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\institution{SIAST Palliser Campus}
\department{Mathematics}
\course{MAT 226}
\unittitle{Trigonometric Identities and Equations}
\unitnumber{1}
\date{January 2011}
\title{STUDY GUIDE}
\author{Robert G. Petry}
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\publisher{the Department of Mathematics, SIAST Palliser Campus}
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\section{Fundamental Identities}
We recall that an \emph{identity} is an equation that is true for all
meaningful values of the variable; for example, $x(x+2)=x^2+2x$ is
true for any value of the variable $x$.  This is to be distinguished
from a \emph{conditional equation} such as $x^2-5x+6=0$ which is true
for only certain values of the variable $x$, here $x=2$ or $x=3$.

\subsection{Reciprocal Relations}
The simplest \emph{trigonometric} identities are the reciprocal
relations involving definitions of the reciprocal functions:

\begin{boxedeqnarray2}
&&\sin\theta=\frac{1}{\csc\theta}~~~~~~\csc\theta=\frac{1}{\sin\theta}~~~~~~  \sin\theta\csc\theta = 1\label{eq:recip1}\\
&&\cos\theta=\frac{1}{\sec\theta}~~~~~~\sec\theta=\frac{1}{\cos\theta}~~~~~  \cos\theta\sec\theta = 1\label{eq:recip2}\\
&&\tan\theta=\frac{1}{\cot\theta}~~~~~\cot\theta=\frac{1}{\tan\theta}~~~~~  \tan\theta\cot\theta = 1\label{eq:recip3}
\end{boxedeqnarray2}

\subsection{Quotient Relations}
An angle $\theta$ in standard position can have its basic trig
functions defined in terms of a point $P$ at $(x,y)$ on the terminal
side as follows:

\begin{minipage}{.53\textwidth}
\[
\sin\theta=\frac{y}{r}~~~~~~\cos\theta=\frac{x}{r}~~~~~~\tan\theta=\frac{y}{x}
\]
where $r$ is the hypotenuse length.  \mbox{Dividing} the left and
right sides of the first two equations proves the first of the
following quotient relations:
\end{minipage}\hspace{.22\textwidth}
\raisebox{-5em}{\includegraphics[width=.25\textwidth,keepaspectratio=,clip=]{Unit_1_fig1.pdf}}

\begin{boxedeqnarray2}
\tan\theta & = & \frac{\sin\theta}{\cos\theta}\label{eq:quotient1}\\
\cot\theta & = & \frac{\cos\theta}{\sin\theta}\label{eq:quotient2}
\end{boxedeqnarray2}

Reciprocating both sides of equation~(\ref{eq:quotient1}) and using
the reciprocal relation~(\ref{eq:recip3}) results in the second
equation.

\subsection{Pythagorean Relations}
Returning to the coordinate trig functions of the previous section,
one has, by the Pythagorean theorem,
\[x^2+y^2=r^2\]
Dividing both sides of this equation by $r^2$ and identifying the trig
functions by their coordinates yields the first of the following
Pythagorean relations.

\begin{boxedeqnarray2}
\sin^2\theta+\cos^2\theta & = & 1 \label{eq:pythag1}\\
1+\tan^2\theta & = & \sec^2\theta \\
1+\cot^2\theta & = & \csc^2\theta
\end{boxedeqnarray2}

The two other Pythagorean relations follow by dividing both sides of
the first equation by $\cos^2\theta$ and $\sin^2\theta$ respectively
and using the reciprocal
relations~(\ref{eq:recip1})-(\ref{eq:recip3}).

We can use our trigonometric identities to write a trigonometric
expression in terms of only $\sin\theta$ and $\cos\theta$ as the
following example shows:

\textbf{Example}

Express the following in terms of only $\sin$ and $\cos$:

$\displaystyle -\tan x\sin x+\sec x$
\clearpage


Our trig identities can be used to \emph{simplify trigonometric
  expressions} as the following examples show.

