The Fantastic New Trigonometric Proof of the Pythagorean Theorem

[@michaelmvnionis]

Absolutly genius! When I first read about the new proof I was trying to underatand what makes it so especial since the Pythagorean theorem is surely the one with the most demostrations of all times. I really liked the new approach, and the animation as well!

[@kirkb2665]

This is just a preexisting geometric series that they found and they put a triangle on top. I know because I found an older version of the exact same series with completed calculations. This is more the product of dumb luck than any real intellectual insight. Look up: "math.stackexchange Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle?"

[@hiderwolf1244]

I love your way of explaining asif your doing it to complete begginers. Im just 13 and have a basic understanding of trigonometry and 2 identities and i had to find a proof of Pythagoras for homework and present it. The other videos skipped key steps that were needed for me to understand for me to understand. Not only did this video help me understand it but also my class understand it better while i was explaining it

[@temmert1106]

Z’s taking high school algebra this year. He just astounded and amazed his teacher with this video :-)

[@aPictureElement]

These type of proofs make you love maths. Awesome video!

[@MathyJaphy]

That’s the effect I’m hoping for. Thank you!

[@DonG-1949]

math is circular logic, not at all real

[@academyofuselessideas]

Pretty awesome... who would've imagine that the geometric series would make a surprise appearance!

[@MathyJaphy]

Yeah, that’s my favorite part!

[@MrSilversMathSheets]

This is a very nice video. Mind Your Decisions had a very similar video out recently, but I think yours was first. That one had a good explanation of why okay to use trig in the proof. The music choice was unfortunate.

[@MathyJaphy]

Thanks. Yes, I saw the Mind Your Decisions video that was posted after mine. He did a good job with it. Sorry you didn't like my music choice. I love how that rockin' tune enhances the coolness of the math, but I know not everyone does. (I assume you meant that *my* music choice was unfortunate, not that Presh Talwakar's choice *not* to use music was unfortunate. :-).

[@adityaprakash4745]

Mind blowing approach man,

[@colinrubow3493]

Very nice and satisfying. Very clear as well. Question: you mentioned the proof is special because it's trigonometric rather than geometric, and then later mention there are other trigonometric proofs. Could you clarify this point?

[@MathyJaphy]

Hi, and thanks for the compliments on my video. Since trigonometric proofs were recently thought to be impossible without circular reasoning, it seems to me that any such proof has a special place in the storied history of the Pythagorean Theorem. In the description, I link to a 2009 paper containing another trigonometric proof which may have been the first to debunk the claim that it couldn't be done.

[@colinrubow3493]

@@MathyJaphy Cool, Thank you.

[@academyofuselessideas]

@@MathyJaphy Thanks for sharing the paper. I will have to read it carefully because it claims that sin^2 (x) + cos^2(x) = 1 is not equivalent to the Pythagorean theorem which I think it is: Assuming the Pythagorean theorem consider a right triangle with one of its angles equal to x and hypotenuse equal to 1. For the other direction, consider a right triangle with sides a, b, and c (c being the hypotenuse), and let x be the angle between the sides a and c. Then, by assuming the trigonometric equality, you get b^2/c^2 + a^2/c^2 = 1, which is the pythagorean theorem... So, maybe i am being silly and there is something i am not understanding... In any case, thanks for sharing that paper!

[@mathemateric]

Very elegant!😊 When I told my teacher I had similar proof in 2022, he didn't show any interest.😢 Almost exactly one and a half year ago. Now, I'm out of high school.

[@MathyJaphy]

Sorry to hear that. I hope you will remain proud of your accomplishment anyway, and that you will continue to share it with others who can appreciate it.

[@h1a8]

Can you please advise me how I can make videos like this with your animations and visuals. 1. What classes do I need to take (video editing, etc)? 2. What software did you use?

[@MathyJaphy]

Thanks for asking! The descriptions of all my videos give credit and links to the software I use. The text animation is done in Apple Keynote. The graphical animations are done with Desmos Graphing Calculator. The resulting video clips are put together with Apple iMovie. It's not trivial, but at least all the software is free and popular, so there are online tutorials that can help you learn how to use each package individually. Making the pieces work together to create a video presentation is where most of the creativity and ingenuity is required. Yes, take a class that teaches how to use video editing software, and any math class that introduces you to Desmos.

