\chapter{Analytic trigonometry}
\section{Sum/difference and double angle identities}
Is there a way to simplify or re-write \(\sin(x+y)\) or \(\sin(2x)\)? The fact that
\(\sin(x+y)\) is \textbf{not} \(\sin x+ \sin y\) is clear as soon as we plug in some angles
for \(x\) and \(y\), as you can easily check. And \(\sin(2x)\) could \textbf{not} possibly be
\(2\sin x\), by a similar simple check.
In this unit we discuss the identities that are instead true
for sums or difference of angles (such as \(\sin(x+y)\), or
\(\cos(x-y)\)) and for double angles (such as \(\sin(2x)\)).
\subsection{The sum/difference identities}
Here are the identities for \(\sin\) and \(\cos\):
\[\fbox{\(\begin{array}{cc}
\displaystyle \sin(x+y) =\sin x \cos y +\cos x \sin y & \hspace{2ex}
\displaystyle \sin(x-y) =\sin x \cos y -\cos x \sin y\\[2ex]
\displaystyle \cos(x+y) =\cos x \cos y -\sin x \sin y & \hspace{2ex}
\displaystyle \cos(x-y) =\cos x \cos y +\sin x \sin y
\end{array}\)}\]
Note how the sine function keeps the addition or subtraction signs on both sides
of the identities, while the cosine function reverses them. Also, the sine function mixes
sine and cosine in each term on the right, while the cosine function does not.
Using these identities, we can find exact trig values for some angles that are not multiples of
\(30^\circ\), \(45^\circ\) or \(60^\circ\).
\begin{example}
Suppose we want to find the exact value of \(\sin(15^\circ)\). Clearly this angle is not a
multiple of \(30^\circ\), \(45^\circ\) or \(60^\circ\). But we can write
\[15=45-30\]
and then use the identity for \(\sin(x-y)\):
\[
\begin{array}{rcll}
\sin(15^\circ) &=& \sin(45^\circ-30^\circ)& \\[2ex]
&=& \sin(45^\circ) \cos(30^\circ)-\cos(45^\circ)\sin(30^\circ)&\mbox{Use identity
for \(\sin(x-y)\)}\\[2ex]
&=& \displaystyle\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot\frac{1}{2}&
\mbox{Use trig values for special angles}\\[2ex]
&=& \displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}&\mbox{Simplify}
\end{array}
\]
\end{example}
\begin{example}
We now find \(\displaystyle \cos\left(\frac{7\pi}{12}\right)\). First we need to split the fraction.
We write:
\[
\begin{array}{rll}
\displaystyle \frac{7}{12}&=& \displaystyle \frac{4+3}{12}\\[2ex]
&=& \displaystyle \frac{4}{12} + \frac{3}{12}\\[2ex]
&=& \displaystyle \frac{1}{3} +\frac{1}{4}
\end{array}
\]
So we find:
\[
\begin{array}{rlll}
\displaystyle \cos\left(\frac{7\pi}{12}\right)&=&
\displaystyle \cos\left(\frac{\pi}{3}+\frac{\pi}{4}\right)&\\[2ex]
&=& \displaystyle \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{4}\right)
- \sin\left(\frac{\pi}{3}\right) \sin\left(\frac{\pi}{4}\right) & \mbox{Use identity for
\(\cos(x+y)\)}\\[2ex]
&=& \displaystyle \frac{1}{2}\cdot\frac{1}{\sqrt{2}} -\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{2}}&
\mbox{Use trig values for special angles}\\[2ex]
&=& \displaystyle \frac{1-\sqrt{3}}{2\sqrt{2}}& \mbox{Simplify}
\end{array}
\]
\end{example}
By using the quotient identity \(\displaystyle \tan \theta=\frac{\sin \theta}{\cos \theta}\),
we can divide
the identity for \(\sin(x+y)\) by the one for \(\cos(x+y)\), and after some simplification
we find the sum/difference identities for the tangent function:
\[\fbox{\(\begin{array}{cc}
\displaystyle \tan(x+y) =\frac{\tan x +\tan y}{1-\tan x \tan y} & \hspace{2ex}
\displaystyle \tan(x-y) =\frac{\tan x -\tan y}{1+\tan x \tan y}
\end{array}\)}\]
\begin{example}
We want to find the exact value of \(\tan(285^\circ)\). This angle is not a
multiple of \(30^\circ\), \(45^\circ\) or \(60^\circ\). But we can write
\[285=240+45\]
and then use the identity for \(\tan(x+y)\):
\[
\begin{array}{rcll}
\tan(285^\circ) &=& \tan(240^\circ+45^\circ)& \\[2ex]
&=& \dfrac{\tan(240^\circ) +\tan(45^\circ)}{1-\tan(240^\circ)\tan(45^\circ)}&\mbox{Use identity
for \(\tan(x+y)\)}\\[2ex]
&=& \displaystyle \dfrac{\sqrt{3}+1}{1-\sqrt{3}\cdot 1}&
\mbox{Reference angle for \(240^\circ\) is \(60^\circ\)}\\
&&&\mbox{ \(240^\circ\) is in Q3, \(\tan(60^\circ)=\sqrt{3}\),
\(\tan(45^\circ)=1\)}\\[3ex]
&=& \displaystyle \frac{\sqrt{3}+1}{1-\sqrt{3}}&\mbox{Simplify}
\end{array}
\]
\end{example}
\subsection{The double angle identities}
If we take \(y=x\) in the identity for \(\sin(x+y)\), we find the
formulas for \(\sin(2x)\):
\[\fbox{\(\displaystyle \sin(2x) =2\sin x \cos x \)}\]
\begin{example}
Suppose we know that \(\sin t=-1/5\), \(t\) is in Q4, and we want to know \(\sin(2t)\).
