\chapter{Trigonometry}
\section{Using inverse trig functions}
In Unit 4.5 we saw how to use the reference angle to find all angles that have a given trig
value that we can \textbf{recognize}
as coming from a 30-60-90 or 45-45-90 triangle, or a quadrantal angle.
If the trig value is \textbf{not} recognizable, we can use the inverse trig functions and a calculator to solve the same type
of problem and find an approximate value for the angle.
\begin{example}
Suppose we know that \(\displaystyle \sin \theta=-\frac{1}{3}\) and \(\theta\) is in the interval \([0,360^\circ]\).
We want to find all possible values of \(\theta\).
Since the sine is negative, the angles must be in Q3 and Q4. But the number 3 is not on the 30-60-90 or
45-45-90 triangle, and the angle surely is not quadrantal. So the best we can do is use a
calculator and get a decimal approximation. We will approximate the value to three decimal places.
\[\theta =\sin^{-1}\left(-\frac{1}{3}\right) \approx -19.471^\circ.\]
This angle is negative and in Q4, because the calculator uses the inverse function \(\sin^{-1}\),
that has range \([-90^\circ,90^\circ]\).
But we are looking for solutions in the interval \([0,360^\circ]\), so we cannot have
negative angles, and we also want the solutions in Q3.
The reference angle is \(\theta_R=19.471^\circ\), and we draw that in Q3 and Q4, where we know
that our solutions must be. Then we find two solutions:
\(\theta_1\approx 180^\circ +19.471^\circ=199.471^\circ\),
and \(\theta_2\approx 360^\circ -19.471\circ =340.529^\circ\) (see the picture):
\[\img{U5_4F8.png}{}{16em}{}\]
\end{example}
\subsection{Applications to right triangles}
In Unit 4.6 we solved the problem of finding all sides of a right triangle when
we are given one side and one angle. We now discuss the reverse problem: find all
angles of a right triangle if we know two sides.
Consider the triangle in the figure below:
\[\img{U4_10F1.png}{}{10em}{}\]
The definitions of the three trig functions given in
Unit 4.3 (using SOH-CAH-TOA) are:
\[\sin\theta =\frac{a}{c}, \hspace{5ex} \cos\theta = \frac{b}{c}, \hspace{5ex}
\tan\theta= \frac{a}{b}.\]
Using the translation formulas from the previous sections, we find three different ways to
solve for the angle \(\theta\):
\[\theta =\sin^{-1}\left(\frac{a}{c}\right), \hspace{5ex}
\theta =\cos^{-1}\left(\frac{b}{c}\right), \hspace{5ex}
\theta= \tan^{-1}\left(\frac{a}{b}\right).\]
So as soon as we have two of the sides, we can find \(\theta\).
\begin{example}
To find angle \(\theta\) in the triangle
\[\img{U4_10F2a.png}{}{15em}{}\]
we are given the Adj and Hyp, so we use the cosine function:
\[\cos\theta =\frac{\mbox{Adj}}{\mbox{Hyp}}=\frac{11}{12},\]
and so
\[\theta=\cos^{-1}\left(\frac{11}{12}\right)\approx 23.55^\circ\]
\end{example}
\subsection{Composition of a trig function with the inverse of the same trig function}
We know from Unit 1.9 that an inverse function undoes the function:
\[\left( f\circ f^{-1}\right)(x)=x, \hspace{5ex} \left( f^{-1}\circ f \right)(x)=x.\]
We would like to apply this to the composition of a trig function with its inverse, such as
\(\sin(\sin^{-1}x)\) and \(\sin^{-1}(\sin x)\).
But we need to be careful with the domain. We know that the original, unrestricted sine
function does \textbf{not} have an inverse. It is only when we restrict the input value \(x\) to the
interval \([-\pi/2,\pi/2]\) (or \([-90^\circ , 90^\circ]\))
that we can define the inverse function \(\sin^{-1}\).
So we cannot expect that if we use an \(x\) value outside of that interval, then \(\sin^{-1}\)
will undo \(\sin\), and in fact it does not, as we show in the examples below. It will be helpful to check
these examples with a calculator.
\begin{example}
Suppose we want to find \(\sin^{-1}(\sin(30^\circ)\). The input value \(30^\circ = \pi/6\) is
in the domain of the \textbf{restricted } sine function, and so \(\sin^{-1}\) will undo it:
\[\sin^{-1}(\sin(30^\circ)=30^\circ.\]
Check with a calculator. In degree mode,
type
\[\fbox{\(\sin^{-1}\)}(\fbox{\(\sin\)}(\fbox{\(30^\circ\)}\fbox{\()\)}\fbox{\()\)}\fbox{\(=\)}\] and you will get back 30.
\end{example}
\begin{example}
But now consider \(\sin^{-1}(\sin(240^\circ)\). The input value is not in the domain
of the restricted sine, and \(\sin^{-1}(\sin(240^\circ)\) is \textbf{not} \(240^\circ\).
(Check with a calculator! What do you get?)
So what is the answer? Once again, we use the reference angle.
