This is not my area of math, but I’m guessing it is.
However, I’m pretty sure it’s not the first trigonometric proof of the Pythagorean Theorem.
First, from their abstract:
In fact, in the book containing the largest known collection of proofs (The Pythagorean Proposition by Elisha Loomis) the author flatly states that “There are no trigonometric proofs, because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean Theorem.”
But Loomis doesn’t prove that statement. It may not even have been true when he wrote it, and that book was published in 1968 which is plenty of time for someone to construct such a proof.
They have proved that Loomis’s statement is false, but a counterexample to his statement may have predated his statement or could have been discovered in the intervening five decades.
Secondly, I recall a proof of the Pythagorean Theorem that was shown to me in the 80’s, that seems like a trigonometric proof in my mind. That is quite simple.
It’s the one on the upper right. They call it the waffle cone. Some of the lengths that make up the geometric series are labeled and the text provides some clues too.
The law of sines is independent of the Pythagorean Theorem. That’s the point of the whole deal.
Loomis, in a book of collected proofs of the Pythagorean theorem, had one throwaway statement equivalent to, “You can’t prove the Pythagorean with trigonometry because trigonometry depends on the the Pythagorean Theorem.” He was wrong about that.
Like I said earlier. It depends upon what you define “trigonometry” as.
The Law of Sines is only called that because one of the key ratios happens to be the same as the sine from trigonometry. If it wasn’t, the law would be called something else.
Not to take away from the work done, but I really think it is a significant stretch to say that have a proof from trigonometry.
Moreover, the claim that this was something considered impossible relies on the definition of trigonometry used when considering the impossibility. Clearly it is not a valid proof if a property proved using Pythagoras was used. That is also clearly what was meant by the point that no such proof was possible. It is perhaps better expressed as not such proof is valid. Law of Sines is not such a property. So the proof is valid. No problem.
I don’t care about the details and applaud these teenagers for their effort. In their own way that is what those who followed up on those details in this thread have done also.
That seems plenty trigonometric, to me. I’m not sure how to define trigonometry any better than “the field of math that involves sines”.
I’m guessing that Loomis included that statement just to explain why he doesn’t include any proof based on the fact that sin^2(\theta)+cos^2(\theta)=1 . It’s really easy to get to the Pythagorean Theorem from that, as countless math students must have discovered over the millennia, but that’s just because that identity comes from the Pythagorean Theorem in the first place. He probably didn’t put any more thought into it than that.
Loomis wasn’t really a mathematician (although he was chair of a college math department for ten years). He mostly taught high school math in his career and most of his writing was about metaphysics and theology.
Also, the idea that the law of sines isn’t really trigonometry is goofyAF. It shows a relationship between the angles and lengths of sides of triangles and does so with a trigonometric function.
I suppose you could start out by proving that \sin\theta = \theta - \frac{\theta^3}{3!} + \cdots (an infinite series!!!), without using the Pythagorean Theorem, and similarly for the cosine and so forth, but that seems like so much more trouble than it is worth that I am not going to try to write down such a proof. Not the sort of thing that would have made it into Euclid’s Book I.
I’m not sure you understand what this thread is about. No one is claiming the proof is elegant.
Like Randi said about bending spoons. If you’re using trigonometry to prove the Pythagorean theorem, you’re doing it the hard way. But that’s not the point.
Don’t misunderstand, I did not say it was not worth doing (what @Chronos suggested, proving that \sin^2a+\cos^2a=1 without using the Pythagorean Theorem, or what was done on those slides) or that it had to be an elegant, textbook-style proof, just that I was too lazy/busy to do it and type it into a post. If I have some free time, that might change.
There are many, many proofs of the Pythagorean Theorem. There are not many proofs of the Pythagorean Theorem that use a fundamental theorem of trigonometry that is not equivalent to or derived from \sin^2\theta + \cos^2\theta = 1.
One can use power series to prove \sin^2\theta + \cos^2\theta = 1 but that doesn’t use a fundamental theorem of trigonometry and \sin^2\theta + \cos^2\theta = 1 is logically equivalent to the Pythagorean Theorem so such a proof would not have proved the Pythagorean Theorem using a fundamental theorem of trigonometry.
The proof in question uses this waffle cone construction to express the the sine of an angle in two ways, and the Pythagorean theorem emerges from setting these two expressions equal to each other. Furthermore, one of the two expressions comes directly from the law of sines which does not depend on the Pythagorean theorem and is not logically equivalent to it.
