\chapter{Analytic trigonometry}
\section{Basic trig identities}
In this unit, we will make heavy use of algebra to simplify expressions or to solve equations
containing trigonometric functions.
As a shortcut, we will always write \(\sin^2 x\) or \(\cos^2 x\)
instead of \((\sin x)^2\) or \((\cos x)^2\), and similarly
for the other trig functions. But the calculator will not understand \(\sin^2 x\).
You will need to type instead \((\sin(x))^2\). On the other hand, WeBWorK will understand both
\(\sin^2 x\) and \((\sin x)^2\).
Make sure not to confuse \(\sin^2 x\) (that means, we are squaring \(\sin x\) )
with \(\sin(x^2)\) (that means, we are taking the sine of \(x^2\).
We begin with a review of some factoring problems.
\subsection{Review of factoring}
Here are a few factoring examples:
\[\begin{array}{rcl}
x^2 -x & =& x(x-1)\\[1ex]
4t^2-1&=&(2t-1)(2t+1)\\[1ex]
9-9y&=& 9(1-y)
\end{array}\]
If instead of a single variable (\(x,t\) or \(y\)) we use a more complicated symbol or
expression, the factoring formulas are the same:
\[\begin{array}{rcl}
\sin^2 x -\sin x & =& \sin x(\sin x-1)\\[1ex]
4\tan^2 \theta-1&=&(2\tan \theta-1)(2\tan \theta+1)\\[1ex]
9 - 9 \cos^2 x&=& 9(1-\cos^2 x)
\end{array}\]
So there is no real difference between doing algebra with single variables like \(x,t,y\)
or with trig functions like \(\sin \theta\), \(\cos \theta\), \(\tan \theta\).
\subsection{The basic trig identities}
We now begin our list of basic trig identities, that we divide into five groups.
\subsubsection{Reciprocal identities}
Recall from the definition of the trig functions using SOH-CAH-TOA in Unit
4.3 that the last three functions \(\csc, \sec, \cot\) are defined as the reciprocal of
the first three \(\sin, \cos, \tan\). This of course also means that the first three
are the reciprocal of the last three. So we have the identities:
\[\fbox{\(\begin{array}{ccc}
\displaystyle \csc \theta =\frac{1}{\sin \theta} & \hspace{2ex}
\displaystyle \sec \theta =\frac{1}{\cos \theta} & \hspace{2ex}
\displaystyle \cot \theta =\frac{1}{\tan \theta} \\[2ex]
\displaystyle \sin \theta =\frac{1}{\csc \theta} & \hspace{2ex}
\displaystyle \cos \theta =\frac{1}{\sec \theta} & \hspace{2ex}
\displaystyle \tan \theta =\frac{1}{\cot \theta}
\end{array}\)}\]
We have in fact already made use of these identites in Unit 4.6, where we used the reciprocal key
\(\fbox{\(x^{-1}\)}\) on a calculator
to compute \(\sec, \csc\)
and \(\cot\).
\begin{example} \
\begin{itemize}
\item
\(\displaystyle \sin\left(\frac{\pi}{4}\right) =\frac{1}{\sqrt{2}}\), so
\(\displaystyle \csc\left(\frac{\pi}{4}\right) =\sqrt{2}\).
\item
\(\displaystyle \cos(60^\circ) =\frac{1}{2}\), so \(\sec(60^\circ) =2\).
\item
\(\displaystyle \tan\left(\frac{\pi}{3}\right) =\sqrt{3}\), so
\(\displaystyle \cot\left(\frac{\pi}{3}\right) =\frac{1}{\sqrt{3}}\).
\end{itemize}
\end{example}
\subsubsection{Co-function identities}
When defining the trig functions using \(\mbox{Opp}\), \(\mbox{Adj}\) and \(\mbox{Hyp}\)
in a right triangle, there are two acute angles that we can consider. These two angles
are \textit{complementary}, meaning that their sum is \(90^\circ\). So if one acute angle is
\(\theta\), then then other is \(90^\circ -\theta\) (or \(\pi/2-\theta\) if we use radians).
