tabular}{|r||r|r|r|r|r|r|}
\hline $x$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ \\
\hline $y$ & $-1$ & $3$ & $7$ & $3$ & $-1$ & $3$ \\
\hline
\end{tabular}
$$

\textbf{Solution:}  First graph the function:

\includegraphics{e2p3.eps}

Now read off the amplitude, average value, period, and phase shift from the
graph and write down
\begin{eqnarray*}
	y &=& \textrm{avg. val.} + \textrm{amplitude} \cdot \sin\left(\frac{2\pi}
		{\textrm{period}}(x - \textrm{phase shift})\right)\\
	y &=& 3 + 4 \sin\left(\frac{2\pi}{4}(x-1)\right)
		= 3 + 4 \sin\left(\frac{\pi}{2}(x-1)\right).
\end{eqnarray*}


%% ----------------------------------------------------------------
\newpage
\textbf{Problem 4.}
Which of the following correctly represents $\cos(-\theta)+\sin(-\theta)$?

\begin{tabular}{llllllllll}
(A) & $ \cos(\theta)+\sin(\theta)$ &
(B) & $ \cos(\theta)-\sin(\theta)$ &
(C) & $-\cos(\theta)+\sin(\theta)$ &
(D) & $-\cos(\theta)-\sin(\theta)$ \\
(E) & None of these.
\end{tabular}

\textbf{Solution:}  Since cosine is an even function and sine is an odd
function, we have, for all $\theta$, $\cos(-\theta)=\cos(\theta)$
and $\sin(-\theta)=-\sin(\theta)$.  So, the answer is B.

\textbf{Problem 5.}
Verify the following trigonometric identity.  (Hint:  use a sum identity and
the Pythagorean identity.)
$$
	\cos(2x) = 1 - 2\sin^2(x).
$$

\textbf{Solution:}
The sum identity for cosine gives us
\begin{eqnarray*}
	\cos(2x) &=& \cos(x+x) \\
	&=& \cos(x)\cos(x) - \sin(x)\sin(x) \\
	&=& \cos^2(x) - \sin^2(x).
\end{eqnarray*}
You're asked to get the right-hand side looking like $1-2\sin^2(x)$, but there
is a $\cos^2(x)$ term in there to be gotten rid of.  Fortunately, we know from
the Pythagorean identity that
$$
	\cos^2(x) + \sin^2(x) = 1.
$$
Using this, we can put
$$
	\cos^2(x) = 1 - \sin^2(x)
$$
into the above, picking up from where we left off, to obtain
\begin{eqnarray*}
	\cos(2x)
	&=& \cos^2(x) - \sin^2(x) \\
	&=& (1-\sin^2(x)) - \sin^2(x) \\
	&=& 1-2\sin^2(x).
\end{eqnarray*}

\textbf{Problem 6.}
Determine the exact value of $\sin(7\pi/12)$ using sum and/or difference
identities.

\textbf{Solution:}  This is perhaps easier in degrees, perhaps not.
Using degrees, we have
$$
	\frac{7\pi}{12} = \frac{7\pi \, 180^\circ}{12 \pi} 
	= 7 \cdot 15^\circ = 105^\circ.
$$
Now, $105^\circ=60^\circ+45^\circ$ so we can take advantage of known values for
sine and cosine at $60^\circ$ and $45^\circ$, along with the sum identity for
sine:
\begin{eqnarray*}
	\sin(60^\circ + 45^\circ)
	&=& \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ) \\
	&=& \frac{\sqrt{3}}{2}\frac{\sqrt 2}{2} + \frac{1}{2}\frac{\sqrt 2}{2}
	= \frac{\sqrt 6 + \sqrt 2}{4}.
\end{eqnarray*}
Using radians, we have
$$
	\frac{7\pi}{12} = \frac{4\pi}{12} + \frac{3\pi}{12}
	= \frac{\pi}{3} + \frac{\pi}{4}
$$
so we can take advantage of known values for sine and cosine at $\pi/3$ and
$\pi/4$, along with the sum identity for sine:
\begin{eqnarray*}
	\sin(\pi/3 + \pi/4)
	&=& \sin(\pi/3)\cos(\pi/4) + \cos(\pi/3)\sin(\pi/4) \\
	&=& \frac{\sqrt{3}}{2}\frac{\sqrt 2}{2} + \frac{1}{2}\frac{\sqrt 2}{2}
	= \frac{\sqrt 6 + \sqrt 2}{4}.
\end{eqnarray*}
Remember that switching between radians and degrees changes the units of the
\emph{input} to the trigonometric functions, but doesn't change the
\emph{output}.  So, we get the same answer either way.

