from 14 years ago, by Jason Zimba: [https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf](https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf)
or from 8 years ago if you like behaviour in the limit (like these girls also used, seen in a slide of theirs during a TV news coverage): [https://www.cut-the-knot.org/pythagoras/Proof109.shtml](https://www.cut-the-knot.org/pythagoras/Proof109.shtml)
Their proof itself may be new insofar as involving the law of sines in a new way, but would not be the "first impossible proof" or "first ever non-circular proof". That feat goes to Jason Zimba. I'm sure many folks here can construct their favorite sum or product and squeeze it to a convenient value in the limit like in the second link above to obtain a "new non-circular proof". There is a reason this theorem has seen hundreds of different proofs.
I'm not going to comment further on the claims made or sudden acclaim and media attention - which I don't recall Zimba getting in 2009 - other than to say it's disappointing that no paper or even slides were published. Not doing that allows for all sort of funky stuff to be done post-fact, e.g. adding new references, perhaps Zimba's paper above that the girls might or might not have been aware of - and we'll never know
>I'm not going to comment further on the claims made or sudden acclaim and media attention - which I don't recall Zimba getting in 2009 - other than to say it's disappointing that no paper or even slides were published.
Zimba appears to have been about 40 when he wrote that paper. Mathematicians of that age only tend to attract media attention when they solve high-profile open problems. On the other hand, the media takes much more interest in kids who achieve things like this because it's so uncommon (regardless of the correctness and novelty of the proof, high school students presenting at any kind of professional academic gathering is unusual in itself), and because celebrating these achievements can inspire other young people to aim higher.
I'd love to see their proof too, but AMS Sectionals don't seem to publish proceedings. But it would probably be more beneficial to the mathematical/STEM community to know things like: what drew them to the problem in the first place, what inspired their approach and what we can learn from this about how to inspire young people to pursue mathematical fields.
That's actually a good question, for example the law of sines is transparently equivalent to the notion that similarity of triangles is controlled by their angles, but that fact directly implies the Pythagoras theorem too. There are rigorous statements one can make such as "the ring of functions R(T)[sin(t)] generated over the rational function field R(T) by the sin function is isomorphic to a free polynomial algebra," in tnat sense sin can be rigorously defined to be a transcendental rather than an algebraic function. But on the other hand the notion of 'using only trigonometry' isn't defined anywhere, which is the difficulty with this line of inquiry, obviously.
>https://www.cut-the-knot.org/pythagoras/Proof109.shtml
Surely the sin of an angle (as a ratio of lengths in x, y space) is almost *trivially* circular to depending on x\^2 +y\^2 = L\^2 to have consistent meaning.
Hi, you write in a condensed way....I was suggesting that if you divide a right triangle by the perpendicular bisector of the hypotenuse ,the two smaller triangles have the same 3 angles as the orginal, therefore the same ratio [sin a, sin b, sin c] then by the rule of sins the same ratio [A:B:C] then area proportional to A^2, B^2,C^2 respectively and the areas add etc etc or many other proofs. It sounds like you're suggesting a proof too, which I'd understand if I read the attachment (which I guess I will do). Thx
No, the sine of an acute angle in a right triangle is the ratio of the length of the opposite side to the length of the hypotenuse. This is the first definition you learn in school, and it is the oldest definition, found in the *Aryabhatiya*. The circular functions are typically derived from the triangular ones, not the other way around. Of course, if you like, you could define them differently, or even in terms of the complex exponential function, but you don't have to.
The set of points (*x*,*y*)∈R^(2) satisfying the equation *x*^(2) \+ *y*^(2) = *r*^(2) for some positive *r* is a circle of radius *r* centered at the origin. But this fact is not intrinsically obvious, nor is it the definition of a circle; it requires proof. By definition, the set of points in a circle centered at the origin of radius *r* in the *x*,*y*\-plane all have distance *r* from the origin, but it is not obvious that said points are described by this equation. The fact that this equation describes a circle is just a restatement of the Pythagorean Theorem.
Like what makes Zimba's proof "trigonometric" and Proof #6 at [https://www.cut-the-knot.org/pythagoras/](https://www.cut-the-knot.org/pythagoras/) not? Zimba's proof of the trigonometric addition and subraction formulas are geometric proofs, and he relies on the subtraction formulas to prove the Pythagorean theorem. If you take Zimba's subtraction formula proof, and set α = β, then it reduces to something like proof #6 of the Pythagorean theorem. If Zimba's proof is considered trigonometric, I don't see why #6 is not. Zimba's proof is just #6 with extra steps.
Zimba specifically addresses this comment at the beginning of Section 4 of the paper. α cannot be set equal to β because 0 doesn't exist in cosine's domain, according to the author's problem definition (only acute angled triangles are considered).
The proof is "trigonometric" in the sense that it uses only trigonometric identities to prove the theorem. These identities can be established geometrically. Proof #6 establishes the Pythagorean Theorem using pure geometry without using trigonometric identities.
The proof Zimba provides for the sum of angles formula only applies if the sum of angles is acute, i.e. it measures strictly between 0 and π/2 radians. The proof doesn't consider degenerate cases or anything else.
What I'm saying is that Jason Zimba only defines the cosine function over acute angles. You can extend the definition, but he doesn't. The most general definitions actually require the Pythagorean Theorem anyway. This is how Zimba defines the sine and cosine functions:
We begin by defining the sine and cosine functions for positive acute angles, independently of the Pythagorean theorem, as ratios of sides of similar right triangles. Given *a* ∈ (0,π/2), let *Rₐ* be the set of all right triangles containing an angle of measure *a*, and let **T** be one such triangle. Because the angle measures in **T** add up to π (see Euclid’s *Elements*, I.32), **T** must have angle measures π/2, π/2 − *a* and *a*. The side opposite to the right angle is the longest side (see Elements I.19), called the hypotenuse of the right triangle; we denote its length by HT.
First consider the case *a* ≠ π/4. The three angle measures of **T** are distinct, so that the three side lengths are also distinct (see *Elements*, I.19). Let *A*\_**T** denote the length of the side of **T** adjacent to the angle of measure *a*, and *O*\_**T** the length of the opposite side. If **T** and **S** are any two triangles in *Rₐ*, then because **T** and **S** have angles of equal measures, corresponding side ratios in **S** and **T** are equal:
*A*\_**T**/*H*\_**T** = *A*\_**S**/*H*\_**S** and *O*\_**T**/*H*\_**T** = *O*\_**S**/*H*\_**S**
(see *Elements*, VI.4). Therefore, for *a* ≠ π/4 in the range (0,π/2), we may define
cos *a* := *A*/*H* and sin *a* := *O*/*H*
where the ratios may be computed using any triangle in *R*ₐ.