See the figure.
\[\img{U5_1F1.png}{}{15em}{}\]
Changing the acute angle under consideration means switching \(\mbox{Opp}\) and \(\mbox{Adj}\).
This in turn means switching sine with cosine, tangent with cotangent, secant with cosecant, and
we have the co-function identities:
\[\fbox{\(\begin{array}{ccc}
\displaystyle \sin(90^\circ-\theta) =\cos \theta & \hspace{2ex}
\displaystyle \cos(90^\circ-\theta) =\sin \theta & \hspace{2ex}
\displaystyle \tan(90^\circ-\theta) =\cot \theta \\[2ex]
\displaystyle \csc(90^\circ-\theta) =\sec \theta & \hspace{2ex}
\displaystyle \sec(90^\circ-\theta) =\csc \theta & \hspace{2ex}
\displaystyle \cot(90^\circ-\theta) =\tan \theta
\end{array}\)}\]
\begin{example} \
\begin{itemize}
\item
\(\displaystyle \sin(30^\circ) = \frac{1}{2}= \cos(60^\circ)\).
\item
\(\displaystyle \tan\left(\frac{\pi}{3}\right) =\sqrt{3} =\cot\left(\frac{\pi}{6}\right)\).
\item
\(\displaystyle \sec\left(20^\circ\right) = \csc\left(70^\circ\right)\).
\end{itemize}
\end{example}
\subsubsection{Quotient identities}
From the \(x,y,r\) definition of \(\sin \) and \(\cos\), we find
\[\frac{\sin \theta}{\cos\theta}=\frac{y/r}{x/r}=\frac{y}{x}=\tan \theta.\]
and similarly for \(\cot \theta\). So we find the quotient identities:
\[\fbox{\(\begin{array}{cc}
\displaystyle \tan \theta =\frac{\sin \theta}{\cos \theta} & \hspace{2ex}
\displaystyle \cot \theta =\frac{\cos \theta}{\sin \theta}
\end{array}\)}\]
\begin{example} \
\begin{itemize}
\item
\(\displaystyle \tan\left(\frac{\pi}{4}\right)=\frac{\sin(\pi/4)}{\cos(\pi/4)}=\frac{1/\sqrt{2}}{1/\sqrt{2}}=1\)
\item
\(\displaystyle \cot\left(\frac{\pi}{3}\right)=\frac{\cos(\pi/3)}{\sin(\pi/3)}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}\)
\end{itemize}
\end{example}
\subsubsection{Even-Odd identities}
This group of identities is obtained by changing \(\theta\) with \(-\theta\). In other words,
we find how \(f(-\theta)\) is related to \(f(\theta)\). Using the
\(x,y,r\) definition, changing \(\theta\) to \(-\theta\) means
leaving \(x\) unchanged, but changing \(y\) to
\(-y\) (see figure):
\[\img{U5_1F2.png}{}{12em}{}\]
So the functions that depend on \(y\) (\(\sin, \tan, \csc, \cot\)) change sign, while the
others do not:
\[\fbox{\(\begin{array}{ccc}
\displaystyle \sin(-\theta) =-\sin \theta & \hspace{2ex}
\displaystyle \cos(-\theta) =\cos \theta & \hspace{2ex}
\displaystyle \tan(-\theta) =-\tan \theta \\[2ex]
\displaystyle \csc(-\theta) =-\csc \theta & \hspace{2ex}
\displaystyle \sec(-\theta) =\sec \theta & \hspace{2ex}
\displaystyle \cot(-\theta) =-\cot \theta
\end{array}\)}\]
\begin{example} \
\begin{itemize}
\item
\(\displaystyle\sin(-45^\circ)=-\sin(45^\circ)=-\frac{1}{\sqrt{2}}\)
\item
\(\displaystyle\cos(-30^\circ)=\cos(30^\circ)=\frac{\sqrt{3}}{2}\)
\item
\(\displaystyle\tan\left(-\frac{\pi}{6}\right)=-\tan\left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}\)
\end{itemize}
\end{example}
\subsubsection{Pythagorean identities}
