Trigonometry (S)

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:

Δ = \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 θ = − sin(−θ),  cos θ = cos(−θ),  sin(θ) = cos(90^{o} − θ).

Addition formulae

sin(A ± B) = sin A cos B ± cos A sin B.
cos(A ± B) = cos A cos B ∓ sin A sin B.

tan(A ± B) = \frac{tan A ± tan B}{1 ∓ tan A tan B} .

Sum to product formulae:

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 (\frac{a + b}{2}) cos (\frac{a − b}{2})
sin A − sin B = 2 sin (\frac{a − b}{2}) cos (\frac{a + b}{2})
cos A + cos B = 2 cos (\frac{a + b}{2}) cos (\frac{a − b}{2})
cos A − cos B = 2 sin (\frac{a + b}{2}) sin (\frac{b − a}{2})

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.

1. Let ABC be a triangle with ∠BAC = 40^{o} and ∠ABC = 60^{o}. Let D and E be the points lying on the sides AC and AB respectively, such that ∠CED = 40^{o} and ∠BCE = 70^{o}. Let F be the point of intersection of the lines BD and CE. Show that the line AF is perpendicular to the line AC.

2. Let ABC be an acute angled triangle with circumcentre O. Let P on BC be the foot of the altitude from A. Suppose that ∠BCA ≥ ∠ABC + 30^{o}. Prove that ∠CAB + ∠COP < 90^{o}.

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

4. . Let ABC be a triangle. Prove that its Euler line is parallel to the side BC if and only if tan B + tan C = 3.