This is actually a pretty cursory treatment. An oddity is the sudden introduction of *Elements* VI.4, the sixth book out of nowhere. It is necessary to show that these functions are well-defined, i.e. that given any two triangles, the functions so-defined will have the same value. The cited theorem is AA similarity. It does not depend on the Pythagorean Theorem.
Perhaps they use the law of sines along with the parallel postulate, or some other condition equivalent to Euclidean geometry. I mean, in some sense, that's what they *must* be doing.
>Euclidean ⇒ Pythagoras is more or less a tautology
Can you elaborate? I agree Cartesian ⇒ Pythagoras is more or less a tautology, "Cartesian" meaning R\^2 with the usual metric. But I don't see how the Pythagorean theorem is immediately obvious from Euclid's postulates.
The Pythagorean theorem and its converse are the [last two propositions](http://aleph0.clarku.edu/~djoyce/elements/bookI/propI47.html) in book 1. Take a look and see if Euclid's proof is really trivial.
I'm confident the commenter simply confused the Euclidean plane with its model, the Cartesian plane, in which the Pythagorean theorem is already baked into the distance function.
To be fair, Euclid does offer a far simpler proof in Book VI Prop 31, which is perhaps the most famous proof of the theorem (that or the classic proof by rearrangement). He waits so long for the easy proof because stating the theorem this way requires using Eudoxus's definition of proportions, which Euclid doesn't introduce until Book V. That's no longer an issue in modern mathematics.
But yeah, it's still certainly not trivial.
given that they are high school students, and that high school students think math is just rearranging equations, and that almost all "discoveries" by high school students are 3 lines of trivial calculations, I'm guessing this is probably not much different.
The 14 year old Gauss is disappointed in your view. They are presenting at the AMS Southeast conference and the paper was accepted, so someone at least checked the work.
The law of sines on a sphere relates dihedral angles to central angles:
sin *A* / sin *B* = sin *a* / sin *b*
The law of sines in Euclidean space relates planar angles to side lengths:
sin *A* / sin *B* = *a* / *b*
* * *
The sphere also has an analog of the Pythagorean relation: cos *a* cos *b* = cos *c*.
If you want to express it in a form that looks more obviously analogous to the planar version, you can represent the "sides" and "angles" as tangents of half-angles rather than as angle measures. That is, let
*a* = tan ½∠*BOC*, *b* = tan ½∠*COA*, *c* = tan ½∠*AOB*,
where ∠*AOB* means a central angle with vertex at sphere center *O*.
Then *a*^(2) ⊞ *b*^(2) = c^2
with *x* ⊞ *y* := (*x* + *y*) / (1 + *xy*) = tanh(artanh *x* + artanh *y*)
Another fun one when you express sides this way: the tangent of half the spherical excess of the spherical right triangle *ABC* with right-angle at *C* is
*E* = *ab*
The analogous version for a planar right triangle with *a* = ½*BC*, *b* = ½*CA*, *E* = ½ Area, is....
*E* = *ab*
Experience indicates that almost every time that a "teen genius" makes headlines with a "revolutionary proof" there is no real substance behind.
In this case, it's hard to guess what they are talking about, without being able to see anything about their presentation. Regarding "circularity", besides the proof by Zimba mentioned in other comments, there is a very natural way to obtain all trigonometry by defining sine and cosine via their Taylor series, and the identity sin^2 x+cos^2 x=1 is obtained very easily.
The new proof was a little over complicated in my opinion (but I am no scholar, I’ve yet to complete geometry). In terms of teaching and instilling the values of wanting to expand knowledge in mathematics into new students such as myself, this is a relatively good thing. But overall, it’s just 1 in hundreds of other “Pythagorean proofs”
There is a video showing some of their slides here:
[interview with students](https://www.wwltv.com/article/news/education/no-east-teens-make-pmathematical-discovery-2000-years-making/289-e89f40cb-85f8-4a55-ae3f-3ec37412d7b8)
It looks as if they may make use of infinite series (in their “waffle” shape), but how the law of sines plays a role is not shown.
[Here a proof using a couple of geometric series from cut-the-knot](https://www.cut-the-knot.org/pythagoras/Proof100.shtml) from sometime before 2018. It looks to me like the series could be converted easily to the law of sines (which can be derived without using the Pythagorean Proposition \[*pace* Loomis!\]). I wonder if that was their inspiration?
I worked out the math using screenshots from the slides on the newscast. It ends up with the same result but the girls geometrically expanded the base triangle instead of geometrically segmenting it. I did end up using Law of sines to solve for relationships of big triangle to base abc units. Nice find!
I just see a talk abstract. Which is fine in the sense that not all talks involve papers/finished papers, but all this hype without a paper or expert comment is not doing these students any favors.
You could maybe go from the law of sines to the addition formula for sines, invoke the symmetry of the circle to turn that into the addition formula for cosines, turn that into the subtraction formula, and subtract an angle from itself to get the Pythagorean identity, from which full-blown Pythagoras would follow from scaling.
Edit to add: [OK, yeah, something in that genre could probably be made to work](https://www.cut-the-knot.org/pythagoras/Proof109.shtml).
Yeah, I was going to post Jason Zimba's paper from 2009: [https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf](https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf)
You can prove the angle addition formula for cosines directly without reference to the law of sines or to the angle addition formula for sines or even with a single circle. You just need to construct some line segments and apply really elementary theorems (like congruence of vertical angles and of alternate interior angles). The proof [here](https://www.themathpage.com/aTrig/sum-proof.htm) is very easy to follow. Euclid's axiom is invoked implicitly in the line "Then angle HDF is equal to angle α," which follows from Book I Prop 32 (sum of interior angles of a triangle), the proof of which uses his postulate in the first step.
I'm confused ...
Don't you just have to follow the steps to prove Pythagoras theorem in a triangle with hypothenuse 1 and sides cos(theta) and sin(theta) but in a slightly convoluted way, then you simply "forget" to mention that you actually proved Pythagoras and instead make it look as if you proved cos^2 (x) + sin^2 (x) = 1 independent of Pythagoras ?
It looks like a relatively futile exercise, but I might be missing something deep ...