The numbers \(x,y,r\) used in the definition of the trig functions are related by the
Pythagorean Theorem:
\[\img{U5_1F3.png}{}{12em}{}\]
\[x^2+y^2=r^2.\]
If we divide through by \(r^2\), we find:
\[\frac{x^2}{r^2} +\frac{y^2}{r^2}=1\]
But the left side is the same as
\[\left(\frac{x}{r}\right)^2 +\left(\frac{y}{r}\right)^2=\cos^2 \theta + \sin^2\theta.\]
So we find the important Pythagorean identity:
\[\fbox{\(\cos^2 \theta + \sin^2 \theta =1\).}\]
This identity is often also used by subtracting either \(\sin^2\) or \(\cos^2\) from both sides:
\[\fbox{\(\sin^2 \theta =1-\cos^2 \theta\)} \hspace{5ex}
\fbox{\(\cos^2 \theta =1-\sin^2 \theta\).}\]
If we divide the identity by \(\cos^2 \theta\) we obtain another identity involving
\(\sec\) and \(\tan\):
\[\fbox{\(1+\tan^2 \theta =\sec^2 \theta\)} \hspace{3ex} \mbox{or} \hspace{3ex}
\fbox{\(\tan^2 \theta =\sec^2 \theta-1\)}\]
and if we divide it by \(\sin^2\theta\) we find an identity involving \(\csc\) and \(\cot\):
\[\fbox{\( \cot^2 \theta +1 =\csc^2 \theta\)} \hspace{3ex} \mbox{or} \hspace{3ex}
\fbox{\(\cot^2 \theta =\csc^2 \theta-1\).}\]
Solving the first Pythagorean identity \(\sin^2 \theta + \cos^2 \theta =1\) for either function we
find
\[\sin\theta = \pm \sqrt{1-\cos^2\theta} \hspace{3ex}\cos \theta = \pm \sqrt{1-\sin^2 \theta}\]
These equations
\textbf{almost} allow us
to find one of \(\sin \theta, \cos \theta\) as soon as we know the other. But in order to
decide between \(+\) or \(-\) we need to have information on the quadrant for \(\theta\).
\begin{example}
Suppose we know that \(\sin x = 0.4\), and \(\theta\) is in Q2. We want to know \(\cos x\).
By the ASTC rule, \(\cos x\) is negative in Q2. Using
the Pythagorean identity
we find
\[
\begin{array}{rcl}
\cos x &=&-\sqrt{ 1-\sin^2 x} \\[1ex]
&=& - \sqrt{1-(0.4)^2}\\[1ex]
&=& - \sqrt{1-0.16}\\[1ex]
&=& - \sqrt{0.84}\\[1ex]
&=& -0.917\ldots
\end{array}\]
\end{example}
\subsubsection{Summary of basic trig identities}
|
\textbf{Reciprocal} | \textbf{Quotient}
|
| \(\begin{array}{ccc}
\displaystyle \csc \theta =\frac{1}{\sin \theta} & \hspace{2ex}
\displaystyle \sec \theta =\frac{1}{\cos \theta} & \hspace{2ex}
\displaystyle \cot \theta =\frac{1}{\tan \theta} \\[2ex]
\displaystyle \sin \theta =\frac{1}{\csc \theta} & \hspace{2ex}
\displaystyle \cos \theta =\frac{1}{\sec \theta} & \hspace{2ex}
\displaystyle \tan \theta =\frac{1}{\cot \theta}
\end{array}\)
|
|
| \(\begin{array}{c}
\displaystyle \tan \theta =\frac{\sin \theta}{\cos \theta}\\[2ex]
\displaystyle \cot \theta =\frac{\cos \theta}{\sin \theta}
\end{array}\)
|
|
\textbf{Co-function}
| \(\begin{array}{ccc}
\displaystyle \sin(90^\circ-\theta) =\cos \theta & \hspace{2ex}
\displaystyle \cos(90^\circ-\theta) =\sin \theta & \hspace{2ex}
\displaystyle \tan(90^\circ-\theta) =\cot \theta \\[2ex]
\displaystyle \csc(90^\circ-\theta) =\sec \theta & \hspace{2ex}
\displaystyle \sec(90^\circ-\theta) =\csc \theta & \hspace{2ex}
