# Rational Trigonometry

I recently had the pleasure to hear the talk NJ Wildberger gave at our university. He was advertising a new way to solve problems in trigonometry. You can find his ideas in his book „Divine Proportions“. You can even download the first chapter of the book for a very good introduction. I will describe only the basics here.

Look at this triangle.

How easy can you compute the height h?

There are several ways to solve this. One idea would be to use the Heron formula to find the area. Or you could use the cosine law to compute the angle gamma, and then apply the sine function to get the height.

Wildberger proposes a way to use the latter approach without cosine or sine. The simple idea is to replace the distance by the distance squared, called the quadrance, and the angle  t by the spread sin(t)^2. In the following plot, I have used quadrances and spreads instead of distances and angles. This is indicated by the rectangle along the line segment, and by a straight line for the angle gamma, instead of an arc.

Surprisingly, all solutions are simple rational fractions. How can we obtain these fractions?

We apply two formulas. The first formula is a translation of the classical cosine law to quadrances and spreads, called the cross law. For the spread of the angle gamma in a triangle with sides □a,□b,□c  it states

$$(square a+square b-square c)^2 = 4 square a square b (1 – square gamma)$$

Remember: all values are quadrances or spreads! To make sure this is obvious, I insert the rectangle in front of the object name. For this blog, I did not dare to use simple variables for values, since the reader might not expect them in terms of rational trigonometry. So I indicate the different meaning with a rectangle everywhere.

Inserting the values □a=49, □b=16, □c=25 from our triangle, we get

$$1600=3136 (1-square gamma)$$

and thus 24/49 for □gamma. That, of course, is the spread of the angle, not the angle in radians or in degrees!

Finally, we get the quadrance of the height easily. The spread of gamma is equal to □h/□b by definition (remember it was sin(gamma)^2). So □h=16*24/49. The true length is the square root of this, and is approximately 2.80.

Computing with angles in terms of their spreads is a bit more complicated than usual. To make up for this, we start with rational values, namely with our spreads. As a reward, we usually get rational fractions.

For a demonstration, let us compute the length of the angle bisector d. For this, we first have to compute the spread delta, which is the bisected angle gamma. Now there is a formula for three spreads, where one spread is the sum or the difference of the other two.

$$(square alpha+squarebeta+squaregamma)^2 = 2 (squarealpha^2+squarebeta^2+squaregamma^2) + 4 squarealpha squarebeta squaregamma$$

This formula is called the triple spread formula. It also applies to three angles in the triangle, since in this case 180° minus one angle, is the sum of the two others, and the spread of 180°-t is the same as t. It replaces the fact that the angles in a triangle add to 180°.

We apply the formula to find the spread of delta, and get the equation

$$(squaregamma+2squaredelta)^2 = 2 (squaregamma^2+2squaredelta^2) + 16 squaregamma squaredelta^2$$

Inserting our value for □gamma, we get the following quadratic equation.

$$-dfrac{96squaredelta^2}{49}+dfrac{96squaredelta}{49}-dfrac{576}{2401}$$

This equation has two solutions for □delta, 1/7 and 6/7, since two lines have two angle bisectors. Note that the spread between lines is the same for each of the four angles we could measure between the lines. The spread depends on the lines, not on rays.

Why do we not simply divide the angle by 2?  If we wanted to do that, we would have to compute the angle gamma itself, which is an expression containing arccos.  This would spoil the idea of staying rational throughout our computations.

Once we have □delta, we need to find the spread □CWB.  Using the cross law in the triangle ABC in the same manner as we did for gamma, we find the spread at beta as 384/1225.

The triple spread rule for the spreads in the triangle CWB has again two solutions. To see this, imagine a second triangle with an angle 180°-delta instead of delta. The correct solution is the larger one. So we find 121/175 for the spread at CWB.

To find the quadrance □d of the angle bisector, we use a third formula, the spread formula. It is the usual sine law, but squared. For a general triangle with sides a,b,c and spreads sa,sb,sc, it reads

$$dfrac{square a}{square alpha} = dfrac{square b}{square beta} = dfrac{square c}{square gamma}$$

Inserting that for our triangle CWB, we get the quadrance of the side d as 2688/121.

All these calculations can be executed with pencil and paper. A computer with a software like Euler or Maxima helps to compute the solutions. Have a look at this page for some examples.

Summarizing, rational trigonometry, as proposed by Wildberger, is a very beautiful and consistent way to get exact results, where classical computations would involve trigonometric functions, together with their inverse functions. The Greeks would have loved this tool!

### Kommentare (9)

1. Dear Professor Grothmann,

Thanks for this very nice presentation of the basic ideas of rational trigonometry.

N J Wildberger

2. If I recall correctly, the intersection of h and c must be a 90 degree angle so one can then apply the law of cosine etc. If correct, the drawing is misleading because it does not look like a right angle.

3. So – what about the case where I have seven points evenly distributed around a circle (of unit radius). What is the quadrance between an adjacent pair of those points, and what is the spread of each of the three angles formed by those two points and the center?

4. You hit the weak spot. Indeed, rational geometry is more pure mathematics in the Greek sense than applied mathematics in the modern, industry oriented sense. Using it, you can find the polynomial you have to solve to find the solution. But the solution turns out not to be expressible in roots. That’s life!

5. Although the answers are pleasnt rational number, the distribution of these quadrances and spread replacements become quadratic and transcendental, instead of distance and angle which have linear distribution but quadratic and transcendental answers. It’s just the same problem, but turned on its head. You can’t avoid it.

6. There is nothing sacred about „rational numbers“, and Wildberger’s attempt to baptist geometry and purge it of the evil transcendental is unnecessary and foolish. Moreover it makes the subject more confusing than it needs to be for the poor students who be forced to learn it.

7. You are both missing the point.

It is so that many geometry problems have simple rational solutions, or at least solutions, which can be expressed by square roots. This follows from the constructions by circle and ruler. Using trigonometric expressions fails to find those simple solutions, unless we use very clever trigonometric identities.

Rational geometry is not good at replacing numerical stuff, since it loses the linearity of angles and lengths, which makes some practical things harder.

For some examples look at

http://euler.rene-grothmann.de/Programs/Examples/Rational%20Trigonometry.html

8. I do not believe that rational trigonometry has the intention to replace the classical trigonometrial, it would be absurd to think so, now that it can complement and some good topics to facilitate some calculations this, no one can deny. However I think it is worth discussing some issues, seems very elegant