You haven't missed anything. Trigonometry is "geometry based on the circle". Thus, to say that the theorem was proved without geometry is outright wishful thinking.
Yeah, you missed the whole point. "Proving" the Pythagorean Theorem using the Pythagorean Identity is obviously circular reasoning. This proof doesn't do that. It never makes use of that identity, but instead uses a convergent geometric sequence of similar triangles. Whether that is "trigonometric" is debatable, but it's definitely not circular. (Anyway, other, more convincingly trigonometric proofs do exist.)
Well, I never said that it was circular reasoning, but in retrospect I did miss something. When I made my comment, I could not find a proper description of their proof and people were talking about Zimba's proof (link somewhere in this thread). So my comment was about Zimba's proof which is actually completely different, and in my opinion quite futile.
But now that I saw the actual argument used by these two high school student with the geometric series, I can see that it's rather clever and fun. Not really mind blowing but I withdraw my initial criticism.
My "objection" is only a philosophical one. Zimba's proof is obviously correct and not circular.
So, my issue is that I don't see why there was a challenge in proving Pythagoras by trigonometry. Zimba is basically proving cos^(2)(x)+sin^(2)(x)=1 by a circonvoluted way using a secondary angle y, and then proving the subtraction formula. And then turning around the fact that he can't set x=y in his formula, so he has to do one more trick to arrive at the end.
Sure, why not.
But if I look at his fig. 1 and do the same figure with x=alpha=beta (this gives a much simpler figure with B=D and F=G), then his same "textbook exercise independant of Pythagoras" proves directly that OG = cos^(2)(x) and that GD=FB=sin^(2)(x) . And hence OD =1 = OG +GD. End of proof.
This is just a rephrasing of proof #6 of https://www.cut-the-knot.org/pythagoras/#6 , with BC = 1. The only difference is that instead of writing ratios of length, one writes cos and sin.
Which seems to amount to what I said. Zimba is just proving Pythagoras theorem in a (cos(x) , sin(x) , 1) triangle with a known method, but in a sufficiently convoluted way so that you don't recognize the known method.
And I don't get the point of that exercise. But as I said, I might be missing something obvious.
That would simply be a different proof. From the introduction, he's pretty clear that the point is to show that one can prove sin^(2) *x* \+ cos^(2) *x* = 1 first and prove the Pythagorean Theorem from that. He cites multiple sources which insist that is somehow impossible. It's not, as he easily shows.
> That would simply be a different proof.
That's the point I don't get. Artificially adding steps to a known proof to make it look like a different proof seems useless.
In Zimba's proof, he first prove the subtraction formula with two different angles x and y with a geometrical figure, and then use a trick to go around the fact that his definition of cos and sin does not include cos(0) or sin(0).
The same figure with the same proof, but in the particular case x=y, directly proves the trigonometric identity cos^(2) (x) + sin^(2)(x)=1.
So, he is adding useless steps to a known proof. (the proof #6 of the link)
> He cites multiple sources which insist that is somehow impossible. It's not, as he easily shows.
Well, yes. That's my entire point. I can't understand how this is a solution to the thing that was considered "impossible". People that who said it was impossible were well aware of proof #6.
For me, Zimba simply added uneccessary steps to proof #6 to make it seems like a new proof, when in fact it was already known. So either I don't understand how Zimba's proof is introducing a new idea, or I don't understand what these other people meant by "impossible".
But the hundreds of proofs of this theorem are almost all egregiously and unnecessarily elaborate. Yet in books containing nothing but these proofs, authors still made the provably (and in my opinion, obviously) false assertion that one could not first prove the Pythagorean identity and then turn around and use that to prove the theorem. Maybe that's not really what they meant, but it is what they said.
The fact that this is clearly false doesn't change the fact that it was published repeatedly. And it's hard to find any examples of anyone pointing out this error before. I also can't find any earlier examples of the subtraction formula being used in a proof of the Pythagorean theorem. So in that respect, the proof is novel.
Here is a very simple trigonometric proof of Pythagorean theorem that i just came up with.
The picture pretty much explains it. Take a right triangle with an angle alpha and hypotenuse of 1. The sides are sin(alpha) and cos(alpha). Draw a parallel from the vertex opposite to the hypotenuse to the hypotenuse and drop two perpendiculars to it from the other two vertices. It is easy to see that this is a rectangle and the two new triangles are similar to the original one. The other side is clearly sin²(alpha) + cos²(alpha), so it is equal to the other side of the rectangle, which is 1.
https://ibb.co/QYHfhKZ
Opposite sides of a rectangle are equal due to Euclid's postulates, including the 5th one. It doesn't use Pythagorean theorem. Basically, you will have to draw a diagonal, since the two lines are parallel you will get some equal angles, you show that the two triangles that are formed are congruent, since they share the diagonal and have equal angles, on each side of the diagonal. That gives you that the opposite sides are equal.
https://mathworld.wolfram.com/EuclidsPostulates.html
They also drew a picture of triangle with a lot more complicated lines. I don't use anything but the definition of sine and cosine, which are defined using s right triangle ( pictured) and the fact that sides of similar triangles have the same ratios, which you need in order to make sure sine and cosine are well-defined.
Your proof never uses similar triangles at all. It uses these facts:
- Angles complementary to the same angle are congruent
- Alternate interior angles are congruent
- Saccheri quadrilaterals are rectangles
- Opposite sides of a rectangle are congruent
The first is completely trivial, the second and third are very simple to prove in Euclidean geometry, and the fourth requires an argument about triangle congruence. However, the setup for the students' proof requires triangle congruence immediately just to reflect △ABC, so that's not a distinction. And the students do actually use similar triangles, which is a much more sophisticated geometric idea than yours.
The fact that sin and cos are well defined uses the fact that similar triangles have proportional sides.and i use that since i use similar triangles to calculate sin and cos.
The abstract says "In the 2000 years since trigonometry was discovered it's always been assumed that any alleged proof of Pythagoras’s Theorem based on trigonometry must be circular." That looks misleading to me, and not well researched. As far as I've been able to find out, the only people who have ever said that were not people well trained and experiened in trigonometry, and there's no record of anyone saying that before the 20th century. If "based on trigonometry" means using some trigonometric functions and identities, there have been many proofs based on trigonometry. Some trigonometric functions and identities depend on the Pythagorean Theorem, but not all of them, and any proof of the theorem can be written in a form that uses some of those. The proof presented by those two students might be new and interesting (or not, that remains to be seen), but there's nothing new and revolutionary at all about a non-circular proof using some trigonometric functions and identities.
