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% Title:  Euclidean Geometry
% Author: Stephen Farrar
% Date:   16/4/2000
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{\footnotesize April 2005}

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\begin{centre}
  \bf \Large
  Trigonometry (S)
\end{centre}

The purpose today is to help develop an alternative method of
attacking geometry problems. Now in an ideal world, every time you
came across a geometry problem, you would come up with an elegant
Euclidean solution for 7 marks. Unfortunately, as I'm sure you
have all experiences, one does come across problems which they are
unable to individually solve via such means. So, an alternative
method has been developed, relying on trigonometry and
computational geometry which can be employed as a strong second
method of attack on geometry problems. Naturally, such solutions
rarely seem as elegant as a Euclidean one, but nevertheless still
earn 7 marks (assuming correctness of course).

The student ready to embark upon the course of solving geometry
problems along a computational line must have an arsenal of
formulae ready to be called upon whenever needed. Since many
problems are stated, or can be expressed in terms of a single
reference triangle, a knowledge of triangle formulae is a must, as
is a knowledge of trigonometric formulae when this extra tool is
added (not all computational solutions require trig.) Of course,
accurate algebraic manipulation is also a necessity.

The most commonly encountered and important formulae, are as
follows.

Triangle formulae:

Area of a triangle:

$$\Delta=\frac{bh}{2}=\frac{ab\sin{C}}{2}=\frac{abc}{4R}=rs=\sqrt{s(s-a)(s-b)(s-c)}$$

Sine rule, Cosine rule:

$$\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R$$

$$a^2=b^2+c^2-2bc\cos{A}$$

$$\tan{(A/2)}=\frac{r}{s-a}$$

Angle Bisector Theorem.

Trig formulae:

Simple 1 variable relations:

$$\sin{\theta}=-\sin(-\theta),\qquad
\cos\theta=\cos(-\theta),\qquad \sin(\theta)=\cos(90^o-\theta).$$

Addition formulae

$$\sin(A\pm B)=\sin A \cos B \pm\cos A \sin B.$$

$$\cos(A\pm B)=\cos A \cos B\mp \sin A\sin B.$$

$$\tan(A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}.$$

Sum to product formulae:

\begin{eqnarray*} 2 \sin A \sin B&=&\cos(A+B)-\cos(A-B) \\
2 \cos A\cos B&=& \cos (A+B)+\cos(A-B)\\
2 \sin A \cos B &=&\sin(A+B)+\sin(A-B) \\
\sin A+\sin
B&=&2\sin\bigg(\frac{a+b}{2}\bigg)\cos\bigg(\frac{a-b}{2}\bigg) \\
\sin A-\sin
B&=&2\sin\bigg(\frac{a-b}{2}\bigg)\cos\bigg(\frac{a+b}{2}\bigg) \\
\cos A+\cos
B&=&2\cos\bigg(\frac{a+b}{2}\bigg)\cos\bigg(\frac{a-b}{2}\bigg) \\
\cos A-\cos
B&=&2\sin\bigg(\frac{a+b}{2}\bigg)\sin\bigg(\frac{b-a}{2}\bigg) \\
\end{eqnarray*}

So how does one proceed?

The first and most important step, is to always start every
geometry problem trying to apply your knowledge of Euclidean
geometry to deduce as much as you can about the diagram. Every
piece of information you can obtain in this way reduces the
effort. There is nothing worse to have spent an hour wading
through algebra to realise that two particular lines are parallel,
when in fact it could've (and should've) been determined within
five minutes with a simple angle chase.

Then, when you attack a geometry problem with a computational
method, one often realises that one needs to restate the problem
in terms of lengths. Trigonometry, as well as triangle formulae,
deal with lengths, so in order to be applicable, the problem needs
to be stated in terms of proving that two particular lengths are
equal. (Or some other result regarding lengths). Often, a
construction may come in handy here.

Less important formulae include:

Appolonius and Stewart's Theorem.

\begin{enumerate}

\item Let $ABC$ be an acute angled triangle with circumcentre $O$.
Let $P$ on $BC$ be the foot of the altitude from $A$. Suppose that
$\angle BCA\geq \angle ABC + 30^o$. Prove that $\angle CAB+\angle
COP <90^o$.

\item In a triangle $ABC$, let $AP$ bisect $\angle BAC$, with $P$
on $BC$, and let $BQ$ bisect $\angle ABC$ with $Q$ on $CA$. It is
known that $\angle BAC=60^o$ and that $AB+BP=AQ+QB$. What are the
possible angles of triangle $ABC$?


\end{enumerate}


\end{document}
