\chapter{Analytic trigonometry}
\section{Verification of trig identities}
First we discuss the difference between an equation and an identity.
\subsection{Equations vs. Identities}
An equation is a statement such as
\[x^2-x-6=0\]
that is true only for some values of the variable (in this case, you can check that it is
only true if \(x=3\) or \(x=-2\)).
On the other hand, an identity is a statement that is true no matter what the variable is, such
as
\[(x+2)^2=x^2+4x+4.\]
To \textit{verify an identity} means to make sure that the two sides are really the same,
no matter what value the variable has. So it would not be enough to check
the identity for many values of the variable, because of course there are infinitely many
such possible values, and we could never check them all!
We begin by naming the two sides LHS and RHS (for Left Hand Side and Right Hand Side). Then we choose
one of the two (usually the one that looks more complicated, or in any case for which we can
see that there is some change we can make) and begin to modify it using algebra, known
identities, simplification, etc.
\begin{example}
We verify the identity
\[(x+2)^2=x^2+4x+4.\]
We work on the LHS:
\[
\begin{array}{rcll}
\mbox{LHS}&=&(x+2)^2 & \\[1ex]
&=&(x+2)(x+2)& \\[1ex]
&=& x^2+2x+2x+4 & \mbox{Expand}\\[1ex]
&=& x^2+4x+4 & \mbox{Simplify}\\[1ex]
&=& \mbox{RHS} & \mbox{Find the RHS}
\end{array}
\]
\end{example}
We now turn to identities that contain trigonometric functions.
\subsection{Verification using the basic trig identities}
There are many trig identities that can be verified using the basic identities of the previous unit.
\begin{example}
We want to verify the identity
\[\sin \theta \cot \theta =\cos \theta.\]
It is a good idea to check some simple values of the variable,
by choosing a value that is in the domain of all trig functions in the identity. This is because if we find a single value of the
variable for which the identity is false, then of course it is useless to try to verify it, and
we just conclude that what we are given is an equation and not an identity. The simplest value to check
is \(\theta=0\), but in this case
we cannot use it because \(\cot 0\) is undefined. So we choose \(\theta=90^\circ\):
\(\sin(90^\circ) \cot(90^\circ) =\cos (90^\circ)\),
or
\((1)(0) =0,\) that is surely true.
If we choose \(\theta=45^\circ\), we find
\(\sin(45^\circ) \cot(45^\circ) =\cos (45^\circ),\)
or
\(\dfrac{1}{\sqrt{2}}(1) =\dfrac{1}{\sqrt{2}},\) that is also true.
The fact that the identity is true for two values of theta encourages us to believe that it may be
true for all other values. But now we need to check it using algebra for all possible values of
\(\theta\).
Often, a good strategy is to use the quotient identities to eliminate \(\tan\) or \(\cot\):
\[
\begin{array}{rcll}
\mbox{LHS}&=&\sin \theta \cot \theta & \\[1ex]
&=&\displaystyle \sin \theta \frac{\cos \theta}{\sin \theta}& \mbox{Use a quotient identity}\\[1ex]
&=& \displaystyle \cancel{\sin \theta} \frac{\cos \theta}{\cancel{\sin \theta}} &
\mbox{Simplify}\\[1ex]
&=& \cos \theta & \\[1ex]
&=& \mbox{RHS} & \mbox{Find the RHS}
\end{array}
\]
Now we have removed every doubt: the identity is surely true for all values of \(\theta\).
