\chapter{Trigonometry}
\section{Inverse trig functions}
When we studied the applications of trig functions in Unit 4.6,
we noted that there are keys on the calculator
marked \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\). These keys are used to find the
\textit{inverse trigonometric functions}, that we are going to study in this unit.
Remember the basic facts about inverse functions (discussed in Unit 1.9):
If the graph of \(f(x)\) satisfies the Horizontal Line Test, then \(f\) has an inverse
\(f^{-1}\) that undoes what \(f\) does, so that
\[(f\circ f^{-1})(x)=x, \hspace{3ex} \mbox{and} \hspace{3ex} (f^{-1}\circ f)(x)=x.\]
Remember also the translation formula: if \(f\) has an inverse, then
\[f(a)=b \hspace{1ex} \Longleftrightarrow \hspace{1ex} f^{-1}(b)=a.\]
Our goal in this unit is to find inverse functions for the three trig functions \(\sin\),
\(\cos\) and \(\tan\). But none of these functions satisfy the
Horizontal Line Test! This is clear from even one cycle of sine or cosine. For the
tangent, it is enough to
look at two cycles to see that the Horizontal Line Test fails.
So what do the \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) keys on the calculator mean?
These keys actually refer to the inverses of what we call
\textit{restricted trig functions}.
The restricted trig functions are obtained from \(\sin\), \(\cos\) and \(\tan\)
by using only a small part
of the original domain, so that the graph of what is left will satisfy the Horizontal Line Test.
We will now discuss how to cut the domain for each function.
\subsection{The inverse sine function }
For \(\sin x\), we restrict the domain as shown in the picture:
\[
\img{U4_9F1.png}{}{18em}{}
\]
So the domain of the restricted sine function is
\[
\fbox{\(\mbox{Restricted sine Domain}=\left[-\frac{\pi}{2},\frac{\pi}{2}\right].\)}
\]
Note that we did not change the range:
\[
\fbox{\(\mbox{Restricted sine Range}=\left[-1,1\right].\)}
\]
The graph of the restricted sine satisfies the Horizontal Line Test. So it has an inverse, that we
denote by
\[
\fbox{\(\sin^{-1} x \hspace{2ex} \mbox{or} \hspace{2ex} \arcsin x.\)}
\]
As for any inverse function,
domain and range of \(\sin^{-1} x\) are found by switching the domain and range of the restricted
sine. Note that the range must be in Quadrant 1 or Quadrant 4:
\[
\fbox{\(\sin^{-1} \mbox{ or } \arcsin: \mbox{Domain }=\left[-1,1\right] \hspace{5ex}
\mbox{Range }=\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), in Q1 or Q4.}
\]
Make a mental picture of where the angle \(\sin^{-1}x\) must be:
\[
\img{U4_9F1a.png}{}{18em}{}
\]
Pay attention to the meaning of the variables. When we write \(y=\sin x\), the variable \(x\)
represents an angle (in radians), and the variable \(y\) represents a number. But if we write
\(y=\sin^{-1}x\), the angle is the output \(y\), and the input \(x\) is a number. When solving
problems involving \(\sin^{-1}x\), it is often a good idea to rename the variables to remind us
which one is the angle. So a useful choice of names is
\[\theta = \sin^{-1}x.\]
We can now state the translation formula for the inverse sine:
\[
\fbox{\(\sin x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\sin^{-1} y.\)}\\[1ex]
or\\[1ex]
\fbox{\(\sin x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\arcsin y.\)}
\]
\begin{example}
What is the exact value of \(\sin^{-1}1\)? First of all we give it a name to remind
us it's an angle:
\[\theta=\sin^{-1}1.\]
Then we use the translation formula:
\[\theta = \sin^{-1}1 \hspace{1ex} \Longleftrightarrow \hspace{1ex} \sin\theta=1.\]
So we need to look for an angle \(\theta\) whose sine is \(1\) (so it must be a quadrantal angle)
\textbf{and is in the right range}
\([-\pi/2,\pi/2]\) for \(\sin^{-1}\). There are three quadrantal angles in this interval:
\(0\), \(\pi/2\) and \(-\pi/2\). The one with the right sine value is \(\pi/2\). So we conclude that
\[
\fbox{\(\displaystyle \sin^{-1}(1)=\frac{\pi}{2}\)}.