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 7.}
If $\cos(\theta)=-4/5$ and $\sin(\theta)>0$, find $\tan(\theta)$.

\textbf{Solution:}  Since $\cos(\theta) = x/r$ is negative, $\theta$ is in
quadrant II or III; since $\sin(\theta) = y/r$ is positive, $\theta$ must be in
quadrant II.  Now we can make a cartoon of what we're looking for, to make it
easier to do the math.  We put the angle $\theta$ in standard position with
a point $(x,y)$ on the terminal edge and also on the unit circle:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,2.1)
%\put(0.0,0.0){\framebox(6,2.1)}
\put(0.80,1.00){\vector(1,0){2.40}}
\put(2.00,0.00){\vector(0,1){2.10}}
\put(2.00,1.00){\line(-4,3){0.80}}
\put(1.20,1.00){\line(0,1){0.60}}
\put(1.50,1.50){$r=1$}
\put(1.25,0.75){$x=-\frac{4}{5}$}
\put(0.50,1.25){$y=?$}
\input{circle.tex}
\end{picture}

Since $x^2+y^2=1$ on the unit circle --- which is the same as
$\cos^2(\theta) + \sin^2(\theta)=1$ --- we can write
\begin{eqnarray*}
	x^2 + y^2 &=& 1 \\
	y^2 &=& 1 - x^2 \\
	&=& 1 - \left(\frac{-4}{5}\right)^2
	= 1 - \frac{16}{25}
	= \frac{9}{25} \\
	y &=& \pm \sqrt{\frac{9}{25}}
	= \pm \frac{3}{5}.
\end{eqnarray*}
Since we are given $\sin(\theta) > 0$, we know to pick $y=3/5$.
Then
$$
	\tan(\theta) = \frac{y}{x} = \frac{3/5}{-4/5} = \frac{-3}{4}.
$$

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 8.}
Which of the following is not an identity?

\begin{tabular}{llllllllll}
(A) & $\sin^2(x)+\cos^2(x)=1$ &
(B) & $\cos(x+y) = \cos(x)\cos(y) + \sin(x)\sin(y)$ \\
(C) & $\sin(2 x) = 2 \sin(x)\cos(x)$ &
(D) & $\cos(\pi/2-x) = \sin(x)$  \\
(E) & These are all identities.
\end{tabular}

\textbf{Solution:}
B is not an identity.  The correct version of the sum identity for cosine
is
$$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y).$$
Answer A is the Pythagorean identity; C is the double-angle identity for sine,
which follows immediately from the sum identity for sine; D is a cofunction
identity.

\textbf{Problem 9.}
Find all solutions for $0 \le x < 2\pi$ (or, if you prefer, do it in degrees
for $0^\circ \le x < 360^\circ$):
$$
	(\sin(x) + 1/2)(\sin(x)-1/2) = 0.
$$

\textbf{Solution:}
Whenever the product of two (or more) real numbers is zero, at least one of
them is zero.  So, we have two cases:

\begin{itemize}
\item If the right-hand factor is zero, then $\sin(x)=1/2$.
There are two points on the unit circle with height $1/2$:  they are
$30^\circ$ and $150^\circ$.
\item If the left-hand factor is zero, then $\sin(x)=-1/2$.
There are two points on the unit circle with height $-1/2$:  they are
$210^\circ$ and $330^\circ$.  (If you're like me, you'd more naturally
think of $330^\circ$ as $-30^\circ$.  But, the problem asked for solutions
between $0^\circ$ and $360^\circ$.)
\end{itemize}
So, we have four solutions: $30^\circ$, $150^\circ$, $210^\circ$, and
$330^\circ$.
In radians, these are
$\pi/6$,
$5\pi/6$,
$7\pi/6$, and
$11\pi/6$.

\textbf{Problem 10.}
Solve for $x$:  $$y=5\,\cos^{-1}(2x)+3.$$

\textbf{Solution:}
\begin{eqnarray*}
	y &=& 5\,\cos^{-1}(2x)+3 \\
	y-3 &=& 5\,\cos^{-1}(2x) \\
	\frac{y-3}{5} &=& \cos^{-1}(2x) \\
	\cos\left(\frac{y-3}{5}\right) &=& 2x \\
	\frac{1}{2}\; \cos\left(\frac{y-3}{5}\right) &=& x.
\end{eqnarray*}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 11.}
Find the exact value of $\cos(\sin^{-1}(-2/3))$.