\displaystyle \cot(90^\circ-\theta) =\tan \theta
\end{array}\)
|
\textbf{Even-Odd}
| \(\begin{array}{ccc}
\displaystyle \sin(-\theta) =-\sin \theta & \hspace{2ex}
\displaystyle \cos(-\theta) =\cos \theta & \hspace{2ex}
\displaystyle \tan(-\theta) =-\tan \theta \\[2ex]
\displaystyle \csc(-\theta) =-\csc \theta & \hspace{2ex}
\displaystyle \sec(-\theta) =\sec \theta & \hspace{2ex}
\displaystyle \cot(-\theta) =-\cot \theta
\end{array}\)
|
\textbf{Pythagorean}
| \(\begin{array}{ccc}
\displaystyle \cos^2 \theta + \sin^2 \theta = 1 & \hspace{2ex}
\displaystyle \sin^2 \theta =1-\cos^2 \theta & \hspace{2ex}
\displaystyle \cos^2 \theta =1-\sin^2 \theta \\
\displaystyle \tan^2\theta = \sec^2\theta -1 & \hspace{2ex}
\displaystyle \tan^2 \theta+1 = \sec^2 \theta\\
\displaystyle \cot^2 \theta = \csc^2\theta -1 & \hspace{2ex}
\displaystyle 1+\cot^2\theta = \csc^2 \theta
\end{array}\)
|
\subsection{Simplification of trigonometric expressions}
Using the basic identities, we can often re-write and greatly
simplify expressions that contain trig functions.
\begin{example}
Suppose we are given the expression
\[\sin \theta +\cos \theta \tan \theta.\]
Using the reciprocal identity for \(\tan \theta\), we find
\[\begin{array}{cl}
& \sin \theta +\cos \theta \tan \theta\\[1ex]
=& \displaystyle \sin \theta +\cos \theta \frac{\sin \theta}{\cos \theta}\\[2ex]
=& \displaystyle \sin \theta +\cancel{\cos \theta} \frac{\sin \theta}{\cancel{\cos \theta}}\\[2ex]
=& \sin \theta + \sin \theta \\[1ex]
=& 2\sin \theta
\end{array}
\]
\end{example}
\begin{example}
We want to simplify
\((1+\tan^2\theta) \cos \theta.\)
\[\begin{array}{rll}
& (1+\tan^2\theta) \cos \theta & \\[1ex]
=& \sec^2 \theta \cos \theta &\mbox{Use a Pythagorean identity}\\[1ex]
=& \displaystyle \sec^2 \theta \frac{1}{\sec \theta} & \mbox{Use a reciprocal identity}\\[1ex]
=& \displaystyle \sec \theta & \mbox{Simplify}
\end{array}
\]
\end{example}
\begin{example}
We now simplify the expression
\(\displaystyle \frac{1}{1-\sin x}+\frac{1}{1+\sin x}.\)
\[\begin{array}{rll}
& \displaystyle \frac{1+\sin x}{(1-\sin x)(1+\sin x)}+\frac{1-\sin x}{(1+\sin x)(1-\sin x)} &
\mbox{Find a common denominator}\\[2ex]
=& \displaystyle \frac{1+\sin x}{1-\sin^2 x}+\frac{1-\sin x}{1-\sin^2x} &\mbox{Expand denominators}\\[2ex]
=& \displaystyle \frac{1+\sin x}{\cos^2x}+\frac{1-\sin x}{\cos^2 x} & \mbox{Use a Pythagorean identity}\\[2ex]
=& \displaystyle \frac{1+\sin x +1-\sin x}{\cos^2 x} & \mbox{Combine the fractions}\\[2ex]
=& \displaystyle \frac{2}{\cos^2 x} & \mbox{Simplify}\\[2ex]
=& \displaystyle 2 \sec^2 x & \mbox{Use a reciprocal identity}
\end{array}
\]
\end{example}
\subsection{Trigonometric substitution}
A \textit{trig substitution} is the re-naming of a variable using a number
times a trig function, such as \(x=3\sin \theta\),
or \(t=5\tan \theta\). We will always assume that \(\theta \) is an acute angle, so
that all trig functions of \(\theta\) will be positive.