\end{example}
In the next example we subtract two fractions and simplify in order to verify the identity.
\begin{example}
We verify the identity
\(\displaystyle \frac{\sec x}{\sin x} -\cot x=\tan x\)
\[
\begin{array}{rcll}
\mbox{LHS}&=&\displaystyle \frac{\sec x}{\sin x} -\cot x & \\[2ex]
&=&\displaystyle \sec x \frac{1}{\sin x} -\cot x & \mbox{Separate the numerator of the fraction}\\[2ex]
&=&\displaystyle \frac{1}{\cos x}\frac{1}{\sin x}-\frac{\cos x}{\sin x} & \mbox{Use reciprocal and quotient identities}\\[2ex]
&=& \displaystyle \frac{1}{\sin x \cos x}-\frac{\cos x \cos x}{\sin x \cos x} &
\mbox{Find a common denominator}\\[2ex]
&=& \displaystyle \frac{1-\cos^2 x}{\sin x \cos x} &
\mbox{Combine the fractions}\\[2ex]
&=& \displaystyle \frac{\sin^2 x}{\sin x\cos x} & \mbox{Use a Pythagorean identity}\\[2ex]
&=& \displaystyle \frac{\sin^{\cancel{2}}\hspace{-0.5ex}x}{\cancel{\sin x}\cos x} & \mbox{Simplify}\\[2ex]
&=& \displaystyle \frac{\sin x}{\cos x} & \\[2ex]
&=& \displaystyle \tan x & \mbox{Use a quotient identity}\\[2ex]
&=& \mbox{RHS} &\mbox{Find the RHS}
\end{array}
\]
\end{example}
In the next example, we begin with the RHS and simplify it until we get the LHS.
\begin{example}
We verify the identity
\( 2\cos^{2}\theta - 1 = \cos^{2}\theta - \sin^{2}\theta\).
\[
\begin{array}{rcll}
\mbox{RHS}&=& \cos^{2}\theta - \sin^{2}\theta & \\[1ex]
&=&\cos^2\theta -(1-\cos^2\theta) & \mbox{Use a Pythagorean identity}\\[1ex]
&=& \cos^2\theta -1 +\cos ^2\theta &
\mbox{Expand}\\[1ex]
&=& 2\cos^2\theta -1&
\mbox{Simplify, and find the LHS}
\end{array}
\]
\end{example}
Sometimes we need to work on both sides of the identity, and simplify both till they become equal.
\begin{example}
We verify the identity:
\(\displaystyle \dfrac{\tan x + \tan y}{1 - \tan x \tan y} = \dfrac{\cot x + \cot y}{\cot x \cot y - 1}\)
\[\begin{array}{rcll}
\mbox{LHS}&=& \dfrac{\tan x + \tan y}{1 - \tan x \tan y} &\mbox{}\\[2ex]
&=& \dfrac{\displaystyle \dfrac{\sin x}{\cos x} + \dfrac{\displaystyle \sin y}{\cos y}}{\displaystyle 1 - \dfrac{\sin x\sin y}{\cos x\cos y}} & \mbox{Quotient} \\[4ex]
&=& \dfrac{\displaystyle \dfrac{\sin x\cos y + \cos x \sin y}{\cos x \cos y}}{\displaystyle \dfrac{\cos x \cos y - \sin x \sin y}{\cos x \cos y}} & \mbox{Combine fractions} \\[4ex]
&=& \dfrac{\sin x\cos y + \cos x \sin y}{\cancel{\cos x \cos y}}\cdot \dfrac{\cancel{\cos x \cos y}}{\cos x \cos y - \sin x \sin y} &\mbox{Flip}\\[3ex]