[@h1a8]

@@MathyJaphy Thank you so much for this. I'm a professional tutor and still training to be a college professor. I would like to utilize technology like this (visual) to teach advanced students better

[@kirkb2665]

You can use the Law of Sines with ANY and ALL proofs of the Pythagorean Theorems. Sine = Opposite/hypotenuse. Law of Sines: a/sin(a) = b/sin(b) = c/sin(c) For any Right triangle: a/sin(a) = a/(a/c) = c b/sin(b) = b/(b/c) = c c/sin(c) = c/(c/c) = c For any Right triangle: a/sin(a) = b/sin(b) = c/sin(c) = c Which means for any Right Triangle: (a/sin(a))^2 = (b/sin(b))^2 = (c/sin(c))^2 = c^2 THEREFORE for any right triangle: a^2 + b^2 = (a/sin(a))^2 = (b/sin(b))^2 = (c/sin(c))^2 = c^2

[@sainiyashraj]

I dont get it can you please elaborate

[@kirkb2665]

@@sainiyashraj Is a Trigonometric Proof Possible for the Theorem of Pythagoras? Michael de Villiers RUMEUS, University of Stellenbosch CONCLUDING COMMENTS To get back to the original question of whether a trigonometric proof for the theorem of Pythagoras is possible, the answer is unfortunately twofold: yes and no. 1) Yes, if we restrict the domain to positive acute angles, any valid similarity proof can be translated into a corresponding trigonometric one, or alternatively, we could use an approach like that of Zimba (2009) or Luzia (2015). 2) No, if we strictly adhere to the unit circle definitions of the trigonometric ratios as analytic functions, since that would lead to a circularity.

[@andyanderson7970]

Nice video! I guess the proof does use a bit of trig, after all (sort of, by letting sin(2α) be the name of the ratio of h/c)! But it’s not necessary, as one can show that in addition to X = x₁/(1 - r) = (2ac/b)/(1 - r) you can calculate the area of the enclosing triangle in the same way: A = ab (1 + r)/(1 - r) and since we also know that A = cX/2 (ac²/b)(1 - r) = ab (1 + r)/(1 - r) c² = b² (1 + a²/b²) c² = a² + b²

[@MathyJaphy]

Yes! Using the area formulas makes for a cleaner proof. One could also argue that trigonometry isn't needed even with the proof as presented here. It's all just ratios of side lengths of similar triangles which we happen to call "sin". This caused me to question what it even means to be a "trigonometric" proof. Even Jason Zimba's proof using the double-angle formula could be questioned (see video description for links if interested). After all, the double-angle formula comes from comparing analogous side lengths in an arrangement of similar triangles!

[@novembern939nn5]

Hi there, any new video in the works??

[@MathyJaphy]

Yes, I have one in progress and a couple more planned!

[@novembern939nn5]

@@MathyJaphy awesome. Can't wait :)

[@fromtheendavi]

Very good

[@calicoesblue4703]

Nice💯💯💯

[@chixenlegjo]

I still think a non-geometric proof of the Pythagorean theorem is still impossible. How do you even state it without geometry?

[@MathyJaphy]

Agreed! My assertion that this proof is "based on trigonometry rather than geometry" was poorly worded, if that's what you're referring to.

[@sharkdavid]

wait aren't there two proofs? I though they each came up with pure trig proofs independently.

[@MathyJaphy]

I don't know about that. From what I've heard, they worked together on this one proof. Apparently, they've come up with more proofs since then.

[@adbellable]

conversely you cant have convergence without euclid

[@acuriousmind6217]

You can't call it a proof since you're using trigonometry which a direct result of the Pythagorean theorem basically trig does not exist without the Pythagorean theorem.

[@MathyJaphy]

Well, you could argue that it's not a trigonometric proof at all, since it only uses the sine function to refer to the ratio between the opposite leg and the hypotenuse. But if you're going to claim that it's not a proof, you'll have to point out where it uses circular reasoning. Also, I would refer you to the link in the description to Jason Zimba's paper which is a bona fide trigonometric proof that does not use circular reasoning.

[@theoremus]

@@MathyJaphy I like Zimba's proof since it does not depend upon the geometric series, which is a limiting process. Zimba's proof depends upon the sine and cosine difference angle formulae, which do not depend on Pythagoras Theorem.

[@ngkktht774]

If you mean the "sin(2a)" and "sin(b)" then those are just names assigned to the ratios between the sides of similar triangles. It's not like the sine was actually "used" as a function. What was really used were just properties of similatr triangles. You can just completely ignore the sin(2b) written in there and consider that the triangles are similar, because angles of both of them are [90deg, beta, 180-90-beta]. And as they are similar, the ratios between their corresponding sides has to be the same. Thus "h/2a=b/c"... The same thing applies to the two "sin(2a)" triangles. And properties of similar triangles can be proved without the pythagorean theorem.

[@ngkktht774]

@@MathyJaphy Mostly true with one exception - the part about the a=b special case @2:55 where you just used sin(90°)=1 like if we did have trigonometry already.