\textbf{Solution:}  Remember that trig functions take angles to numbers, while
their inverse functions take numbers to angles.  So, $\sin^{-1}(-2/3)$ is an
angle.  Which angle is it?  The $\sin^{-1}$ function has a range of $-90^\circ$
to $90^\circ$; $-2/3$ is negative so we are looking for an angle $\theta$ with
negative height, where $\sin(\theta)=-2/3$.  We can draw a cartoon to help us
get the math right:

\setlength{\unitlength}{1.0in}
\begin{picture}(6,2.2)
%\put(0.0,0.0){\framebox(6,2.2)}
\put(0.80,1.00){\vector(1,0){2.40}}
\put(2.00,0.00){\vector(0,1){2.10}}
\put(2.00,1.00){\line(4,-3){0.80}}
\put(2.80,1.00){\line(0,-1){0.60}}
\put(2.10,0.50){$r=1$}
\put(3.00,0.75){$y=-\frac{2}{3}$}
\put(2.25,1.10){$x=?$}
\input{circle.tex}
\end{picture}

On the unit circle, $r=1$ and so $\cos(\theta)=x/r=x/1=x$.  We can use
the Pythagorean theorem, $x^2+y^2=1$, to find
\begin{eqnarray*}
	x^2 + \left(\frac{-2}{3}\right)^2 &=& 1 \\
	x^2 + \frac{4}{9} &=& 1 \\
	x^2 &=& 1 -  \frac{4}{9}
	= \frac{5}{9} \\
	x &=& \pm \frac{\sqrt 5}{3}.
\end{eqnarray*}
All angles given back by the sine function are between $-90^\circ$ and
$90^\circ$ and so are in quadrants I and IV, where $x$ is non-negative.
So, we choose the positive square root to get
$
	x = \frac{\sqrt 5}{3}.
$

\textbf{Problem 12.}
Find the exact value of $\arcsin(\sin(135^\circ))$.

\textbf{Solution:}  From chapter 1, $\sin(135^\circ)$ is one of our well-known
values, namely, $\sqrt{2}/2$.  Now remember that trig functions map angles to
numbers, while their inverse functions map numbers to angles.  There are, of
course, \emph{many} angles whose sine is $\sqrt 2/2$, and functions (in order
to be functions) can have only one output value.  We define the inverse
sine function to map numbers from $-1$ to $1$ to angles between $-90^\circ$
and $90^\circ$.  So, $\arcsin(\sqrt 2/2) = 45^\circ$.

\textbf{Problem 13.}
Writing question:  Why is it that $\tan^{-1}(2)$ is defined, while
$\sin^{-1}(2)$ and $\cos^{-1}(2)$ are undefined?

\textbf{Solution:}  The domain of the tangent function is $(-\infty,+\infty)$
while the domain of the sine and cosine functions is $[-1,1]$; 2 is in the
former but not the latter.

The above sentence is an acceptable answer, of acceptable length.  For more
explanation of why that, in turn, is true, though:
\begin{itemize}

\item Remember that trig functions map angles to numbers, and their inverse
functions map numbers to angles.

\item Also remember that when we draw an angle in standard position, then pick
a point $(x,y)$ which is on the terminal edge of the angle and also on the unit
circle, the $y$ coordinate (the height) of that point is the sine of the angle.
Likewise, the $x$ coordinate of that point is the cosine of the angle.  The
tangent of the angle is $y/x$ which is the slope of the hypotenuse.  The $y$
and $x$ coordinates are on the unit circle, so they're confined between $-1$
and $1$.  The slope, on the other hand, can get as steep as we want.

\item Last, remember that when we compute $\sin^{-1}$ of a number, we're
asking:  \emph{What's our favorite angle whose sine is that number?} Similarly
for the $\cos^{-1}$ and $\tan^{-1}$ functions.

\end{itemize}

So, since sine and cosine don't have value 2 for any angle, we won't be able to
find an angle with sine or cosine equal to 2.  On the other hand, since the
tangent function takes on all real values, we will be able to find an angle
with tangent equal to 2.

%% ================================================================
\end{document}
