WildTrig2: Quadrance via Pythagoras and Archimedes

We show how the basic notion of rational trigonometry---quadrance---arises from the geometry of the ancient Greeks. The little-known sister theorem to Pythagoras features prominently, and is closely related to a theorem of Archimedes.

Category:

Science & Technology

Tags:

• quadrance
• trigonometry
• rational
• triple
• quad
• formula
• Archimedes
• Pythagoras
• geometry

[DaveKotschessa]

I'm a U.S. student and wondering if I can pursue this Trigonometry as part of my undergraduate research. Any advice on how I might go about that?

[njwildberger]

@DaveKotschessa I would look for an interested advisor at your institution, show him/her my book and ask if you could do some kind of project from it. Even better, you could ask that they consider offering a course in Rational Trigonometry and Universal Geometry. Combined with Universal Hyperbolic Geometry, this could be a very nice course, I have taught something in this direction here at UNSW to 3rd year students.

[Toxie207]

I really like this form of trigonometry. It has taken me months, but I am getting my head round it now, despite 20 years of having classic trigonometry hammered into my brain! I am a head of a Mathematics department at a High school in the UK and would love to introduce these concepts of Rational Trigonometry. Unfortunately, the exam process really requires my students to use classic trigonometry, as the questions often involve trig proofs using the classic trig functions. Dogma prevails! :(

[njwildberger]

@Toxie207 Thanks for the comment. I am very glad that you are taking the time to learn this approach. Perhaps you can think about offering it as an extra topic to some of your better students?

[Ohcacful]

Lol. Took me 15 minutes to prove the symmetrical relation and I'm a mathematics major.

[njwildberger]

@Ohcacful Don't be dismayed. Many maths students have insufficient experience working and manipulating with algebraic expressions. It is a key stumbling block to progress in maths, so take every opportunity to practice. Rational Trigonometry provides quite a few chances to improve your skills, as you shall see as you work through the series.

[horstmueller1000]

It is simple to get the symmetric formular - just expand 2( Q1^2 + Q2^2 + Q3^2) into Q1^2 + Q2^2 + Q3^2 + Q1^2 + Q2^2 + Q3^2 - Wow !!! it remains to prove that 2( Q1Q2 + Q2Q3 + Q1Q3) = Q1^2 + Q2^2 + Q3^2 - UFF ! We know that (Q3-Q2-Q1)^2 = 4 Q1Q2. So now calculate the left hand side again to obtain (Q3-Q2-Q1)^2 = Q1^2 + Q2^2 + Q3^2 + terms. Meaning that Q1^2 + Q2^2 + Q3^2 = (Q3-Q2-Q1)^2 - terms. Now combine this: 2( Q1Q2 + Q2Q3 + Q1Q3) = triple quad formular - the terms. That's it.

[asnothe]

I'm having trouble with the pythagorean theorem proof. How can we conclude that the white area at the end is Q3, when we haven't yet proved that is in fact a square shape (perpendicular lines)? Do we need to use the fact that both *angles* in a triangle, and *angles* along a straight line, total 180 degrees? I'm really enjoying these series of Prof Wildberger's.

[freezzertime]

this is how math should be taught in schools! Why isn't it?

[freezzertime]

you're pyth. thm. was excellent!!! *tears*