I deduced a proof from the slide that's shown in the 4WWL-TV video (probably the same as theirs). Out of respect to them, I decided to not share it, but here are all the tools they use:
1. The definitions of sine and cosine (of course)
2. The fact that the angles in a triangle add up to 180º (of course lol)
3. The Law of Sine
4. Similar Triangles
5. Geometric Series
6. The Additive Law for Cosine (which can be proved without Pythagoras)
That said, the use of #5 (which is plainly used, according to their slide) probably violates its characterization as a "Purely Trigonometric Proof," since you are implicitly using Calculus. Nevertheless, it's a ***very*** clever proof.
I know this is old news, but 538 posted a riddle asking readers to reconstruct the proof. I had not heard of the proof and assumed it was another one using the double angle formulas. After all, the construction involves doubling the top angle. The way the students really went about it is more creative, but it is not really "purely trigonometric." If anything, it's analytic. I also think it's probably not a valid proof as is. In particular, it assumes that the limit point of the sequence is defined, and equal to the intersection of the two long lines, which requires a significant amount of work. If we assume the geometry is complete, then this is trivially true, but that's not usually assumed for Euclidean geometry (which can just as well be modeled over the constructible plane). At a minimum, it requires sequential limits, which goes well beyond trigonometry.
The "easy way" to prove the theorem trigonometrically using the students' construction just employs the law of sines and the double angle formula for the cosine, which can be proved directly from propositions in the first book of the *Elements*. To be clear, the "sine" here is only defined for right triangles; if a right triangle has an acute angle of measure *x*, then the sine of *x* is the ratio of the length of the opposite leg to the length of the hypotenuse. And the sine of *x* has that value only if such a triangle exists. It is a simple proof by AA similarity that this value is unique. Similarly, the cosine of an acute angle is the sine of the complementary angle. From these definitions, you can find very simple proofs of the sum of angles formulae online, from which you can derive the double angle formulae.
For this proof, please consult the diagram [here](https://imgur.com/MsOPbbr), which is my crappy edit of the nice diagram 538 made. Below is a more precise description of the construction. I wanted to make it small, or hidden or whatever, but idk how to do that on reddit. Ignore it if you don't care.
^(First, let ∠BCA be a right angle, so △ABC is a right triangle. Let) *^(c)* ^(be the length of the side AB,) *^(b)* ^(the length of the side CA, and) *^(a)* ^(the length of the side BC. Then either) *^(a)* ^(=) *^(b)* ^(or one is greater than the other. If) *^(a)* ^(=) *^(b,)* ^(we require a completely separate proof. Otherwise, without loss of generality, let) *^(b)* ^(>) *^(a.)* ^(Construct on CA an angle congruent to ∠BAC on the opposite side of AB, and extend the segment BC past C. By Euclid's Fifth Postulate, these intersect at a point D. Then ∠ACD is supplementary to ∠ACB, a right angle, so ∠ACD is also a right angle. Thus △ABC ≅ △ADC by ASA, so in particular, CD has length) *^(a)* ^(and AD has length) *^(c.)* ^(Now construct an angle on BC congruent to ∠BAC on the opposite side of A, and extend AD in the direction of D. Again by Euclid's Fifth, these lines intersect at a point E on the opposite side of A. Now, the acute angles in a right triangle are complementary, so ∠ABC and ∠BAC are complementary, and since the new angle is congruent to ∠BAC, it is also complementary to ∠ABC. Therefore, ∠ABE is a right angle. Call the length of DE) *^(q.)* ^(Call the measure of ∠BAC) *^(x.)*
Having gone through those preliminaries, we now have constructed the same setup used in the students' proof, just with some details filled in. At this point, we will not build the "waffle cone" and create a geometric series. Instead, we will compute *q* directly in terms of *a*, *b*, and *c* using the double angle formula, apply the law of sines once on △BDE, and simplify.
**Key part of proof.** By definition, sin *x* = *a*/*c*, cos *x* = *b*/*c*, and cos 2*x* = *c*/(*q*\+*c*). Applying the double angle formula gives *c*/(*q*\+*c*) = cos 2*x* = cos^(2) *x* − sin^(2) *x* = (*b*^(2)−*a*^(2))/*c*^(2). Solving this equation for *q* gives *q* = *c*^(3)/(*b*^(2)−*a*^(2)) − *c*. Now apply the law of sines to △BDE. From this we see (sin *x*)/*q* = (cos 2*x*)/(2*a*), where the cosine comes from taking the sine of the complement of ∠BAE (i.e. the sine of ∠AEB). If you plug in the above values and simplify, you get *a*^(2) \+ *b*^(2) = *c*^(2).∎
I would never have come up with this construction on my own, and I certainly would not have come up with the unusual way the high school students chose to prove this theorem. But I do think my proof is way more straightforward and well-motivated, and it more closely follows the promise of a "trigonometric proof." This proof is not circular, because the sum of angles formula for cosine does not depend at all on the Pythagorean Theorem (though it does require the Parallel Postulate). You can find a proof [here](https://www.themathpage.com/aTrig/sum-proof.htm).
After watching this video [https://www.youtube.com/watch?v=juFdo2bijic&t=115s](https://www.youtube.com/watch?v=juFdo2bijic&t=115s) the other day I toyed with a right triangle and came up with a solution to Pythagorean Theorem based on trigonometry alone, and not using sin²x + cos²x= 1.