&=& \dfrac{\sin x\cos y + \cos x \sin y}{\cos x \cos y - \sin x \sin y} & \mbox{Simplify}
\end{array}
\]
\[\begin{array}{rcll}
\mbox{RHS}&=&\dfrac{\cot x + \cot y}{\cot x \cot y - 1}\\[2ex]
&=& \cfrac{\displaystyle \cfrac{\cos x}{\sin x} + \cfrac{\displaystyle \cos y}{\sin y}}{\displaystyle \cfrac{\cos x\cos y}{\sin x\sin y} - 1} \\[5ex]
&=& \dfrac{\displaystyle \dfrac{\cos x\sin y + \cos y \sin x}{\sin x \sin y}}{\displaystyle \dfrac{\cos x \cos y - \sin x \sin y}{\sin x \sin y}} \\[4ex]
&=& \displaystyle \dfrac{\cos x\sin y + \cos y \sin x}{\cancel{\sin x \sin y}} \cdot \dfrac{\cancel{\sin x \sin y}}{\cos x \cos y - \sin x \sin y} \\[3ex]
&=& \dfrac{\cos x\sin y + \cos y \sin x}{\cos x \cos y - \sin x \sin y} \\[2ex]
&=& \mbox{Same as LHS!}
\end{array}\]
\end{example}
Problems
\problem
Verify the trig identity:
\[\frac{\tan^2 \theta +1}{\csc^2 \theta}=\tan^2\theta\]
\begin{sol}
We begin with the left side because it looks more complicated:
\[
\begin{array}{rcll}
\mbox{LHS}&=&\displaystyle \frac{\tan^2 \theta +1}{\csc^2 \theta} & \\[2ex]
&=&\displaystyle \frac{\sec^2\theta}{\csc^2\theta}& \mbox{Use a Pythagorean identity}\\[2ex]
&=& \displaystyle \sec^2 \theta \frac{1}{\csc^2\theta}&
\mbox{Re-write}\\[2ex]
&=& \displaystyle \frac{1}{\cos^2 \theta}\sin^2\theta &
\mbox{Use reciprocal identities}\\[2ex]
&=& \displaystyle \frac{\sin^2\theta}{\cos^2\theta} & \mbox{Re-write}\\[2ex]
&=& \displaystyle \tan^2 \theta &
\mbox{Use a quotient identity}\\[2ex]
&=& \displaystyle \mbox{RHS} & \mbox{Find the RHS}
\end{array}
\]
\end{sol}
\mproblem
Verify the trig identity:
\[\frac{\sec^2 \theta -1}{\sin^2 \theta}=\sec^2\theta\]
\problem
Verify the trig identity:
\[\sec x=\cos x + \sin x \tan x.\]
\begin{sol}
We begin with the right side:
\[
\begin{array}{rcll}
\mbox{RHS}&=&\displaystyle \cos x + \sin x \tan x& \\[1ex]
&=&\displaystyle \cos x +\sin x \frac{\sin x}{\cos x}& \mbox{Use a quotient identity}\\[2ex]
&=& \displaystyle \cos x +\frac{\sin ^2 x}{\cos x}&
\mbox{Re-write}\\[2ex]
&=& \displaystyle \frac{\cos x}{1}\frac{\cos x}{\cos x}+\frac{\sin^2 x}{\cos x} &
\mbox{Find a common denominator}\\[2ex]
&=& \displaystyle \frac{\cos^2 x+\sin^2 x}{\cos x} & \mbox{Add the fractions}\\[1ex]
&=& \displaystyle \frac{1}{\cos x} &
\mbox{Use a Pythagorean identity}\\[2ex]
&=& \displaystyle \sec x & \mbox{Use a reciprocal identity}\\[1ex]
&=& \mbox{LHS} & \mbox{Find the LHS}.