\]
We can also use the \(\sin^{-1}\) key on a calculator to get a decimal approximation. In radians
mode we find
\[
\fbox{\(\sin^{-1}\)}(\hspace{1ex}
\fbox{1} \hspace{1ex} \fbox{)}
\hspace{1ex} \fbox{\(=\)} \hspace{1ex} 1.57079\ldots
\]
which is the decimal approximation for \(\pi/2\).
Make sure the calculator is in the right mode. If the mode is set to degree, you will get
\(\sin^{-1}(1)=90^\circ\).
\end{example}
\begin{example}
What is the exact value of \(\sin^{-1}(-1/2)\)?
We write:
\[\theta = \sin^{-1}\left(-\frac{1}{2}\right) \hspace{1ex} \Longleftrightarrow \hspace{1ex}
\sin \theta = -\frac{1}{2}.\]
So we need to find the angle \(\theta\) whose sine is \(-1/2\) \textbf{and is in the right range}
for \(\sin^{-1}\). This last condition is essential: there are many angles whose sine
is \(-1/2\), but only one of them will be in the interval \([-\pi/2,\pi/2]\).
The numbers \(1\) and \(2\) remind us of the \(30\)-\(60\)-\(90\) triangle. After drawing it,
we find that the reference angle is \(\pi/6\), or \(30^\circ\). There are two angles in
\([-\pi/2,\pi/2]\) with this
reference angle: \(\pi/6\) in Q1, and \(-\pi/6\) in Q4. But we need to get a negative sine,
so (by the ASTC rule) it must be the angle in Q4:
\[
\fbox{\(\displaystyle \sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)}.
\]
We can check with a calculator (in degree mode):
\[\fbox{\(\sin^{-1}\)}(\hspace{1ex}
\fbox{\(-\)} \hspace{1ex} \fbox{\(1\)} \hspace{1ex} \fbox{/}\hspace{1ex}
\fbox{\(2\)} \hspace{1ex} \fbox{)}
\hspace{1ex} \fbox{\(=\)} \hspace{1ex} -30.\]
\end{example}
\subsection{The inverse cosine function}
To define the inverse of \(\cos x\), we proceed in a similar way. First we restrict the domain
as shown:
\[
\img{U4_9F2.png}{}{18em}{}
\]
Domain and Range of the restricted cosine function are:
\[
\fbox{\(\mbox{Restricted cosine Domain}=[0,\pi],
\hspace{3ex}\mbox{Restricted cosine Range}=\left[-1,1\right].\)}
\]
The graph of the restricted cosine satisfies the Horizontal Line Test, and the inverse is:
\[
\fbox{\(\cos^{-1} x \hspace{2ex} \mbox{or} \hspace{2ex} \arccos x.\)}
\]
The range must be in Quadrant 1 or Quadrant 2:
\[\fbox{\(\cos^{-1} \mbox{ or } \arccos: \mbox{Domain }=\left[-1,1\right] \hspace{5ex}
\mbox{Range }=\left[0,\pi\right]\), in Q1 or Q2 .}\]
Make a mental picture of where the angle \(\cos^{-1}x\) must be:
\[
\img{U4_9F2a.png}{}{18em}{}
\]
The translation formula for the inverse cosine is:
\[
\fbox{\(\cos x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\cos^{-1} y.\)}\\[1ex]
or\\[1ex]
\fbox{\(\cos x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\arccos y.\)}
\]
\begin{example}
What is \(\cos^{-1}(-1)\)? Write:
\[\theta=\cos^{-1}(-1)\]
and use the translation formula:
\[\theta = \cos^{-1}(-1) \hspace{1ex} \Longleftrightarrow \hspace{1ex} \cos\theta=-1.\]
So we need a quadrantal angle in \([0,\pi]\) whose cosine is \(-1\). That angle is \(\pi\).
