IB Math Visualized [Part 2]

By Christopher Yu

(Math and Physics tutor at The Edge Learning Center)

In the previous blog, I showed you three examples of visualizing a mathematical formula or theorem that you will come across in the IBDP Mathematics HL/SL curriculums. In this blog, I would like to continue the discussion and show you how visual thinking can help you solve mathematical problems.

The Problem

Let’s just try to evaluate

which involves the inverse trigonometric function , without using a calculator.  The challenge here is that the two arguments are not some “nice numbers” such as 1 or, in which case we can just recall the sides of the two special triangles – 45°-45° and 30°-60°.

To begin with, let’s get some idea about the possible value of the summands. The domain and range of are R and  respectively. At this point, if you realize that you don’t remember the range and domain, chances are you don’t have the following image in your mind:

IB math graph

(Source: Wolfram MathWorld)

The Visual Idea

However, the values at which is evaluated are positive real numbers . The restricted range of for , which corresponds to the possible size of the two acute angles in a right triangle. So, let’s think in terms of right triangle(s)!

Now, if we let , the original problem becomes evaluating α + β. Keep this in mind.

On the other hand, it follows that

Based on the above expressions, we can come up with the following two right triangles:

IB math triangle

IB math triangle

As a reminder, the two right triangles result from reading the two tangent values in the context of right triangles as the ratio.

Notice that all sides adjacent to the right angles of the two triangles are integers. We can thus align the longest sides of the two right triangles with two adjacent sides of a 3 by 2 grid of unit squares as follows:
IB math triangles

This is indeed the key and creative step. The purpose of forming the rectangular grid is to relate the sum α+β with some known (and convenient) angle within the grid. In particular, the right angle ∠BAD is the sum of the three angles α, β, and γ. If we know the angle γ=∠PAQ, we can then compute α+β in radians by

To find the angle γ=∠PAQ, let’s look at the triangle ∆PAQ.
IB math triangles

What kind of triangle is it? We notice that, after connecting the points P and Q, ∆PCQ and ∆ABP are congruent. Moreover, ∆PCQ and ∆PRA are also congruent (as ∆ABP and ∆PRA are congruent), and ∆PCQ can be obtained by rotating ∆PRA about the point P through radians in the clockwise direction. We can thus deduce that ∠APQ=  in radians, and hence that ∆APQ is also a right triangle. Furthermore, it is a 45°-45° right triangle because AP and PQ are corresponding sides of two congruent triangles. Finally, we conclude that γ=∠PAQ= in radians, and hence α+β==, that is,

Wrapping Up

To see is to believe, which is especially true for the visual learners. Even if you are not one of those, it does no harm for you to adopt it as an alternative way to solve mathematical problems.
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