\textbf{Examples}

Simplify the following:
\begin{enumerate}
\item
~~~~~~~~$\displaystyle \frac{\tan\theta(\csc^2\theta-1)}{\cos\theta\cot\theta+\sin\theta}$
\vspace{3in}

\item
~~~~~~~~$\displaystyle \frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}$
\end{enumerate}
\clearpage
Trig identities can be used to \emph{prove} whether other trig
identities are true.  This is done by using known identities to
transform one side of the equation until it is identical to the other
side.  Often converting expressions to $\sin\theta$ or $\cos\theta$
can help.

\textbf{Example}

Prove the following identity:

$\displaystyle \frac{\sin\theta+\tan\theta}{1+\cos\theta}=\tan\theta$

\textassignment{Sec. 18-1}{Ex. 1 (P. 466) \# 1-49 (odd)}

\section{Sum or Difference of Angles}

\subsection{Sine and Cosine of $\mathbf \alpha\pm\beta$}
A geometric consideration of the combination of two angles (see
Text~p.~468) results in the following formulas for the sine and cosine
of the \emph{sum of two angles}:
\begin{eqnarray}
\sin(\alpha+\beta) & = & \sin\alpha\cos\beta+\cos\alpha\sin\beta\nonumber\\
\cos(\alpha+\beta) & = & \cos\alpha\cos\beta-\sin\alpha\sin\beta\nonumber
\end{eqnarray}

For $\beta$ in the first quadrant, $-\beta$ is in the fourth quadrant
so sine, but not cosine will change sign:
\[\cos(-\beta)=\cos\beta~~~~~~~~~\sin(-\beta)=-\sin\beta\]
Using these results and replacing $\beta\rightarrow -\beta$ in our sum
formulas above gives the difference formulas:
\begin{eqnarray}
\sin(\alpha-\beta) & = & \sin\alpha\cos\beta-\cos\alpha\sin\beta\nonumber\\
\cos(\alpha-\beta) & = & \cos\alpha\cos\beta+\sin\alpha\sin\beta\nonumber
\end{eqnarray}

The sum and difference formulas can be written simply as:

\begin{boxedeqnarray2}
\sin(\alpha\pm\beta) & = & \sin\alpha\cos\beta\pm\cos\alpha\sin\beta\label{eq:sinepm}\\
\cos(\alpha\pm\beta) & = & \cos\alpha\cos\beta\mp\sin\alpha\sin\beta\label{eq:cospm}
\end{boxedeqnarray2}

\subsection{Tangent of $\alpha\pm\beta$}
Since, by (\ref{eq:quotient1}),
\[\tan(\alpha\pm\beta)=\frac{\sin(\alpha\pm\beta)}{\cos(\alpha\pm\beta)}\]
one can use the results (\ref{eq:sinepm}) and (\ref{eq:cospm}) to
arrive at the sum and difference formulas for tangent:

\begin{boxedequation}
\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}
\label{eq:tanpm}
\end{boxedequation}
(Here we divided both the numerator and denominator by $\cos\alpha\cos\beta$.)

These new trigonometric identities allow the simplification of more
expressions as well as the proof of further identities as the
next examples show.

\clearpage
\textbf{Examples}
\begin{enumerate}
\item Expand:~~~~ $\displaystyle \sin(\alpha+\beta+\gamma)$

\vspace{.2\textheight}

\item Simplify:~~~~ $\displaystyle \sin 5x\cos 2x - \cos 5x\sin2x$
\vspace{.1\textheight}

\item Prove:~~~~ $\sin(\theta+\phi)+\sin(\theta-\phi)\displaystyle=2\sin\theta\cos\phi$
\vspace{.2\textheight}

\item Prove:~~~~ $\displaystyle 0=\tan\left(\theta-\frac{\pi}{4}\right)+\cot\left(\theta+\frac{\pi}{4}\right)$
\end{enumerate}
\clearpage

\subsection{Adding a Sine Wave and Cosine Wave of the Same Frequency}
A function of the form 

\[A\sin\omega t\]

is a \emph{sine wave}.  The constants $A$ and $\omega$ are called
the \emph{amplitude} and \emph{\mbox{(angular)} frequency} respectively. As
time changes the wave oscillates between $+A$ and $-A$ with a frequency
$\omega/2\pi$.  A \emph{cosine wave} may be similarly defined:

\[B\cos\omega t\]

This oscillates in the same way as a sine wave except the peaks of
$\pm B$ occur at the times where the cosine wave vanishes and
vice-versa.