Using the double angle identity we find
\[
\begin{array}{rcl}
\sin(2t)&=& 2\sin t \cos t\\[1ex]
&=& \displaystyle -2\cdot \frac{1}{5}\cos t\\[1ex]
&=& \displaystyle -\frac{2}{5}\cos t.
\end{array}
\]
So we need to find \(\cos t\) to finish this problem. We know that \(t\) is in Q4, so \(\cos t\)
is positive. Using the \(x,y,r\) definitions,
\[\sin t =\frac {y}{r} = -\frac{1}{5},\] so \(y=-1\) and \(r=5\). Then we solve
\(x^2+(-1)^2=5^2\) and get \(x=\pm \sqrt{24}=\sqrt{24}\) (Quadrant 4). So \(\cos t =\sqrt{24}/5 =2\sqrt{6}/5\).
So we can now finish the problem and find
\[
\begin{array}{rcl}
\sin(2x)&=& \displaystyle -\frac{2}{5}\cos x\\[2ex]
&=& \displaystyle -\frac{2}{5}\frac{2\sqrt{6}}{5}\\[2ex]
&=& \displaystyle -\frac{4\sqrt{6}}{25}.
\end{array}
\]
\end{example}
In the same way, taking \(y=x\) in the identity for \(\cos(x+y)\), we find a formula
for \(\cos(2x)\). In this case, we can also use the Pythagorean identities to find two more
identities:
\[\fbox{\(\begin{array}{cc}
\displaystyle \cos(2x) =\cos^2 x -\sin^2 x\\[1ex]
\displaystyle \cos(2x) =1-2\sin^2 x\\[1ex]
\displaystyle \cos(2x) =2\cos^2 x-1
\end{array}\)}\]
Note that the first identity for \(\cos(2x)\) involves both \(\sin\) and \(\cos\), the
second involves only \(\sin\), and the third involves only \(\cos\). Each of these
can be useful in different situations.
\begin{example}
Suppose we know that \(\displaystyle \sin x=\frac{1}{3}\), and we want to know \(\cos(2x)\).
Using the second identity, we find
\[
\begin{array}{rcl}
\cos(2x)&=& 1-2\sin^2 x\\[1ex]
&=& \displaystyle 1-2\left(\frac{1}{3}\right)^2\\[1ex]
&=& \displaystyle 1-\frac{2}{9}\\[1ex]
&=& \displaystyle\frac{7}{9}
\end{array}
\]
\end{example}
\begin{example}
Suppose now we know \(\displaystyle \cos x=-\frac{3}{5}\), and we want to find \(\cos(2x)\).
This time we use the third identity:
\[
\begin{array}{rcl}
\cos(2x)&=& 2\cos^2 x-1\\[1ex]
&=& \displaystyle 2\left(-\frac{3}{5}\right)^2-1\\[2ex]
&=& \displaystyle 2\cdot\frac{9}{25}-1\\[2ex]
&=& \displaystyle\frac{18}{25}-\frac{25}{25}\\[2ex]
&=& \displaystyle -\frac{7}{25}.