These substitutions are quite useful when working with certain types of square root expressions,
because they allow us to rationalize the expression (meaning that we can eliminate the
square root). In these problems, we always assume that the angle is acute.
\begin{example}
Suppose we are given the expression
\[\sqrt{9-x^2}\]
Using the trig substitution \(x=3\sin \theta\), we find
\[\begin{array}{clll}
9-x^2&=& \displaystyle 9-(3\sin \theta )^2 & \mbox{Use the trig substitution}\\[1ex]
&=& \displaystyle 9-9 \sin^2 \theta & \\[1ex]
&=& 9(1-\sin^2\theta) & \mbox{Factor out the \(9\)}\\[1ex]
&=& 9 \cos ^2 \theta & \mbox{Use a Pythagorean identity}
\end{array}
\]
Note that both \(9\) and \(\cos^2\theta\) are squares.
So when we substitute in the original square root expression, we can simplify:
\[\begin{array}{cll}
\sqrt{9-x^2}&=& \sqrt{9\cos ^2 \theta}\\[1ex]
&=& 3\cos\theta
\end{array}
\]
\end{example}
\begin{example}
We now rationalize the expression
\[\sqrt{4+x^2}\]
This time we use the trig substitution \(x=2\tan \theta\).
\[\begin{array}{clll}
4+x^2&=& \displaystyle 4+(2\tan \theta )^2 & \mbox{Use the trig substitution}\\[1ex]
&=& \displaystyle 4+ 4 \tan^2 \theta & \\[1ex]
&=& 4(1+\tan^2\theta) & \mbox{Factor out the \(4\)}\\[1ex]
&=& 4 \sec ^2 \theta & \mbox{Use a Pythagorean identity}
\end{array}
\]
Substituting inside the square root, we find
\[\begin{array}{cll}
\sqrt{4+x^2}&=& \sqrt{4\sec ^2 \theta}\\[1ex]
&=& 2\sec\theta
\end{array}
\]
\end{example}
\begin{example}
As last example of trig substitution, we rationalize the expression
\[\sqrt{x^2-16}\]
using the trig substitution \(x=4\sec \theta\).
\[\begin{array}{clll}
x^2-16&=& \displaystyle (4\sec\theta)^2 -16 & \mbox{Use the trig substitution}\\[1ex]
&=& \displaystyle 16\sec^2 \theta -16 & \\[1ex]
&=& 16(\sec^2\theta-1) & \mbox{Factor out the \(16\)}\\[1ex]
&=& 16\tan^2\theta & \mbox{Use a Pythagorean identity}
\end{array}
\]
Substituting inside the square root, we find
\[\begin{array}{cll}
\sqrt{x^2-16}&=& \sqrt{16\tan ^2 \theta}\\[1ex]
&=& 4\tan\theta
\end{array}
\]
\end{example}
\subsubsection{Summary of trigonometric substitutions}
|
\(\displaystyle \sqrt{a^2-x^2}\) |
| \(x=a\sin \theta\) |
| |
\(\displaystyle \sqrt{a^2+x^2}\) |
| \(x=a\tan \theta\) |
| |
\(\displaystyle \sqrt{x^2-a^2}\) |
| \(x=a\sec \theta\) |
|
Problems
\problem
Simplify the expressions:
\begin{enumerate}
\item
\(\sec \theta \cot \theta +\csc \theta\)
\item
\((\sec^2 x-1)\cot x\)
\item
\(\sin \theta - \sin \theta \cos^2\theta\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\[
\begin{array}[t]{rcll}