[@MathyJaphy]

I completely agree. That’s what I was saying in my first reply in this thread.

[@PrithwirajSen-nj6qq]

Here algebra also used. So may not say it a trigonometric proof. But a combination of both of the two. Thanks.

[@MathyJaphy]

That's true, and geometry is used as well!

[@thomasolson7447]

Hyperbollic geometry, but there is a curve ball here. f:=(x,y)-> , normalized R^2 vector. f(x^2-y^2, 2*x*y)=, f(x,y) forms an angle with , this doubles that angle. f(x^2-(i*y)^2, 2*x*i*y)=, inserting an i*y instead of y. My equations and whatever that thing is are very close, but not the same. There is an extra 'a' and the 'y' is negative. More than likely an artifact of presentation and my simplification.

[@kirkb2665]

Their proof is just a preexisting geometric series that they found and they put a triangle on top. I know because I found an older version of the exact same series with completed calculations. This is more the product of dumb luck than any real intellectual insight. Look up: "math.stackexchange Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle?"

[@gmnboss]

Mainly on trig plus a bit calculus...infinite series

[@MathyJaphy]

Point taken. And there’s still some geometry in there too.

[@kirkb2665]

Is a Trigonometric Proof Possible for the Theorem of Pythagoras? Michael de Villiers RUMEUS, University of Stellenbosch CONCLUDING COMMENTS To get back to the original question of whether a trigonometric proof for the theorem of Pythagoras is possible, the answer is unfortunately twofold: yes and no. 1) Yes, if we restrict the domain to positive acute angles, any valid similarity proof can be translated into a corresponding trigonometric one, or alternatively, we could use an approach like that of Zimba (2009) or Luzia (2015). 2) No, if we strictly adhere to the unit circle definitions of the trigonometric ratios as analytic functions, since that would lead to a circularity.

[@MathyJaphy]

Thanks for referring that paper, I hadn't seen it before. I agree completely that this proof uses trig only as terminology for ratios of sides of similar triangles. So, calling it a "trigonometric" proof is a bit weak. The paper is thoughtful and well written, and its only flaw is that it mentions two KZbin videos, neither of which is mine. :-)

[@kidredglow2060]

WOW

[@theoremus]

Your video animations are cool but your derivation of X/Y is inordinately complex.

[@MathyJaphy]

Thank you! I had fun making those animations. And you're right, my derivation of X/Y is complicated. I like how the complexity dissolves away by the end, leaving the Pythagorean simplicity. However, a good proof remains straightforward throughout. I think your Variation #1 is the winner. Much prettier than mine! By the way, thanks for the mention in your latest Variation video. I am honored!

[@theoremus]

@@MathyJaphy Overall, I think that Jason Zimba's proof is the best trig proof. It relies on the difference angle formulae. It does not rely on a limiting process.

[@anfarahat]

Unnecessarily complex.

[@BuleriaChk]

c=a+b c^2 = (a+b)^2 = [a^2 + b^2] + [2ab] (binomial expansion) c^2 a^2+b^2 The "proof" in the video is only valid in the imagination. (Pythagoras was also confused).

[@MathyJaphy]

In a triangle, the sum of any two side lengths is greater than the length of the third side. So your presumption, c=a+b, is incorrect.

[@BuleriaChk]

@@MathyJaphy The Pythagorean calculation does not include the area even if I assign c' = a + b. The point is that the sides of the triangle are affine without the area; one leg can be on earth, one inthe middle of the 'andromeda cluster, and one at the bottom of the ocean. The equaation c^2 = a^2 + b^2 obtains only if one omits the product ab (i.e., multiplication, where the area of the triangle is A = 2ab. The can only be obtained by using imaginary number where the product +/- iab is eliminatd by complex conjugation psipsi*

[@BuleriaChk]

# = a+b #^2 = [a^2 + b^2] +[2ab] # = 7 = 3 + 4 #^2 = 7^2 = 49 = [25] +[24] 25 (count is preserved under multiplication) That is, c:= a + ib c* = a - ib cc* = [a^2 + b^2] + a(ib) -a(ib) (b is imaginary) (a+b) ^2 = [a^2 + b^2] +[2ab] (binomial expansion); Fermat's Last Theorem for the case n=2

[@BuleriaChk]

@@MathyJaphy That is, Fermat's Last Theorem is valid for the case n=2 for all positive real numbers c^2 a^2 + b^2 since in second order (I repeat, sigh. ad infinitum, ad nauseam) c= a + b c^2 = [a^2 + b^2] + [2ab] (Binomial Expansion, proved by Newton) [a^2 + b^2] (why) figure it out and you will be enlightened....

[@BuleriaChk]

@@MathyJaphy 2nd order equations relate areas, not lengths,