A common problem is to combine a sine wave and a cosine wave of the
\emph{same} frequency~$\omega$.  If one writes $A$ and $B$ in terms of
two new constants $R$ and $\phi$ as follows:
\begin{boxedequation}
\label{eq:phase1}
A=R\cos\phi ~~~~~~~~~~B=R\sin\phi
\end{boxedequation}
then
\begin{eqnarray}
\lefteqn{A\sin\omega t+B\cos\omega t} \nonumber\\
& = & R\sin\omega t\cos\phi+R\cos\omega t\sin\phi \nonumber\\
& = & R\left(\sin \omega t\cos\phi + \cos\omega t\sin\phi\right)\nonumber
\end{eqnarray}
Using the sine angle addition formula (\ref{eq:sinepm}) on the last line gives:
\begin{boxedequation}
A\sin\omega t+B\cos\omega t=R\sin\left(\omega t+\phi\right)
\end{boxedequation}
Thus the new wave has the same frequency $\omega$ as the original
waves but now peaks with an amplitude $R$ at times shifted by a phase
angle $\phi$.  These values are found by solving~(\ref{eq:phase1})
for $R$ and $\phi$ to get:
\begin{boxedequation}
\label{eq:phase2}
R=\sqrt{A^2+B^2}~~~~~~~~~~\phi=\arctan\frac{B}{A}
\end{boxedequation}
as done for converting between polar and Cartesian forms of a
vector.\newline
(Or see diagram in text page 472.)

\clearpage
\textbf{Example}

Write as a single sine function:

$\displaystyle y=1.91\sin\omega t+2.87\cos\omega t$

\textassignment{Sec. 18-2}{Ex. 2 (P. 473) \# 1-35 (odd)}
\clearpage

\section{Double Angle Formulas}
Trigonometric functions of $2\alpha$ can be reduced to functions of
$\alpha$. Setting $\beta=\alpha$ in the trig addition formulas
(\ref{eq:sinepm})-(\ref{eq:tanpm}) the left-hand side of these
equations become functions of $2\alpha$.  For sine one has

\begin{boxedequation}
\sin 2\alpha=2\sin\alpha\cos\alpha \label{eq:doublesine}
\end{boxedequation}

For cosine one arrives at the first of the following identities:

\begin{boxedeqnarray2}
\cos 2\alpha & = & \cos^2\alpha-\sin^2\alpha \nonumber \\
\cos 2\alpha & = & 1-2\sin^2\alpha \label{eq:doublecos} \\
\cos 2\alpha & = & 2\cos^2\alpha-1 \nonumber
\end{boxedeqnarray2}

The latter two identities for $\cos 2\alpha$ are found by using the
Pythagorean identity (\ref{eq:pythag1}) on the first one to simplify
it further to be in terms of either just sine or cosine.

Tangent of $2\alpha$ may similarly be found by setting $\beta=\alpha$
in (\ref{eq:tanpm}):

\begin{boxedequation}
\tan 2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}
\end{boxedequation}

\textbf{Examples}
\begin{enumerate}
\item Simplify: ~~~~$\displaystyle 2\sin 3x\cos 3x$%Answer \sin 6x
\clearpage

\item Simplify: ~~~~$\displaystyle \frac{2-\sec^2\theta}{\sec^2\theta}$% Answer \cos 2x
\vspace{.25\textheight}

\item Prove:~~~~$\displaystyle \tan x=\frac{\sin x+\sin 2x}{\cos 2x+\cos x+1}$
\vspace{.25\textheight}

\item If $\theta$ and $\phi$ are two acute angles in a right triangle prove:
\[-\cot2\theta=\tan(\theta-\phi)\]
\end{enumerate}
\textassignment{Sec. 18-3}{Ex. 3 (P. 476) \# 1-19 (odd)}
\clearpage

\section{Half-Angle Formulas}
Formulas for trig functions of $\alpha/2$ in terms of $\alpha$ can be
found by manipulating the double-angle formulas of the previous
section.  Replacing $\alpha\rightarrow\alpha/2$ in the second identity
of (\ref{eq:doublecos}) and solving for $\sin(\alpha/2)$ gives:

\begin{boxedequation}
\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}}\label{eq:sinehalf}
\end{boxedequation}

Here (and below) the choice of $+$ or $-$ must be determined by the
quadrant in which $\alpha/2$ is found. (Recall the CAST prescription.)