So we conclude that
\[\fbox{\(\cos^{-1}(-1)=\pi\)}.\]
\end{example}
\begin{example}
We want to find the exact value of \(\cos^{-1}( -1/2 )\). We write:
\[\theta=\cos^{-1}\left(\frac{1}{2}\right).\]
Then we use the translation formula:
\[\theta = \cos^{-1}\left(-\frac{1}{2}\right) \hspace{1ex} \Longleftrightarrow
\hspace{1ex} \cos\theta=\frac{1}{2}.\]
So we need to look for an angle \(\theta\) whose cosine is \(-1/2\),
\textbf{and is in the right range}
\([0,\pi]\) for \(\cos^{-1}\).
After drawing the \(30\)-\(60\)-\(90\) triangle, we find that the reference angle
is \(\pi/3\), or \(60^\circ\), and there are two angles in \([0,\pi]\)
with that reference angle: \(\pi/3\) in Q1 and \(2\pi/3\) in Q2 . But (by the ASTC rule) the
angle must be in Q2 to have a negative cosine. So we conclude that
\[
\fbox{\(\displaystyle \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}\)}.
\]
\subsection{The inverse tangent function}
We now look at the tangent function. This time we can make the graph satisfy the Horizontal Line
Test by simply taking one cycle between the two vertical asymptotes at \(-\pi/2\) and \(\pi/2\):
\[
\img{U4_9F3.png}{}{12em}{}
\]
Domain and Range of the restricted tangent function are:
\[
\fbox{\(\displaystyle \mbox{Restricted tangent Domain}=\left(-\frac{\pi}{2},\frac{\pi}{2}\right),
\mbox{Restricted tangent Range}=(-\infty,\infty).\)}
\]
Note that the domain is the same as that of the restricted sine, but without the endpoints.
The graph of the restricted tangent satisfies the Horizontal Line Test, and the inverse is:
\[
\fbox{\(\tan^{-1} x \hspace{2ex} \mbox{or} \hspace{2ex} \arctan x.\)}
\]
The range of \(\tan^{-1}\) must be in Quadrant 1 or Quadrant 4:
\[
\fbox{\(\displaystyle \tan^{-1} \mbox{ or } \arctan: \mbox{Domain }=(-\infty,\infty) \hspace{5ex}
\mbox{Range }=\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), in Q1 or Q4.}
\]
So the output of \(\tan^{-1}\) will be in the same quadrant as for \(\sin^{-1}\):
\[
\img{U4_9F3a.png}{}{18em}{}
\]
The translation formula for the inverse tangent is:
\[
\fbox{\(\tan x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\tan^{-1} y.\)}\\[1ex]
or\\[1ex]
\fbox{\(\tan x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\arctan y.\)}
\]
\begin{example}
To find the exact value of \(\tan^{-1} 1 \), we write:
\[\theta=\tan^{-1}1.\]
Then we use the translation formula:
\[\theta = \tan^{-1}1 \hspace{1ex} \Longleftrightarrow \hspace{1ex} \tan\theta=1.\]
So we need to look for an angle \(\theta\) whose tangent is \(1\), \textbf{and is in the right range}
\((-\pi/2,\pi/2)\) for \(\tan^{-1}\). The reference angle is \(\pi/4\), that is already in the
right interval. So
\[
\fbox{\(\displaystyle \tan^{-1}1=\frac{\pi}{4}\)}.