We need to change the input angle \(\theta\) with
another angle that has the same reference angle as \(\theta\), keeps the same sign for
the trig function, and is in the right domain \([-90^\circ,90^\circ]\).
So we do it in three steps:
\begin{enumerate}
\item Find the reference angle: for \(\theta=240^\circ\), the reference angle is
\(\theta' = 60^\circ\).
\item
Determine the sign of the given trig value: \(\sin(240^\circ)\) will be negative,
by the ASTC rule, because \(240^\circ\) is in Q3.
\item Choose the angle in the interval
\([-90^\circ,90^\circ]\) that will keep the sine negative, and has the same reference
angle \(60^\circ\). The interval \([-90^\circ,90^\circ]\) is either Q1 or Q4,
and we want to the sine to be negative, so we must use Q4, and we choose the angle \(-60^\circ\).
\end{enumerate}
So we can substitute the original angle \(240^\circ\) with \(-60^\circ\):
\[\sin^{-1}(\sin(240^\circ))=\sin^{-1}(\sin(-60^\circ)).\]
But now \(-60^\circ\) is in the right domain for the restricted sine function, and we so
it will be undone by \(\sin^{-1}\):
\[\sin^{-1}(\sin(240^\circ))=\sin^{-1}(\sin(-60^\circ))=-60^\circ.\]
\end{example}
The simplification of \(\sin(\sin^{-1}x) \) is easier: either \(x\) is in the right domain
for \(\sin^{-1}\), and then the expression simplifies to \(x\), or it is not, and then the
expression is undefined. The same is true for \(\cos^{-1}\). The inverse function
\(\tan^{-1}\) is even easier, because it is never undefined.
\begin{example}
\begin{enumerate}
\item
To simplify \(\sin(\sin^{-1}(-0.45))\), we note that \(-0.45\) is in \([-1,1]\), the domain
of \(\sin^{-1}\). So
\[\sin(\sin^{-1}(-0.45))=-0.45.\]
\item In the expression \(\cos(\cos^{-1} 2)\), the input value is not in the domain of
\(\cos^{-1}\). So
\[\cos(\cos^{-1} 2) \mbox{ is undefined}.\]
\item
The expression \(\tan(\tan^{-1} x)\) is never undefined, because the domain of \(\tan^{-1}\)
is \((-\infty,\infty)\). So
\[\tan(\tan^{-1} x) = x \mbox{ for all reall numbers } x.\]
\end{enumerate}
\end{example}
\subsection{Composition of a trig function with the inverse of a different trig function}
Sometimes we need to simplify expressions that contain a trig function composed with the inverse of a different
trig function.
\begin{example}
We want to find the exact value of \(\sin(\tan^{-1}(-2/3))\). First of all remember that
\(\tan^{-1}(-2/3)\) is an angle, so we give it the name \(\theta\):
\[\theta = \tan^{-1}\left(-\frac{2}{3}\right).\]
Then use the translation formula to get:
\[\tan \theta = -\frac{2}{3}.\]
So the tangent is negative, and the angle \(\theta\) must be in Q2 or Q4. But
\(\theta\) is in the range of \(\tan^{-1}\), that is \((-\pi/2,\pi/2)\), so Q2 is impossible, and
\(\theta\) must be in Q4. Now using the \(x,y,r\) definition, write
\[\tan \theta =-\frac{2}{3} = \frac{y}{x}.\]
Since we are in Q4, \(x\) is positive and \(y\) is negative. So \(x=3\), \(y=-2\). Now find
\(r\):
\[\begin{array}{rcl}
3^2+(-2)^2 & = & r^2\\[1ex]
9+4&=& r^2 \\[1ex]
13 &=& r^2\\[1ex]
r&=& \sqrt{13}
\end{array}\]
Since \(\sin \theta = y/r\), we conclude that
\[\sin \theta = \sin\left(\tan^{-1}\left(-\frac{2}{3}\right)\right)=-\frac{2}{\sqrt{13}}.\]
\end{example}
If an expression involving a trig and an inverse trig function contains a variable, we can often
simplify it, but we always need to keep track of the sign, as in the following example.
\begin{example}
We want to simplify the expression
\(\cos(\tan^{-1} t)\).
As before, we write
\(\theta = \tan^{-1} t\)
and use the translation formula: \(\tan \theta = t\). Using the \(x,y,r\) definition,
\[\tan \theta =\frac{y}{x} =\frac{t}{1}.\]
At this point we need to think about the sign of the variables. Since the range of \(\tan^{-1}\) is in Q1 or Q4, we know that
that \(x\) will be positive, so we can take \(x=1\) and \(y=t\), and we find \(r=\sqrt{x^2+y^2}=\sqrt{1+t^2}\).