Similarly one may replace $\alpha\rightarrow\alpha/2$ in the third
identity of (\ref{eq:doublecos}) and solve for $\cos(\alpha/2)$ to
get the following cosine identity:
\begin{boxedequation}
\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}\label{eq:cosehalf}
\end{boxedequation}

Three different identities for $\tan\alpha/2$ can be obtained:
\begin{boxedeqnarray2}
\tan\frac{\alpha}{2} & = & \frac{1-\cos\alpha}{\sin\alpha} \nonumber \\
\tan\frac{\alpha}{2} & = & \frac{\sin\alpha}{1+\cos\alpha} \label{eq:tanhalf}\\
\tan\frac{\alpha}{2} & = & \pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}} \nonumber
\end{boxedeqnarray2}
The last of these is obtained by noting that by quotient relation
(\ref{eq:quotient1}) that

\[\tan\frac{\alpha}{2}=\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}~~~~~~~(*)\]

into which one may substitute (\ref{eq:sinehalf}) and
(\ref{eq:cosehalf}).  The first identity of (\ref{eq:tanhalf}) may be
found by multiplying the numerator and denominator of ($*$) by
$2\sin(\alpha/2)$ and simplifying the numerator using the second
identity of (\ref{eq:doublecos}) and the denominator using
(\ref{eq:doublesine}).  Finally the second identity of
(\ref{eq:tanhalf}) may be shown by multiplying the numerator and
denominator of the first identity of (\ref{eq:tanhalf}) by
\mbox{$1+\cos\alpha$} and simplifying the new numerator with the
Pythagorean relation~(\ref{eq:pythag1}).

\clearpage
\textbf{Examples}
\begin{enumerate}
\item Prove:~~~~$\displaystyle \frac{\cos x-\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}}=-1$
\vspace{.3\textheight}

\item In right triangle ABC show that:
\[\frac{b+c}{a}=\cot\frac{A}{2}\]
\end{enumerate}
\textassignment{Sec. 18-4}{Ex. 4 (P. 480) \# 1-13 (odd)}
\clearpage

\section{Trigonometric Equations}

\subsection{Single Trigonometric Function with a Single Angle}
Unlike an identity, a conditional trigonometric equation is true for
only certain values of the angle.  The simplest trigonometric
equations involve a single trig function with a single angle.

In general there are an infinite number of solutions due to the
periodicity of the trig function
(e.g. $\sin(20^{\circ})=\sin(20^{\circ}+360^{\circ})$~) so we will
restrict ourselves to finding \emph{all non-negative solutions less
  than $360^{\circ}$}.

\textbf{Examples}

Solve for all $\theta$ with  $0\le\theta<360^\circ$:
\begin{enumerate}
\item ~~~~$\displaystyle 2\cos\theta+\sqrt{3}=0$
\vspace{.2\textheight}

\item ~~~~$\displaystyle 2\cos 3\theta=\frac{1}{2}$
\end{enumerate}
\clearpage
Often trigonometric identities can be used to convert a more
complicated equation into one involving a single trigonometric
function of a single angle.

\textbf{Example}~~~ Solve for all $x$ with  $0\le x<360^\circ$:

~~~~ $\displaystyle \sqrt{3}\cos x=\sin x$
\vspace{.25\textheight}

If the equation involves a single trig function that is squared we
have an equation in \emph{quadratic form}.  In this case we can use
our general techniques for solving a quadratic equation followed by
solving the trigonometric equations generated by equating the given
trig function with the quadratic solutions. As above, trig identities
may be first used to put the equation into a quadratic form as shown
in the following example.

\textbf{Example}~~~ Solve for all $\theta$ with  $0\le \theta<360^\circ$:

~~~~ $\displaystyle 3+\cos\theta=-2\sec\theta$

\clearpage
\subsection{Equations with One Angle and More Than One Function}
If trig identities cannot reduce an equation to a single trig
function, one can try transposing all terms to one side of the
equation and then factor it.  Set each factor to zero and solve as
before.