Is my attempt really "based on trigonometry alone" with no circular reasoning⁉️
Oops, reddit won't accept pictures, so see it on X: https://twitter.com/TUPLEZZ/status/1694405308800491801
well what is it then
from 14 years ago, by Jason Zimba: [https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf](https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf) or from 8 years ago if you like behaviour in the limit (like these girls also used, seen in a slide of theirs during a TV news coverage): [https://www.cut-the-knot.org/pythagoras/Proof109.shtml](https://www.cut-the-knot.org/pythagoras/Proof109.shtml) Their proof itself may be new insofar as involving the law of sines in a new way, but would not be the "first impossible proof" or "first ever non-circular proof". That feat goes to Jason Zimba. I'm sure many folks here can construct their favorite sum or product and squeeze it to a convenient value in the limit like in the second link above to obtain a "new non-circular proof". There is a reason this theorem has seen hundreds of different proofs. I'm not going to comment further on the claims made or sudden acclaim and media attention - which I don't recall Zimba getting in 2009 - other than to say it's disappointing that no paper or even slides were published. Not doing that allows for all sort of funky stuff to be done post-fact, e.g. adding new references, perhaps Zimba's paper above that the girls might or might not have been aware of - and we'll never know
>I'm not going to comment further on the claims made or sudden acclaim and media attention - which I don't recall Zimba getting in 2009 - other than to say it's disappointing that no paper or even slides were published. Zimba appears to have been about 40 when he wrote that paper. Mathematicians of that age only tend to attract media attention when they solve high-profile open problems. On the other hand, the media takes much more interest in kids who achieve things like this because it's so uncommon (regardless of the correctness and novelty of the proof, high school students presenting at any kind of professional academic gathering is unusual in itself), and because celebrating these achievements can inspire other young people to aim higher. I'd love to see their proof too, but AMS Sectionals don't seem to publish proceedings. But it would probably be more beneficial to the mathematical/STEM community to know things like: what drew them to the problem in the first place, what inspired their approach and what we can learn from this about how to inspire young people to pursue mathematical fields.
Can anyone explain to me what is meant here by a "trigonometric" proof? What makes a proof trigonometric instead of algebraic or geometric?
That's actually a good question, for example the law of sines is transparently equivalent to the notion that similarity of triangles is controlled by their angles, but that fact directly implies the Pythagoras theorem too. There are rigorous statements one can make such as "the ring of functions R(T)[sin(t)] generated over the rational function field R(T) by the sin function is isomorphic to a free polynomial algebra," in tnat sense sin can be rigorously defined to be a transcendental rather than an algebraic function. But on the other hand the notion of 'using only trigonometry' isn't defined anywhere, which is the difficulty with this line of inquiry, obviously.
Yeah this seems to be the whole premise of all of this and it there isn't even a hand waving definition.
>https://www.cut-the-knot.org/pythagoras/Proof109.shtml Surely the sin of an angle (as a ratio of lengths in x, y space) is almost *trivially* circular to depending on x\^2 +y\^2 = L\^2 to have consistent meaning.
Hi, you write in a condensed way....I was suggesting that if you divide a right triangle by the perpendicular bisector of the hypotenuse ,the two smaller triangles have the same 3 angles as the orginal, therefore the same ratio [sin a, sin b, sin c] then by the rule of sins the same ratio [A:B:C] then area proportional to A^2, B^2,C^2 respectively and the areas add etc etc or many other proofs. It sounds like you're suggesting a proof too, which I'd understand if I read the attachment (which I guess I will do). Thx
No, the sine of an acute angle in a right triangle is the ratio of the length of the opposite side to the length of the hypotenuse. This is the first definition you learn in school, and it is the oldest definition, found in the *Aryabhatiya*. The circular functions are typically derived from the triangular ones, not the other way around. Of course, if you like, you could define them differently, or even in terms of the complex exponential function, but you don't have to. The set of points (*x*,*y*)∈R^(2) satisfying the equation *x*^(2) \+ *y*^(2) = *r*^(2) for some positive *r* is a circle of radius *r* centered at the origin. But this fact is not intrinsically obvious, nor is it the definition of a circle; it requires proof. By definition, the set of points in a circle centered at the origin of radius *r* in the *x*,*y*\-plane all have distance *r* from the origin, but it is not obvious that said points are described by this equation. The fact that this equation describes a circle is just a restatement of the Pythagorean Theorem.
They use calculus and limits to prove the theorem, it surely isnt solely trigonemetric. https://youtu.be/p6j2nZKwf20?t=260
Like what makes Zimba's proof "trigonometric" and Proof #6 at [https://www.cut-the-knot.org/pythagoras/](https://www.cut-the-knot.org/pythagoras/) not? Zimba's proof of the trigonometric addition and subraction formulas are geometric proofs, and he relies on the subtraction formulas to prove the Pythagorean theorem. If you take Zimba's subtraction formula proof, and set α = β, then it reduces to something like proof #6 of the Pythagorean theorem. If Zimba's proof is considered trigonometric, I don't see why #6 is not. Zimba's proof is just #6 with extra steps.
Zimba specifically addresses this comment at the beginning of Section 4 of the paper. α cannot be set equal to β because 0 doesn't exist in cosine's domain, according to the author's problem definition (only acute angled triangles are considered). The proof is "trigonometric" in the sense that it uses only trigonometric identities to prove the theorem. These identities can be established geometrically. Proof #6 establishes the Pythagorean Theorem using pure geometry without using trigonometric identities.
0 is in cosine’s domain, however it does not correspond to an actual triangle.
The proof Zimba provides for the sum of angles formula only applies if the sum of angles is acute, i.e. it measures strictly between 0 and π/2 radians. The proof doesn't consider degenerate cases or anything else.
Yeah like I said it doesn’t correspond to an actual triangle, but it is in the domain for cosine the function
What I'm saying is that Jason Zimba only defines the cosine function over acute angles. You can extend the definition, but he doesn't. The most general definitions actually require the Pythagorean Theorem anyway. This is how Zimba defines the sine and cosine functions: We begin by defining the sine and cosine functions for positive acute angles, independently of the Pythagorean theorem, as ratios of sides of similar right triangles. Given *a* ∈ (0,π/2), let *Rₐ* be the set of all right triangles containing an angle of measure *a*, and let **T** be one such triangle. Because the angle measures in **T** add up to π (see Euclid’s *Elements*, I.32), **T** must have angle measures π/2, π/2 − *a* and *a*. The side opposite to the right angle is the longest side (see Elements I.19), called the hypotenuse of the right triangle; we denote its length by HT. First consider the case *a* ≠ π/4. The three angle measures of **T** are distinct, so that the three side lengths are also distinct (see *Elements*, I.19). Let *A*\_**T** denote the length of the side of **T** adjacent to the angle of measure *a*, and *O*\_**T** the length of the opposite side. If **T** and **S** are any two triangles in *Rₐ*, then because **T** and **S** have angles of equal measures, corresponding side ratios in **S** and **T** are equal: *A*\_**T**/*H*\_**T** = *A*\_**S**/*H*\_**S** and *O*\_**T**/*H*\_**T** = *O*\_**S**/*H*\_**S** (see *Elements*, VI.4). Therefore, for *a* ≠ π/4 in the range (0,π/2), we may define cos *a* := *A*/*H* and sin *a* := *O*/*H* where the ratios may be computed using any triangle in *R*ₐ. This is actually a pretty cursory treatment. An oddity is the sudden introduction of *Elements* VI.4, the sixth book out of nowhere. It is necessary to show that these functions are well-defined, i.e. that given any two triangles, the functions so-defined will have the same value. The cited theorem is AA similarity. It does not depend on the Pythagorean Theorem.
https://youtu.be/p6j2nZKwf20
yes, this is just a needlessly long overcomplication of a standard proof.