\sec \theta \cot \theta +\csc \theta&=&\displaystyle \frac{1}{\cos \theta}
\frac{\cos \theta}{\sin \theta} +\csc\theta & \mbox{Use reciprocal and quotient identities}\\[2ex]
&=& \displaystyle \frac{1}{\cancel{\cos \theta}}
\frac{\cancel{\cos \theta}}{\sin \theta}+\csc \theta &\mbox{Simplify}\\[2ex]
&=& \displaystyle
\frac{1}{\sin \theta} +\csc \theta&\\[2ex]
&=& \csc \theta +\csc \theta& \mbox{Use a reciprocal identity}\\[1ex]
&=& \fbox{\(2\csc \theta\)}
\end{array}
\]
\item
\[
\begin{array}[t]{rcll}
(\sec^2 x-1)\cot x&=&\displaystyle \tan^2 x \cot x & \mbox{Use a Pythagorean identity}\\[1ex]
&=& \displaystyle \tan^2 x
\frac{1}{\tan x} &\mbox{Use a reciprocal identity}\\[1ex]
&=& \displaystyle
\tan^{\cancel{2}} x \frac{1}{\cancel{\tan x}} &\mbox{Simplify}\\[1ex]
&=& \fbox{\(\tan x\)}&
\end{array}
\]
\item
\[
\begin{array}[t]{rcll}
\sin \theta - \sin \theta \cos^2\theta&=&\sin \theta (1-\cos^2 \theta) &
\mbox{Factor \(\sin \theta\)}\\[1ex]
&=& \sin \theta \sin^2 \theta &\mbox{Use a Pythagorean identity}\\[1ex]
&=& \displaystyle
\fbox{\(\sin^3 \theta\)} &\mbox{Simplify}
\end{array}
\]
\end{enumerate}
\end{sol}
\mproblem
Simplify the expressions:
\begin{enumerate}
\item
\(\csc \theta \tan \theta +\sec \theta\)
\item
\((1-\cos^2x)\csc x\)
\item
\(\cos x + \cos x \tan^2 x\)
\end{enumerate}
\problem
Use the trig substitution \(x=5\sin \theta\) to rationalize \(\sqrt{25-x^2}\).
\begin{sol}
\[\begin{array}{clll}
25-x^2&=& \displaystyle 25 -(5\sin \theta)^2 & \mbox{Use the trig substitution}\\[1ex]
&=& \displaystyle 25 - 25 \sin ^2 \theta & \\[1ex]
&=& 25(1-\sin^2\theta) & \mbox{Factor out the \(25\)}\\[1ex]
&=& 25\cos^2\theta & \mbox{Use a Pythagorean identity}
\end{array}
\]
Substituting inside the square root, we find
\[\begin{array}{cll}
\sqrt{25-x^2}&=& \sqrt{25\cos ^2 \theta}\\[1ex]
&=& \fbox{\(5\cos\theta\)}
\end{array}
\]
\end{sol}
\mproblem
Use the trig substitution \(x=2\cos \theta\) to rationalize \(\sqrt{4-x^2}\).
\problem
Use the trig substitution \(x=4\tan \theta\) to rationalize \(\sqrt{x^2+16}\).
\begin{sol}
\[\begin{array}{clll}
x^2+16&=& \displaystyle (4\tan \theta)^2 +16 & \mbox{Use the trig substitution}\\[1ex]
&=& \displaystyle 16 \tan ^2 \theta+16 & \\[1ex]
&=& 16(\tan^2\theta+1) & \mbox{Factor out the \(16\)}\\[1ex]
&=& 16\sec^2\theta & \mbox{Use a Pythagorean identity}
\end{array}
\]
Substituting inside the square root, we find
\[\begin{array}{cll}
\sqrt{x^2+16}&=& \sqrt{16\sec ^2 \theta}\\[1ex]
&=& \fbox{\(4\sec\theta\)}
\end{array}
\]
\end{sol}
\mproblem
Use the trig substitution \(x=6\sec \theta\) to rationalize \(\sqrt{x^2 -36}\).