Since \(\cos \theta =x/r\),
we find
\[\cos \theta = \cos(\tan^{-1} t)=\frac{1}{\sqrt{1+t^2}}.\]
\end{example}
Problems
\problem
Find the degree measure of the angle in the picture, approximated to two decimal places:
\[\img{U4_10F3a.png}{}{15em}{}\]
\begin{sol}
Since
\[\sin \theta =\frac{\mbox{Opp}}{\mbox{Hyp}}=\frac{7}{15},\]
we find
\[
\fbox{\(\displaystyle \theta =\sin^{-1}\left(\frac{7}{15}\right)\approx 27.82^\circ\)}
\]
\end{sol}
\mproblem
Find the degree measure of the angle in the picture, approximated to two decimal places:
\[
\img{U4_10F4.png}{}{15em}{}
\]
\problem
Find the exact value of the following:
\begin{enumerate}
\item
\(\cos^{-1}(\cos(-30^\circ))\).
\item
\(\sin(\sin^{-1} 3)\)
\item
\(\displaystyle \tan^{-1}\left(\tan \left(\frac{2\pi}{3}\right)\right)\)
\item
\(\displaystyle \cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right)\)
\item
\(\cos(\cos^{-1}(-0.34))\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The angle \(-30^\circ\) is not in \([0,\pi]\), the domain for the restricted cosine.
The reference angle is \(30^\circ\), and the quadrant for \(-30^\circ\) is Q4, where
cosine is positive. So we need an angle in \([0,\pi]\) with reference angle \(30^\circ\)
and positive cosine. That angle is \(30^\circ\). So
\[\fbox{\(\cos^{-1}(\cos(-30^\circ))=30^\circ\)}.\]
\item
The domain of \(\sin^{-1}\) is \([-1,1]\), so
\[\fbox{\(\sin(\sin^{-1} 3)\) is undefined}.\]
\item
The angle \(2\pi/3\) is not in the interval \((-\pi/2,\pi/2)\), the domain of the
restricted tangent. The reference angle is \(\pi/3\), and the quadrant is Q2, where the
tangent is negative. So we need an angle in \((-\pi/2,\pi/2)\) with reference angle
\(\pi/3\) and negative tangent. That angle is \(-\pi/3\). So
\[\fbox{\(\displaystyle \tan^{-1}\left(\tan \left(\frac{2\pi}{3}\right)\right)=-\frac{\pi}{3}
\)}.\]
\item
The angle \(5\pi/6\) is in \([0,\pi]\), the domain of the restricted cosine. So
\[\fbox{\(\displaystyle \cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right)
=\frac{5\pi}{6}\)}.\]
\item
The domain of \(\cos^{-1}\) is \([-1,1]\). So
\[\fbox{\(\cos(\cos^{-1}(-0.34))=-0.34\)}.\]
\end{enumerate}
\end{sol}
\mproblem
Find the exact value of the following:
\begin{enumerate}
\item
\(\sin^{-1}(\sin(-30^\circ))\).
\item
\(\tan(\tan^{-1} 12)\)
\item
\(\displaystyle \cos^{-1}\left(\cos \left(\frac{5\pi}{3}\right)\right)\)
\item
\(\displaystyle \sin^{-1}\left(\sin\left(\frac{5\pi}{6}\right)\right)\)
\item
\(\sin(\sin^{-1}(-2))\)
\end{enumerate}
\problem \
\begin{enumerate}
\item Find the exact value:
\(\displaystyle \cos\left(\tan^{-1}\left(-\frac{12}{5}\right)\right)\)
\item
Simplify the expression:
\(\displaystyle \tan\left(\sin^{-1}t\right)\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
Write
\[\theta =\tan^{-1}\left(-\frac{12}{5}\right), \hspace{3ex}
\tan \theta = -\frac{12}{5}=\frac{y}{x}.\]
Since the tangent is negative, the angle must be in Q2 or Q4. But Q2 is impossible, because
the range of \(\tan^{-1}\) is \((-\pi/2,\pi/2)\). So \(\theta\) must be in Q4, where \(x\) is
positive and \(y\) is negative. So we use \(x=5\), \(y=-12\), and then
\(r=\sqrt{x^2+y^2}=\sqrt{25+144}=13\). We need \(\cos \theta =x/r\), so we find
\[\fbox{\(\displaystyle \cos\theta = \cos\left(\tan^{-1}\left(-\frac{12}{5}\right)\right)
= \frac{5}{13}\)}.\]
\item
Write
\[\theta=\sin^{-1}t , \hspace{3ex} \sin \theta = t=\frac{t}{1}=\frac{y}{r},\]
and since \(r\) must be positive, use \(r=1\), \(y=t\), so \(x=\sqrt{r^2-y^2}=\sqrt{1-t^2}\),
and since \(\tan \theta = y/x\),
\[\fbox{\(\displaystyle \tan \theta = \tan \left(\sin^{-1}t\right)=
\frac{t}{\sqrt{1-t^2}}\)}.\]
\end{enumerate}
\end{sol}
\mproblem \
\begin{enumerate}
\item Find the exact value:
\(\displaystyle \sin\left(\tan^{-1}\left(-\frac{4}{3}\right)\right)\)
\item Simplify the expression:
\(\displaystyle \cos\left(\tan^{-1}t\right)\)
\end{enumerate}