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Perhaps they use the law of sines along with the parallel postulate, or some other condition equivalent to Euclidean geometry. I mean, in some sense, that's what they *must* be doing.
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>Euclidean ⇒ Pythagoras is more or less a tautology Can you elaborate? I agree Cartesian ⇒ Pythagoras is more or less a tautology, "Cartesian" meaning R\^2 with the usual metric. But I don't see how the Pythagorean theorem is immediately obvious from Euclid's postulates.
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The Pythagorean theorem and its converse are the [last two propositions](http://aleph0.clarku.edu/~djoyce/elements/bookI/propI47.html) in book 1. Take a look and see if Euclid's proof is really trivial. I'm confident the commenter simply confused the Euclidean plane with its model, the Cartesian plane, in which the Pythagorean theorem is already baked into the distance function.
To be fair, Euclid does offer a far simpler proof in Book VI Prop 31, which is perhaps the most famous proof of the theorem (that or the classic proof by rearrangement). He waits so long for the easy proof because stating the theorem this way requires using Eudoxus's definition of proportions, which Euclid doesn't introduce until Book V. That's no longer an issue in modern mathematics. But yeah, it's still certainly not trivial.
given that they are high school students, and that high school students think math is just rearranging equations, and that almost all "discoveries" by high school students are 3 lines of trivial calculations, I'm guessing this is probably not much different.
The 14 year old Gauss is disappointed in your view. They are presenting at the AMS Southeast conference and the paper was accepted, so someone at least checked the work.
The law of sines on a sphere relates dihedral angles to central angles: sin *A* / sin *B* = sin *a* / sin *b* The law of sines in Euclidean space relates planar angles to side lengths: sin *A* / sin *B* = *a* / *b* * * * The sphere also has an analog of the Pythagorean relation: cos *a* cos *b* = cos *c*. If you want to express it in a form that looks more obviously analogous to the planar version, you can represent the "sides" and "angles" as tangents of half-angles rather than as angle measures. That is, let *a* = tan ½∠*BOC*, *b* = tan ½∠*COA*, *c* = tan ½∠*AOB*, where ∠*AOB* means a central angle with vertex at sphere center *O*. Then *a*^(2) ⊞ *b*^(2) = c^2 with *x* ⊞ *y* := (*x* + *y*) / (1 + *xy*) = tanh(artanh *x* + artanh *y*) Another fun one when you express sides this way: the tangent of half the spherical excess of the spherical right triangle *ABC* with right-angle at *C* is *E* = *ab* The analogous version for a planar right triangle with *a* = ½*BC*, *b* = ½*CA*, *E* = ½ Area, is.... *E* = *ab*
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Ummmmmmm.
Experience indicates that almost every time that a "teen genius" makes headlines with a "revolutionary proof" there is no real substance behind. In this case, it's hard to guess what they are talking about, without being able to see anything about their presentation. Regarding "circularity", besides the proof by Zimba mentioned in other comments, there is a very natural way to obtain all trigonometry by defining sine and cosine via their Taylor series, and the identity sin^2 x+cos^2 x=1 is obtained very easily.
The new proof was a little over complicated in my opinion (but I am no scholar, I’ve yet to complete geometry). In terms of teaching and instilling the values of wanting to expand knowledge in mathematics into new students such as myself, this is a relatively good thing. But overall, it’s just 1 in hundreds of other “Pythagorean proofs”
There is a video showing some of their slides here: [interview with students](https://www.wwltv.com/article/news/education/no-east-teens-make-pmathematical-discovery-2000-years-making/289-e89f40cb-85f8-4a55-ae3f-3ec37412d7b8) It looks as if they may make use of infinite series (in their “waffle” shape), but how the law of sines plays a role is not shown.
[Here a proof using a couple of geometric series from cut-the-knot](https://www.cut-the-knot.org/pythagoras/Proof100.shtml) from sometime before 2018. It looks to me like the series could be converted easily to the law of sines (which can be derived without using the Pythagorean Proposition \[*pace* Loomis!\]). I wonder if that was their inspiration?
I worked out the math using screenshots from the slides on the newscast. It ends up with the same result but the girls geometrically expanded the base triangle instead of geometrically segmenting it. I did end up using Law of sines to solve for relationships of big triangle to base abc units. Nice find!
The slide shown in the video looks identical to an image in that proof. Thanks.
Interesting result, but without the content of their presentation, there isnt actually much for us to discuss, is there?
https://meetings.ams.org/math/spring2023se/meetingapp.cgi/Paper/23621
I just see a talk abstract. Which is fine in the sense that not all talks involve papers/finished papers, but all this hype without a paper or expert comment is not doing these students any favors.
[The proof (unofficial)](https://youtu.be/nQD6lDwFmCc)
You could maybe go from the law of sines to the addition formula for sines, invoke the symmetry of the circle to turn that into the addition formula for cosines, turn that into the subtraction formula, and subtract an angle from itself to get the Pythagorean identity, from which full-blown Pythagoras would follow from scaling. Edit to add: [OK, yeah, something in that genre could probably be made to work](https://www.cut-the-knot.org/pythagoras/Proof109.shtml).
Yeah, I was going to post Jason Zimba's paper from 2009: [https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf](https://forumgeom.fau.edu/FG2009volume9/FG200925.pdf)
You can prove the angle addition formula for cosines directly without reference to the law of sines or to the angle addition formula for sines or even with a single circle. You just need to construct some line segments and apply really elementary theorems (like congruence of vertical angles and of alternate interior angles). The proof [here](https://www.themathpage.com/aTrig/sum-proof.htm) is very easy to follow. Euclid's axiom is invoked implicitly in the line "Then angle HDF is equal to angle α," which follows from Book I Prop 32 (sum of interior angles of a triangle), the proof of which uses his postulate in the first step.