\end{array}
\]
\end{example}
We gather together here all the identities of this unit for quick reference:
\textbf{Sum/difference identities}
| \(\begin{array}{cc}
\displaystyle \sin(x+y) =\sin x \cos y +\cos x \sin y & \hspace{2ex}
\displaystyle \sin(x-y) =\sin x \cos y -\cos x \sin y\\[2ex]
\displaystyle \cos(x+y) =\cos x \cos y -\sin x \sin y & \hspace{2ex}
\displaystyle \cos(x-y) =\cos x \cos y +\sin x \sin y
\end{array}\) |
\(\begin{array}{cc}
\displaystyle \tan(x+y) =\frac{\tan x +\tan y}{1-\tan x \tan y} & \hspace{2ex}
\displaystyle \tan(x-y) =\frac{\tan x -\tan y}{1+\tan x \tan y}
\end{array}\) |
\textbf{Double angle identities}
| \(\displaystyle \sin(2x) =2\sin x \cos x \) |
| \(\cos(2x) = \begin{cases} \cos^2 x -\sin^2 x\\[1ex]
1-2\sin^2 x\\[1ex]
2\cos^2 x-1
\end{cases}\) |
Problems
\problem
Find the exact value of the following:
\begin{enumerate}
\item
\(\displaystyle\cos\left(\frac{\pi}{12}\right)\)
\item
\( \sin(105^\circ )\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
Split the fraction \(1/12\) as:
\[\begin{array}{rcl}
\displaystyle\frac{1}{12}&=&\displaystyle\frac{3-2}{12}\\[2ex]
&=& \displaystyle\frac{3}{12}-\frac{2}{12}\\[2ex]
&=& \displaystyle\frac{1}{4}-\frac{1}{6}.
\end{array}
\]
Then use the identity for \(\cos(x-y)\):
\[
\begin{array}{rlll}
\displaystyle \cos\left(\frac{\pi}{12}\right)&=&
\displaystyle \cos\left(\frac{\pi}{4}-\frac{\pi}{6}\right)&\\[2ex]
&=& \displaystyle \cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right)
+ \sin\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{6}\right) & \mbox{Use identity for
\(\cos(x-y)\)}\\[2ex]
&=& \displaystyle \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} +\frac{1}{\sqrt{2}}\cdot \frac{1}{2}&
\mbox{Use trig values for special angles}\\[2ex]
&=& \fbox{\(\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}\)}& \mbox{Simplify}
\end{array}
\]
\item
Write \(105=60+45\). Then
\[
\begin{array}{rcll}
\sin(105^\circ) &=& \sin(60^\circ+45^\circ)& \\[1ex]
&=& \sin(60^\circ) \cos(45^\circ)+\cos(60^\circ)\sin(45^\circ)&\mbox{Use identity
for \(\sin(x+y)\)}\\[1ex]
&=& \displaystyle\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}+\frac{1}{2}\cdot\frac{1}{\sqrt{2}}&
\mbox{Use trig values for special angles}\\[1ex]
&=& \fbox{\(\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}\)}&\mbox{Simplify}
\end{array}
\]
\end{enumerate}
\end{sol}
\mproblem
Find the exact value of the following:
\begin{enumerate}
\item
\(\displaystyle\sin\left(\frac{\pi}{12}\right)\)
\item
\( \displaystyle\cos(255^\circ)\)
\end{enumerate}
\problem
Suppose that \(\sin \theta=-0.27\), and \(\theta\) is in Q3. Find \(\sin(2\theta)\), approximated
to two decimals.
\begin{sol}
Use the double angle identity for the sine function:
\[
\begin{array}{rcl}
\sin(2\theta)&=& 2\sin \theta \cos \theta\\[1ex]
&=& \displaystyle 2(-0.27)\cos \theta\\[1ex]
&=& \displaystyle -0.54\cos \theta.
\end{array}
\]
Since \(\theta \) is in Q3, \(\cos\theta\) is negative.
So
\[
\begin{array}{rcl}
\cos \theta &=&-\sqrt{1-\sin^2 \theta}\\[1ex]
&=& \displaystyle -\sqrt{1-(-0.27)^2}\\[1ex]
&=& \displaystyle -\sqrt{1-0.0729}\\[1ex]
&=& \displaystyle -\sqrt{0.9271}\\[1ex]
&=& \displaystyle -0.9628\ldots
\end{array}.
\]
Now we can find \(\sin(2\theta)\):
\[
\begin{array}{rcl}
\sin(2\theta)&=& \displaystyle -0.54(-0.9628)\\[1ex]
&\approx&\fbox{\( 0.520\)}
\end{array}
\]
\end{sol}
\mproblem
Suppose that \(\cos \theta=0.15\), and \(\theta\) is in Q4. Find \(\sin(2\theta)\), approximated
to two decimals.
\problem
Suppose that \(\displaystyle \sin x=-\frac{2}{3}\). Find the exact value of \(\cos(2x)\).
\begin{sol}
Since we are given the value of \(\sin x\), we use the identity for \(\cos (2x)\) involving
\(\sin x\):
\[
\begin{array}{rcl}
\cos(2x)&=& 1-2\sin^2 x\\[1ex]
&=& \displaystyle 1-2\left(-\frac{2}{3}\right)^2\\[1ex]
&=& \displaystyle 1-2\cdot \frac{4}{9}\\[1ex]
&=& \displaystyle 1-\frac{8}{9}\\[1ex]
&=& \fbox{\(\displaystyle \frac{1}{9}\)}
\end{array}
\]
\end{sol}
\mproblem
Suppose that \(\displaystyle \cos x=-\frac{1}{4}\). Find the exact value of \(\cos(2x)\).