\]
\end{example}
\begin{example}
To find the exact value of \(\tan^{-1}(-\sqrt{3})\), we write:
\[\theta=\tan^{-1}\left(-\sqrt{3}\right)\]
and use the translation formula:
\[\theta = \tan^{-1}\left(-\sqrt{3}\right) \hspace{1ex} \Longleftrightarrow
\hspace{1ex} \tan\theta=-\sqrt{3}.\]
So we need an angle in \((-\pi/2,\pi/2)\) whose tangent is \(-\sqrt{3}\). The reference angle is
\(\pi/3\) or \(60^\circ\), and there are two possible angles: \(\pi/3\) in Q1 and \(-\pi/3\) in
Q4. But by the ASTC rule the angle must be in Q4 to have a negative tangent, so
\[\fbox{\(\displaystyle \tan^{-1}\left(-\sqrt{3}\right)=-\frac{\pi}{3}\)}.\]
\end{example}
\subsection{Calculator approximations for inverse trig functions}
Of course there are many values of the inverse trig functions \(\sin^{-1}, \cos^{-1}, \tan^{-1}\)
that we cannot find exactly, and we use a calculator for a decimal approximation.
\begin{example}
Using a calculator in degree mode, we find:
\[\sin^{-1}(1/3)\approx 19.4712^\circ, \hspace{5ex} \cos^{-1}(-0.8)\approx 143.1301 ^\circ,
\hspace{5ex} \tan^{-1}(-15) \approx -86.1859 ^\circ.\]
If we try to evaluate \(\sin^{-1}\) for an input value that is not in the domain \([-1,1]\),
we will get an error message:
\[\sin^{-1}(2) \mbox{ ERROR}\]
Naturally, that is because there is no angle \(\theta\) whose sine is \(2\).
\end{example}
\subsection{Inverse functions and the ASTC rule}
Using the range of the inverse trig functions and the ASTC rule,
we can determine the quadrant for any angle given as an inverse trig function value without
having to actually calculate its value.
\begin{example}
Suppose we are given
\[\theta =\sin^{-1}\left(-\frac{1}{3}\right).\]
This angle must be in the interval \([-\pi/2,\pi/2]\), because that is the range
of \(\sin^{-1}\). So only Q1 and Q4 are possible.
The translation formula gives us
\[\sin \theta = -\frac{1}{3},\]
so the sine is negative and by the ASTC rule the angle must be in Q4.
\end{example}
\begin{example}
If we are given
\[\theta=\cos^{-1}(-0.83)\]
we know that the angle must be in the interval \([0\pi]\), so only Q1 and Q2 are possible.
The translation formula \(\cos \theta= - 0.83\) says that the cosine is negative and so by
the ASTC rule the angle is is Q2 .
\end{example}
Domain and Range of the inverse trig functions
| \(\begin{array}{lllll}
\sin^{-1}: & D=[-1,1], \hspace{2ex} &\displaystyle R=\left[-\frac{\pi}{2},\frac{\pi}{2}\right], &
\mbox{in Q1 or Q4} & \img{U4_9F4.png}{-5em}{12em}{}
\\[4ex]
\cos^{-1}: & D=[-1,1], \hspace{2ex} & \displaystyle R=\left[0,\pi\right], &
\mbox{in Q1 or Q2 } & \img{U4_9F5.png}{-5em}{12em}{} \\[4ex]
\tan^{-1}: & D=(-\infty,\infty), \hspace{2ex} & \displaystyle
R=\left(-\frac{\pi}{2},\frac{\pi}{2}\right), & \mbox{in Q1 or Q4} & \img{U4_9F4.png}{-5em}{12em}{}.
\end{array}\) |
Translation formulas for the inverse trig functions
| \(\sin x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\sin^{-1} y.\) |
| \(\cos x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\cos^{-1} y.\) |
| \(\tan x = y \hspace{1ex} \Longleftrightarrow \hspace{1ex} x=\tan^{-1} y.\) |
Problems
\problem
Find the exact value of each of the following:
|
a. \(\sin^{-1} 0\) | b. \(\displaystyle \cos^{-1} \left(\frac{1}{2}\right)\) |
c. \(\displaystyle \sin^{-1} \left(-\frac{\sqrt{3}}{2}\right)\) | d. \(\displaystyle \tan^{-1} \left(-1\right)\)
|
\begin{sol}
\begin{enumerate}
\item
\(\theta = \sin^{-1}0 \hspace{1ex} \Longleftrightarrow \hspace{1ex} \sin\theta=0.\)
The only angle in \([-\pi/2,\pi/2]\) whose sine is \(0\) is \(\theta=0\). So
\[\fbox{\(\sin^{-1}0 =0\)}\]
\item
\(\displaystyle \theta = \cos^{-1}\left(\frac{1}{2}\right) \hspace{1ex} \Longleftrightarrow \hspace{1ex} \cos\theta=\frac{1}{2}.\)
The angle in \([0,\pi]\) whose cosine is \(1/2\) is \(\theta=\pi/3\). So
\[
\fbox{\( \displaystyle \cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)}.