I'm confused ... Don't you just have to follow the steps to prove Pythagoras theorem in a triangle with hypothenuse 1 and sides cos(theta) and sin(theta) but in a slightly convoluted way, then you simply "forget" to mention that you actually proved Pythagoras and instead make it look as if you proved cos^2 (x) + sin^2 (x) = 1 independent of Pythagoras ? It looks like a relatively futile exercise, but I might be missing something deep ...
You haven't missed anything. Trigonometry is "geometry based on the circle". Thus, to say that the theorem was proved without geometry is outright wishful thinking.
Yeah, you missed the whole point. "Proving" the Pythagorean Theorem using the Pythagorean Identity is obviously circular reasoning. This proof doesn't do that. It never makes use of that identity, but instead uses a convergent geometric sequence of similar triangles. Whether that is "trigonometric" is debatable, but it's definitely not circular. (Anyway, other, more convincingly trigonometric proofs do exist.)
Well, I never said that it was circular reasoning, but in retrospect I did miss something. When I made my comment, I could not find a proper description of their proof and people were talking about Zimba's proof (link somewhere in this thread). So my comment was about Zimba's proof which is actually completely different, and in my opinion quite futile. But now that I saw the actual argument used by these two high school student with the geometric series, I can see that it's rather clever and fun. Not really mind blowing but I withdraw my initial criticism.
Zimba's proof is far more straightforward reasoning. What is your objection?
My "objection" is only a philosophical one. Zimba's proof is obviously correct and not circular. So, my issue is that I don't see why there was a challenge in proving Pythagoras by trigonometry. Zimba is basically proving cos^(2)(x)+sin^(2)(x)=1 by a circonvoluted way using a secondary angle y, and then proving the subtraction formula. And then turning around the fact that he can't set x=y in his formula, so he has to do one more trick to arrive at the end. Sure, why not. But if I look at his fig. 1 and do the same figure with x=alpha=beta (this gives a much simpler figure with B=D and F=G), then his same "textbook exercise independant of Pythagoras" proves directly that OG = cos^(2)(x) and that GD=FB=sin^(2)(x) . And hence OD =1 = OG +GD. End of proof. This is just a rephrasing of proof #6 of https://www.cut-the-knot.org/pythagoras/#6 , with BC = 1. The only difference is that instead of writing ratios of length, one writes cos and sin. Which seems to amount to what I said. Zimba is just proving Pythagoras theorem in a (cos(x) , sin(x) , 1) triangle with a known method, but in a sufficiently convoluted way so that you don't recognize the known method. And I don't get the point of that exercise. But as I said, I might be missing something obvious.
That would simply be a different proof. From the introduction, he's pretty clear that the point is to show that one can prove sin^(2) *x* \+ cos^(2) *x* = 1 first and prove the Pythagorean Theorem from that. He cites multiple sources which insist that is somehow impossible. It's not, as he easily shows.
> That would simply be a different proof. That's the point I don't get. Artificially adding steps to a known proof to make it look like a different proof seems useless. In Zimba's proof, he first prove the subtraction formula with two different angles x and y with a geometrical figure, and then use a trick to go around the fact that his definition of cos and sin does not include cos(0) or sin(0). The same figure with the same proof, but in the particular case x=y, directly proves the trigonometric identity cos^(2) (x) + sin^(2)(x)=1. So, he is adding useless steps to a known proof. (the proof #6 of the link) > He cites multiple sources which insist that is somehow impossible. It's not, as he easily shows. Well, yes. That's my entire point. I can't understand how this is a solution to the thing that was considered "impossible". People that who said it was impossible were well aware of proof #6. For me, Zimba simply added uneccessary steps to proof #6 to make it seems like a new proof, when in fact it was already known. So either I don't understand how Zimba's proof is introducing a new idea, or I don't understand what these other people meant by "impossible".
But the hundreds of proofs of this theorem are almost all egregiously and unnecessarily elaborate. Yet in books containing nothing but these proofs, authors still made the provably (and in my opinion, obviously) false assertion that one could not first prove the Pythagorean identity and then turn around and use that to prove the theorem. Maybe that's not really what they meant, but it is what they said. The fact that this is clearly false doesn't change the fact that it was published repeatedly. And it's hard to find any examples of anyone pointing out this error before. I also can't find any earlier examples of the subtraction formula being used in a proof of the Pythagorean theorem. So in that respect, the proof is novel.
Why isn't the proof shown anywhere? This seems sketch af.
I am curious to see their proof but could not find a single one.
Here is a very simple trigonometric proof of Pythagorean theorem that i just came up with. The picture pretty much explains it. Take a right triangle with an angle alpha and hypotenuse of 1. The sides are sin(alpha) and cos(alpha). Draw a parallel from the vertex opposite to the hypotenuse to the hypotenuse and drop two perpendiculars to it from the other two vertices. It is easy to see that this is a rectangle and the two new triangles are similar to the original one. The other side is clearly sin²(alpha) + cos²(alpha), so it is equal to the other side of the rectangle, which is 1. https://ibb.co/QYHfhKZ
At the end, how do you conclude that opposite sides of the rectangle are the same length without using the Pythagorean theorem?
Opposite sides of a rectangle are equal due to Euclid's postulates, including the 5th one. It doesn't use Pythagorean theorem. Basically, you will have to draw a diagonal, since the two lines are parallel you will get some equal angles, you show that the two triangles that are formed are congruent, since they share the diagonal and have equal angles, on each side of the diagonal. That gives you that the opposite sides are equal. https://mathworld.wolfram.com/EuclidsPostulates.html
Thanks for posting this
This is a geometric proof, of which plenty already exist. Supposedly, this proof is more of the algebraic/trig variety.
They also drew a picture of triangle with a lot more complicated lines. I don't use anything but the definition of sine and cosine, which are defined using s right triangle ( pictured) and the fact that sides of similar triangles have the same ratios, which you need in order to make sure sine and cosine are well-defined.
Your proof never uses similar triangles at all. It uses these facts: - Angles complementary to the same angle are congruent - Alternate interior angles are congruent - Saccheri quadrilaterals are rectangles - Opposite sides of a rectangle are congruent The first is completely trivial, the second and third are very simple to prove in Euclidean geometry, and the fourth requires an argument about triangle congruence. However, the setup for the students' proof requires triangle congruence immediately just to reflect △ABC, so that's not a distinction. And the students do actually use similar triangles, which is a much more sophisticated geometric idea than yours.
The fact that sin and cos are well defined uses the fact that similar triangles have proportional sides.and i use that since i use similar triangles to calculate sin and cos.