\textbf{Example}~~~ Solve for all $x$ with  $0\le x<360^\circ$:

~~~~ $\displaystyle 0=2\tan x+3\sin x\tan x$

\textassignment{Sec. 18-5}{Ex. 5 (P. 486) \# 1-25 (odd)}
\clearpage

\section{Inverse Trigonometric Functions}

We recall from MAT120 that for a given function $f(x)$ an inverse
function $f^{-1}(x)$ may be defined.  For the trigonometric functions
$sine$, $cosine$, and $tangent$ the inverses are given the special
names $\arcsin$, $\arccos$, and $\arctan$ respectively.  Because they
are inverses they are also often written $\sin^{-1}$, $\cos^{-1}$, and
$\tan^{-1}$, especially on calculators.

If for some value of $x$ and $y$ one has
\[y=\sin x\]
then one can, by definition of the inverse, define the $\arcsin$ of
the value $y$ to be $x$:
\[x=\arcsin y\]

The problem however is that there are (infinitely) many angles $x$
which have the same value of $\sin$.  For instance if
\[1=\sin x\]
then the following are all solutions:
\[x=90^{\circ},(90+360)^{\circ},(90-360)^{\circ},(90+720)^{\circ},(90-720)^{\circ},\ldots\]

A function can have only a \emph{single} value.  To make the inverse a
\emph{function} we restrict the range of angles the inverse is allowed
to take.  For instance for $\arcsin$ we force the angle to be between
$-90^{\circ}$ and $+90^{\circ}$, so, in the above example,
$\arcsin{1}=90^{\circ}$. In radians the range is $-\pi/2$ to $\pi/2$.
The numbers in this range are called the \emph{principal values} of
$\arcsin$.  These are the values which are given by a calculator.

For $\arctan$ the principal values lie in the same range as $\arcsin$.
For $\arccos$, however, the principal values are chosen to lie in the
range $0^{\circ}$ to $180^{\circ}$ ($0$ to $\pi$ radians).

The range of principal values must be chosen to include a quadrant for
which the trig function is positive and one for which it is negative.
Remembering the CAST mnemonic, this explains why the range of
principal values for $\arccos$ must differ from the other two
functions, since cosine is positive in both quadrant $I$ and $IV$.

Further graphs and analysis of the inverse trig functions may be found
in the textbook pages 487-9.

The inverse trig functions of the reciprocal functions $\csc$, $\sec$,
and $\cot$ are called $\mathrm{arccsc}$, $\mathrm{arcsec}$, and
$\mathrm{arccot}$ respectively (or $\csc^{-1}$, $\sec^{-1}$, and
$\cot^{-1}$).  They have the same principal value ranges as $\arcsin$,
$\arccos$, and $\arctan$ respectively.  The inverse reciprocal
functions are evaluated on a calculator by first reciprocating the
argument and then taking the corresponding inverse trig function.
\clearpage

\textbf{Examples} ~~Evaluate the following in degrees:
\begin{enumerate}
\item ~~~~$\displaystyle \sin^{-1}(-0.819)$
\vspace{.3\textheight}

\item ~~~~$\displaystyle \sec^{-1}(2.91)$
\end{enumerate}

\textassignment{Sec. 18-6}{Ex. 6 (P. 489) \# 1-9 (odd)}
\clearpage
\clearpage
\subsection*{Review Problems: Trigonometric Identities \& \mbox{Equations}}
\begin{enumerate}
\item Simplify:
$\displaystyle\frac{(\sec\theta+1)\tan{\frac{\theta}{2}}}{\sec\theta}$
\item Prove the following identities:
\begin{enumerate}
\item $\displaystyle \cos \theta=\sec \theta - \sec \theta \sin^2 \theta$
\item $\displaystyle \frac{1+\cos \theta}{1-\cos \theta} = \csc^2\frac{\theta}{2}-1$