It is not enough to just prove \sin^2\theta + \cos^2\theta = 1 by any method and then say you can get to the Pythagorean Theorem from there and have thus proved the Pythagorean Theorem using trigonometry, because \sin^2\theta + \cos^2\theta = 1 is equivalent to the Pythagorean Theorem. Once you get there, you’re done, and if you got there without trigonometry you don’t have a proof that relies on a fundamental theorem of trigonometry.
\sin\theta is just shorthand for \frac{b}{c} and \cos\theta is shorthand for \frac{a}{c} (using a as the length of the adjacent leg, b as the length of the opposite leg, and c as the length of the hypotenuse).
So when you use power series (or any non-trig method) to prove \sin^2\theta + \cos^2\theta = 1 you have proved \frac{a^2}{c^2} + \frac{b^2}{c^2} = 1.
You are not using trig when you go from:
\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1
to
a^2 + b^2 = c^2.
Even though \sin^2\theta + \cos^2\theta = 1 is a trig identity.
So if you didn’t use trig to get \sin^2\theta + \cos^2\theta = 1 you haven’t proved the Pythagorean Theorem using trig.
Uh, no, that’s exactly how proofs work: You show that the thing you’re starting with is equivalent to the thing you’re trying to prove. Or at least that the thing you have implies the thing you’re trying to prove: You don’t usually care whether the thing you’re trying to prove implies the thing you already have, but if it does, then they’re equivalent.
That said, it’s still nontrivial to prove that \sin^2\theta + \cos^2\theta = 1 without the Pythagorean Theorem. You can do it with the power series for sin and cos, but then you also need to show that those are the power series for sin and cos, and I suspect that the process of showing that, itself, involves the Pythagorean Theorem.
a^2 + b^2 = c^2 and \sin^2\theta + \cos^2\theta = 1 are both the Pythagorean Theorem. They are two trivially different presentations of the same fundamental fact.
There are many, many ways to prove \sin^2\theta + \cos^2\theta = 1 without assuming the Pythagorean theorem (including with power series which I have linked to in this thread). Using any one of those methods and then taking the trivial step from \frac{a^2}{c^2} + \frac{b^2}{c^2} = 1 to a^2 + b^2 = c^2 is not what is meant by proving the Pythagorean with trigonometry.
If that were the case, you could take essentially any proof of the Pythagorean Theorem, relabel/rewrite everything using sines and cosines, and voila you now have a proof of the Pythagorean Theorem using trig. That is obviously not what this thread is about or what these girls did.
Perhaps it is not completely clear what is “allowed”. I feel silly talking about a hypothetical proof I have not supplied (to be clear once more, I am not referring to whatever the girls in the OP did), but suppose somewhere I use the double-angle formula for sines, for example. Not trigonometric? Or too close to dancing around simply proving the Pythagorean identity?
It is crystal clear what is allowed. Anything that is independent of the Pythagorean Theorem. It is also required that some fundamental result of trigonometry be used.
These girls used the the Law of Sines which is definitely trigonometry (despite @Francis_Vaughan 's somewhat weird objection). Furthermore it is well known that the Law of Sines is independent of the Pythagorean Theorem. On top of that, these girls show that the Law of Sines does not depend on the Pythagorean Theorem as part of their presentation.
I provide a proof in post 52. It is an extremely well known exercise in manipulation of power series. It isn’t, however, a way to prove the Pythagorean Theorem without using trig for reasons I have laid out in previous posts.
The double angle formula for sines is inarguably a fundamental result of trigonometry and it can be proved independent of the Pythagorean Theorem (cite provided upon request). Thus, if you can prove any formulation of the Pythagorean Theorem using the double angle formula for sines you have indeed provided a trigonometric proof of the Pythagorean Theorem.
I just realized that in the proof I supplied above using the law of sines, you can use the double angle formula for sines at the exact same step with extremely minimal changes.
I typed “trigonometric proof of the Pythagorean theorem” into a search engine, and literally the first PDF argued (AFAIK not originally, either) something like: start with
\begin{align}
\cos y &=\\
&= \cos\bigl(x-(x-y)\bigr) \\
&= \cos x \cos(x-y) + \sin x \sin(x-y) \\
&= \cos x (\cos x \cos y + \sin x \sin y) + \sin x (\sin x \cos y - \sin y \cos x) \\ &= \cos y \cdot (\cos^2x+\sin^2x).
\end{align}