\]
\item
\(\displaystyle \theta = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \hspace{1ex} \Longleftrightarrow \hspace{1ex} \sin\theta=-\frac{\sqrt{3}}{2}.\)
The reference angle is \(\pi/3\), and by the ASTC rule, the sine is negative in Q2 I and Q4. But the
range of \(\sin^{-1}\) is \([-\pi/2,\pi/2]\), so \(\theta\) must be in Q4:
\[
\fbox{\(\displaystyle \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}\)}.
\]
\item
\(\displaystyle \theta = \tan^{-1} \left(-1\right) \hspace{1ex} \Longleftrightarrow \hspace{1ex} \tan\theta=\left(-1\right).\)
The reference angle is \(\pi/4\), and by the ASTC rule the angle must be in Q4. So
\[
\fbox{\(\displaystyle \tan^{-1}\left(-1\right)=-\frac{\pi}{4}\)}.
\]
\end{enumerate}
\end{sol}
\mproblem
Find the exact value of each of the following:
|
a. \(\displaystyle \cos^{-1} 0\) | b.
\(\displaystyle \sin^{-1} (-1)\) | c. \(\displaystyle \cos^{-1} (-\sqrt{3}/2)\) | d. \(\tan^{-1}(1/\sqrt{3})\) |
.
\problem
Use the translation formulas to re-write each equation:
|
a. \(\sin \theta = 0.1\) | b. \(\cos^{-1}(1-t) = a\) | c. \(\tan(-2x^2)=b+1\).
|
\begin{sol}
\begin{enumerate}
\item \(\theta = \sin^{-1}(0.1)\)
\item \(1-t=\cos a\)
\item \(-2x^2 = \tan^{-1}(b+1)\).
\end{enumerate}
\end{sol}
\mproblem
Use the translation formulas to re-write each equation:
|
a. \(\sin^{-1}(1-x^2) = 3t\) | b. \(\cos(t+1) = x\) | c. \(\tan^{-1}(5x)=3y\).
|
\problem
Find the quadrant of the given angles without using a calculator.
|
a. \(\sin^{-1}(-0.2)\). | b. \(\tan^{-1}(10)\) | c. \(\cos^{-1}(1/5)\) | d. \(\cos^{-1}(-0.9)\)
|
\begin{sol}
\begin{enumerate}
\item
The angle is in \([-\pi/2,\pi/2]\), so only Q1 and Q4 are possible. The sine is negative, so
by the ASTC rule the angle is in Q4.
\item
The angle is in \((-\pi/2,\pi/2)\), so only Q1 and Q4 are possible. The tangent is positive,
so by the ASTC rule the angle is in Q1.
\item
The angle is in \([0,\pi]\), so only Q1 and Q2 are possible. The cosine is positive,
so by the ASTC rule the angle is in Q1.
\item
The angle is in \([0,\pi]\), so only Q1 and Q2 are possible. The cosine is negative,
so by the ASTC rule the angle is in Q2 .
\end{enumerate}
\end{sol}
\mproblem
Find the quadrant of the given angles without using a calculator.
|
a. \(\cos^{-1}(-1/4)\) | b. \(\tan^{-1}(-0.7)\) | c. \(\sin^{-1}(-1/5)\) | d. \(\sin^{-1}(0.4)\)
|