\end{enumerate}
\item Solve for all $\theta$, $0^\circ\le \theta < 360^\circ$:
\begin{enumerate}
\item $\displaystyle\tan \theta+5.00=0$
\item $\displaystyle\sec 2\theta=2$
\item $\displaystyle 2\sin^2 \theta+\sin \theta = 0$
\end{enumerate}
\item In right triangle $ABC$ show that acute angle $A$ satisfies:
\[\sin(2A)=\frac{2ab}{c^2}\]
where $c$ is the hypoteneuse length.
\item Evaluate in degrees:\ \  $\displaystyle \csc^{-1}(1.495)$
\end{enumerate}
\clearpage
\subsection*{Solutions}
\begin{description}
\item [1.] $\displaystyle \sin\theta$
\item [3.(a)\,] $\displaystyle \theta=101^\circ,281^\circ$
\item [\,~~(b)] $\displaystyle \theta=30^\circ,150^\circ,210^\circ,330^\circ$
\item [\,~~(c)] $\displaystyle \theta=0^\circ,180^\circ,210^\circ,330^\circ$
\item [5.] $\displaystyle 41.98^\circ$
\end{description}
\clearpage
\section*{Formulas}
\setcounter{equation}{0}

\textbf{Reciprocal Relations}
\begin{boxedeqnarray2}
\vspace{-5em}&&\sin\theta=\frac{1}{\csc\theta}~~~~~~\csc\theta=\frac{1}{\sin\theta}~~~~~~  \sin\theta\csc\theta = 1\\
&&\cos\theta=\frac{1}{\sec\theta}~~~~~~\sec\theta=\frac{1}{\cos\theta}~~~~~  \cos\theta\sec\theta = 1\\
&&\tan\theta=\frac{1}{\cot\theta}~~~~~\cot\theta=\frac{1}{\tan\theta}~~~~~  \tan\theta\cot\theta = 1
\end{boxedeqnarray2}

\textbf{Quotient Relations}
\begin{boxedeqnarray2}
\tan\theta & = & \frac{\sin\theta}{\cos\theta}\\
\cot\theta & = & \frac{\cos\theta}{\sin\theta}
\end{boxedeqnarray2}

\textbf{Pythagorean Relations}
\begin{boxedeqnarray2}
\sin^2\theta+\cos^2\theta & = & 1 \\
1+\tan^2\theta & = & \sec^2\theta \\
1+\cot^2\theta & = & \csc^2\theta
\end{boxedeqnarray2}

\textbf{Sum or Difference of Angles}

\begin{boxedeqnarray2}
\sin(\alpha\pm\beta) & = & \sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\
\cos(\alpha\pm\beta) & = & \cos\alpha\cos\beta\mp\sin\alpha\sin\beta\\
\tan(\alpha\pm\beta) & = & \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}
\end{boxedeqnarray2}

\textbf{Adding a Sine Wave and Cosine Wave of the Same Frequency}

\begin{boxedequation}
A\sin\omega t+B\cos\omega t=R\sin\left(\omega t+\phi\right)
\end{boxedequation}

\begin{boxedequation}
A=R\cos\phi ~~~~~~~~~~B=R\sin\phi
\end{boxedequation}

\begin{boxedequation}
R=\sqrt{A^2+B^2}~~~~~~~~~~\phi=\arctan\frac{B}{A}
\end{boxedequation}
\clearpage
\textbf{Double Angle Formulas}

\begin{boxedequation}
\sin 2\alpha=2\sin\alpha\cos\alpha
\end{boxedequation}

\begin{boxedeqnarray2}
\cos 2\alpha & = & \cos^2\alpha-\sin^2\alpha \nonumber \\
\cos 2\alpha & = & 1-2\sin^2\alpha \\
\cos 2\alpha & = & 2\cos^2\alpha-1 \nonumber
\end{boxedeqnarray2}

\begin{boxedequation}
\tan 2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}
\end{boxedequation}

\textbf{Half-Angle Formulas}

\begin{boxedequation}
\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}}
\end{boxedequation}

\begin{boxedequation}
\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}
\end{boxedequation}

\begin{boxedeqnarray2}
\tan\frac{\alpha}{2} & = & \frac{1-\cos\alpha}{\sin\alpha} \nonumber \\
\tan\frac{\alpha}{2} & = & \frac{\sin\alpha}{1+\cos\alpha} \\
\tan\frac{\alpha}{2} & = & \pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}} \nonumber
\end{boxedeqnarray2}
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