\documentclass[11pt]{article}
\usepackage{precalc}
\usepackage{graphicx}
\newcommand{\invsin}{\calc{sin^{-1}}}
\newcommand{\invcos}{\calc{cos^{-1}}}
\newcommand{\invtan}{\calc{tan^{-1}}}
\newcommand{\calc}[1]{\ensuremath{\mathsf{#1}}}
\begin{document}
\assigntitle{11}{Inverse trigonometric relations}
\begin{problem}
\topic{a shady character}
A shady-looking character slinks up next to you and whispers,
``Pssst. Hey kid. I'll tell you a real number $x$, and I'll bet you
\$50 you can't tell me an angle whose sine is $x$.'' Should you
take the bet?
\end{problem}
\begin{problem}
What if the shady character promises to tell you a real number
between $0$ and $1$?
\end{problem}
\begin{problem}
Now the shady character wants to bet you \$500 that if he gives you
a real number $x$ between $-1$ and $1$, you can't tell him
\emph{fifty different} angles whose cosine is $x$. Should you take
the bet this time?
\end{problem}
\section{Inverse trig relations}
\label{sec:inverse}
\topic{reversing trig operations}
Now that we know how to find the sine or cosine of an angle (or
tangent, secant, cosecant, or cotangent), we often want to do the
reverse: that is, given some number $x$, find an angle whose sine or
cosine is $x$. In other words, we want to find \emph{inverse
functions} for sine and cosine. However, there are a few problems
to think about first.
\begin{problem}
\begin{subproblems}
\item Clearly, the domain of the sine function is $\R$, the set of all
real numbers, since we know how to find the sine of \emph{any}
angle. Suppose we specify that the codomain of sine is also $\R$.
In this case, is sine surjective (onto)? Why or why not?
\item Now suppose we specify the codomain of sine to be the interval
$[-1,1]$. Now is sine surjective?
\item Is the sine function injective (one-to-one)?
\item What can you conclude about the existence of an inverse sine
function?
\end{subproblems}
\end{problem}
\topic{inverse relations, not functions}
Because of this, instead of inverse \emph{functions}, we talk about
inverse \emph{relations} for sine and cosine; for a given real number
$x \in [-1,1]$, these inverse relations produce a \emph{set} of angles
(an inverse \emph{function} would only be allowed to produce a single
angle), all of whose sine (or cosine, or tangent, \emph{etc.}) is $x$.
\section{The inverse cosine relation}
\label{sec:arccos}
\topic{arccos}
The inverse cosine relation is usually called arccosine and
abbreviated $\arccos$. (When writing arccos in an equation, be sure
to use the \LaTeX\ command \verb|\arccos| so that it is typeset
upright, like this: $\arccos(0.5)$ instead of like this:
$arccos(0.5)$. The same goes for arcsin and arctan.) For any input
real number $x \in [-1,1]$, it outputs the set of all angles $\theta$
for which $\cos(\theta) = x$.
To figure out how arccosine should work, let's take as an example $x =
1/2$.
\begin{problem}
Find an angle $\theta$ for which $cos(\theta) = 1/2$. That is, find
\emph{one of} the angles output by $\arccos(1/2)$.
\end{problem}
\begin{problem}
Given the angle $\theta$ you found in the previous problem, what's
one quick way to find a different angle which also has a
cosine of $1/2$? (Hint: see problem 8a from Assignment 9!)
\end{problem}
\begin{problem}
Given the same $\theta$ again, what's \emph{another} quick way to
find a different angle with a cosine of $1/2$? (Hint: see problems 6
and 7 from Assignment 9!)
\end{problem}
\topic{computing arccos} It turns out that this is enough. In order
to compute $\arccos(x)$, you only need to somehow find a single angle
$\theta$ for which $\cos(\theta) = x$; all the rest can be found by
applying the above two tricks. Putting it all together:
\begin{equation} \label{eq:arccos}
\arccos x = \pm \theta + k \cdot 2\pi
\end{equation}
where $\cos(\theta) = x$ and $k \in \Z$.
That is, given one angle whose cosine is $x$, we can (optionally) take
the negative of it, and add or subtract $2\pi$ as many times as we
want; all the angles we get in this way will have a cosine of $x$.
\begin{problem}
List six different angles in $\arccos(\sqrt{2}/2)$.
\end{problem}
\begin{problem}
Find all the angles between $2\pi$ and $3\pi$ which have a cosine of $-1/2$.
\end{problem}
\section{The inverse sine relation}
\label{sec:arcsin}
\topic{arcsin}
The inverse sine relation, as you might have guessed, is called
arcsine (abbreviated $\arcsin$).