\end{array}
\]
\end{sol}
\mproblem
Verify the trig identity:
\[\sin \theta +\cot \theta \cos \theta= \csc\theta\]
\problem
Verify the identity:
\[\frac{\cos^2 \theta}{\sin \theta}=\csc \theta -\sin \theta\]
\begin{sol}
In this identity there are three different trig functions, so we reduce the number by substituting
for \(\cos^2\theta\) using a Pythagorean identity:
\[
\begin{array}{rcll}
\mbox{LHS}&=&\displaystyle \frac{\cos^2 \theta}{\sin \theta}& \\[2ex]
&=&\displaystyle \frac{1-\sin^2 \theta}{\sin \theta}& \mbox{Use a Pythagorean identity}\\[2ex]
&=& \displaystyle \frac{1}{\sin \theta}-\frac{\sin^2 \theta}{\sin \theta}&
\mbox{Separate the fraction}\\[2ex]
&=& \displaystyle \csc \theta-\sin \theta &
\mbox{Use a reciprocal identity, simplify}\\[2ex]
&=& \displaystyle \mbox{RHS} & \mbox{Find the RHS}
\end{array}
\]
\end{sol}
\mproblem
Verify the identity:
\[\frac{\sin^2 \theta}{\cos \theta}=\sec \theta -\cos \theta\]
\problem
Verify the identity:
\[\frac{-\csc(-x)}{\tan(90^\circ -x)}=\sec x\]
\begin{sol}
\[
\begin{array}{rcll}
\mbox{LHS}&=&\displaystyle \frac{-\csc(-x)}{\tan(90^\circ -x)}& \\[2ex]
&=&\displaystyle \frac{\csc x}{\cot x}& \mbox{Use even-odd and co-function identities}\\[2ex]
&=& \displaystyle \csc x \frac{1}{\cot x}&
\mbox{Re-write}\\[2ex]
&=& \displaystyle \csc x \tan x &
\mbox{Use a reciprocal identity}\\[2ex]
&=& \displaystyle \frac{1}{\sin x}\frac{\sin x}{\cos x} & \mbox{Use reciprocal and quotient identities}\\[2ex]
&=& \displaystyle \frac{1}{\cancel{\sin x}}\frac{\cancel{\sin x}}{\cos x} &\mbox{Simplify}\\[2ex]
&=& \sec x & \mbox{Use a reciprocal identity}\\[2ex]
&=& \mbox{RHS} & \mbox{Find the right side}
\end{array}
\]
\end{sol}
\mproblem
Verify the identity:
\[\frac{\sec(-x)}{\cot(90^\circ -x)}=\csc x\]
\problem
Verify the identity
\[\sec^2 \theta \cot \theta -\cot \theta = \tan \theta\]
\begin{sol}
\[
\begin{array}{rcll}
\mbox{LHS}&=&\displaystyle \sec^2 \theta \cot \theta -\cot \theta & \\[2ex]
&=&\displaystyle \cot \theta (\sec^2 \theta -1) & \mbox{Factor \(\cot \theta\)}\\[2ex]
&=& \displaystyle \cot \theta \tan^2 \theta&
\mbox{Use a Pythagorean identity}\\[2ex]
&=& \displaystyle \frac{1}{\tan\theta} \tan^2 \theta &
\mbox{Use a reciprocal identity}\\[2ex]
&=& \displaystyle \frac{1}{\cancel{\tan\theta}} \tan^{\cancel{2}} \theta & \mbox{Simplify}\\[2ex]
&=& \displaystyle\tan \theta &\\[2ex]
&=& \mbox{RHS} & \mbox{Find the right side}
\end{array}
\]
\end{sol}
\mproblem
Verify the identity
\[\csc^2 \theta \tan \theta - \cot \theta =\tan \theta\]
\problem
Verify the identity
\[\cos x \tan^2 x + \cos x = \sec x\]
\begin{sol}
\[
\begin{array}{rcll}
\mbox{LHS}&=&\displaystyle \cos x \tan^2 x + \cos x = \sec x & \\[1ex]
&=&\displaystyle \cos x (\tan^2 x+1) & \mbox{Factor \(\cos x\)}\\[1ex]
&=& \displaystyle \cos x \sec^2 x&
\mbox{Use a Pythagorean identity}\\[1ex]
&=& \displaystyle \frac{1}{\sec x} \sec^2 x &
\mbox{Use a reciprocal identity}\\[1ex]
&=& \displaystyle \frac{1}{\cancel{\sec x}} \sec^{\cancel{2}} x & \mbox{Simplify}\\[1ex]
&=& \displaystyle\sec x &\\[1ex]
&=& \mbox{RHS} & \mbox{Find the right side}
\end{array}
\]
\end{sol}
\mproblem
Verify the identity
\[\cos \theta \cot \theta = \csc \theta -\sin\theta\]