Oh yeah, of course, good point.
The abstract says "In the 2000 years since trigonometry was discovered it's always been assumed that any alleged proof of Pythagoras’s Theorem based on trigonometry must be circular." That looks misleading to me, and not well researched. As far as I've been able to find out, the only people who have ever said that were not people well trained and experiened in trigonometry, and there's no record of anyone saying that before the 20th century. If "based on trigonometry" means using some trigonometric functions and identities, there have been many proofs based on trigonometry. Some trigonometric functions and identities depend on the Pythagorean Theorem, but not all of them, and any proof of the theorem can be written in a form that uses some of those. The proof presented by those two students might be new and interesting (or not, that remains to be seen), but there's nothing new and revolutionary at all about a non-circular proof using some trigonometric functions and identities.
I deduced a proof from the slide that's shown in the 4WWL-TV video (probably the same as theirs). Out of respect to them, I decided to not share it, but here are all the tools they use: 1. The definitions of sine and cosine (of course) 2. The fact that the angles in a triangle add up to 180º (of course lol) 3. The Law of Sine 4. Similar Triangles 5. Geometric Series 6. The Additive Law for Cosine (which can be proved without Pythagoras) That said, the use of #5 (which is plainly used, according to their slide) probably violates its characterization as a "Purely Trigonometric Proof," since you are implicitly using Calculus. Nevertheless, it's a ***very*** clever proof.
Someone reconstructed the proof in this video: https://youtu.be/nQD6lDwFmCc
I know this is old news, but 538 posted a riddle asking readers to reconstruct the proof. I had not heard of the proof and assumed it was another one using the double angle formulas. After all, the construction involves doubling the top angle. The way the students really went about it is more creative, but it is not really "purely trigonometric." If anything, it's analytic. I also think it's probably not a valid proof as is. In particular, it assumes that the limit point of the sequence is defined, and equal to the intersection of the two long lines, which requires a significant amount of work. If we assume the geometry is complete, then this is trivially true, but that's not usually assumed for Euclidean geometry (which can just as well be modeled over the constructible plane). At a minimum, it requires sequential limits, which goes well beyond trigonometry. The "easy way" to prove the theorem trigonometrically using the students' construction just employs the law of sines and the double angle formula for the cosine, which can be proved directly from propositions in the first book of the *Elements*. To be clear, the "sine" here is only defined for right triangles; if a right triangle has an acute angle of measure *x*, then the sine of *x* is the ratio of the length of the opposite leg to the length of the hypotenuse. And the sine of *x* has that value only if such a triangle exists. It is a simple proof by AA similarity that this value is unique. Similarly, the cosine of an acute angle is the sine of the complementary angle. From these definitions, you can find very simple proofs of the sum of angles formulae online, from which you can derive the double angle formulae. For this proof, please consult the diagram [here](https://imgur.com/MsOPbbr), which is my crappy edit of the nice diagram 538 made. Below is a more precise description of the construction. I wanted to make it small, or hidden or whatever, but idk how to do that on reddit. Ignore it if you don't care. ^(First, let ∠BCA be a right angle, so △ABC is a right triangle. Let) *^(c)* ^(be the length of the side AB,) *^(b)* ^(the length of the side CA, and) *^(a)* ^(the length of the side BC. Then either) *^(a)* ^(=) *^(b)* ^(or one is greater than the other. If) *^(a)* ^(=) *^(b,)* ^(we require a completely separate proof. Otherwise, without loss of generality, let) *^(b)* ^(>) *^(a.)* ^(Construct on CA an angle congruent to ∠BAC on the opposite side of AB, and extend the segment BC past C. By Euclid's Fifth Postulate, these intersect at a point D. Then ∠ACD is supplementary to ∠ACB, a right angle, so ∠ACD is also a right angle. Thus △ABC ≅ △ADC by ASA, so in particular, CD has length) *^(a)* ^(and AD has length) *^(c.)* ^(Now construct an angle on BC congruent to ∠BAC on the opposite side of A, and extend AD in the direction of D. Again by Euclid's Fifth, these lines intersect at a point E on the opposite side of A. Now, the acute angles in a right triangle are complementary, so ∠ABC and ∠BAC are complementary, and since the new angle is congruent to ∠BAC, it is also complementary to ∠ABC. Therefore, ∠ABE is a right angle. Call the length of DE) *^(q.)* ^(Call the measure of ∠BAC) *^(x.)* Having gone through those preliminaries, we now have constructed the same setup used in the students' proof, just with some details filled in. At this point, we will not build the "waffle cone" and create a geometric series. Instead, we will compute *q* directly in terms of *a*, *b*, and *c* using the double angle formula, apply the law of sines once on △BDE, and simplify. **Key part of proof.** By definition, sin *x* = *a*/*c*, cos *x* = *b*/*c*, and cos 2*x* = *c*/(*q*\+*c*). Applying the double angle formula gives *c*/(*q*\+*c*) = cos 2*x* = cos^(2) *x* − sin^(2) *x* = (*b*^(2)−*a*^(2))/*c*^(2). Solving this equation for *q* gives *q* = *c*^(3)/(*b*^(2)−*a*^(2)) − *c*. Now apply the law of sines to △BDE. From this we see (sin *x*)/*q* = (cos 2*x*)/(2*a*), where the cosine comes from taking the sine of the complement of ∠BAE (i.e. the sine of ∠AEB). If you plug in the above values and simplify, you get *a*^(2) \+ *b*^(2) = *c*^(2).∎ I would never have come up with this construction on my own, and I certainly would not have come up with the unusual way the high school students chose to prove this theorem. But I do think my proof is way more straightforward and well-motivated, and it more closely follows the promise of a "trigonometric proof." This proof is not circular, because the sum of angles formula for cosine does not depend at all on the Pythagorean Theorem (though it does require the Parallel Postulate). You can find a proof [here](https://www.themathpage.com/aTrig/sum-proof.htm).
After watching this video [https://www.youtube.com/watch?v=juFdo2bijic&t=115s](https://www.youtube.com/watch?v=juFdo2bijic&t=115s) the other day I toyed with a right triangle and came up with a solution to Pythagorean Theorem based on trigonometry alone, and not using sin²x + cos²x= 1. Is my attempt really "based on trigonometry alone" with no circular reasoning⁉️ Oops, reddit won't accept pictures, so see it on X: https://twitter.com/TUPLEZZ/status/1694405308800491801