We won't go through the details, but a process very similar to the one
for arccosine can be used to find angles which all have the same
sine. The only difference is that instead of taking the negative of
$\theta$ (an angle and its negative never have the same sine), we can
subtract $\theta$ from $\pi$ to get another angle with the same sine.
That is:
\begin{equation}
\label{eq:arcsin}
\arcsin x = \theta + 2k\pi \text{ OR } (\pi - \theta) + 2k\pi
\end{equation}
where $\sin(\theta) = x$ and $k \in \Z$.
\begin{problem}
Find all the angles between $0$ and $4\pi$ which have a sine of $1/2$.
\end{problem}
\section{Inverse trig relations and your calculator}
\label{sec:calculator}
\topic{calculators don't like printing infinite things}
So, where are arccos and arcsin on your calculator? Look hard as you
may, you will not find them. This is for a very practical reason:
having a button which makes your calculator print out an infinite
number of answers would be, shall we say\dots Not Very Useful.
\topic{inverse trig functions}
Instead, your calculator has buttons labeled \invsin,
\invcos, and \invtan, which are similar but slightly different. Let's
figure out what they do.
\begin{problem}
Try typing $\calc{\invsin(1/2)}$ into your calculator. What
answer do you get? What is this in terms of $\pi$ (hint: try
dividing by $\pi$)?
\end{problem}
\begin{problem}
Use your calculator to compute the \invsin\ of a bunch of other
values. All the answers should fall within a certain range. What do
you think the range of possible answers is?
\end{problem}
In other words, \invsin\ is actually an inverse \emph{function} for
sine on a very restricted domain, and similarly for the others.
But this is very useful: given $x$, it lets you compute \emph{one}
angle whose sine is $x$. This is all we really need. As we know from
equation \eqref{eq:arcsin}, once we know one such angle $\theta$,
figuring out all the others is simple.
\section{Beware!}
\label{sec:beware}
\begin{center}
\topic{beware!}
{\huge Beware!}
\end{center}
\invsin\ (and likewise \invcos\ and \invtan) are quite possibly
the \emph{most horribly chosen mathematical notation} in the
\emph{history of the world}.\footnote{like, seriously.} Let me explain why.
\begin{itemize}
\item Normally, raising something to the power of negative one means
to take the reciprocal. For example, $5^{-1} = 1/5$.
\item So, you \emph{might think} that $\sin^{-1} x$ means $1/(\sin
x)$, but if you thought that you would be HORRIBLY WRONG.
$\sin^{-1}$ is \emph{just a notation} for the inverse sine function;
it has \emph{nothing to do with raising anything to the negative first
power}!. It might as well be $\sin^{\oplus} x$ or
$\sin^{\rightarrow \bot \#} x$.
\item In particular, $\sin^{-1} x$ and $1/(\sin x) = \csc x$ are
\emph{not the same thing}.
\item To make things even worse, as you will learn next week, $\sin^2
x$ \emph{is} actually used as a notation for the square of $\sin
x$. In other words:
\begin{gather}
\sin^2 x = (\sin x)^2 \\
\intertext{but}
\sin^{-1} x \neq (\sin x)^{-1}
\end{gather}
\item I would like to find whoever made up this notation for inverse
trig functions and make them pay \$10 to every student who has ever
been confused by it. The only problem is they probably don't have
A GAZILLION DOLLARS.
\end{itemize}
\section{Inverse trig relations and problem-solving}
\label{sec:problem-solving}
\topic{solving equations with arctrig relations}
We can use arccos and arcsin to solve problems involving sine and
cosine. If you have an equation of the form \[ \sin \theta = x, \]
you can solve for $\theta$ by taking the arcsin: \[ \theta = \arcsin
x. \] Of course, the same goes for cosine. You can then use your
calculator to compute $\invsin x$ or $\invcos x$, and then use
equations \eqref{eq:arcsin} and \eqref{eq:arccos} to find other
solutions as necessary.
\begin{problem}
Find all angles $\theta \in [0,2\pi)$ for which \[ 3
(\cos(\theta))^2 + 2 = 4. \]
\end{problem}
\begin{problem}
The triangle shown in \pref{fig:prob1} is a right triangle with legs
of length $3$ and $4$. What is $\theta$? (Give your answer in
radians, rounded to three decimal places.)
\begin{figure}[htp]
\centering
\includegraphics{diagrams/arctrig-prob-1.eps}
\caption{A triangle.}
\label{fig:prob1}
\end{figure}
\end{problem}
\end{document}