Nice little visual proof, but I think that the "algebraic manipulation" that change a sin into a cos would have been worth explaining. In fact, it feels like the heavy lifting of this proof is in the algebraic manipulation.
The sine doesn’t become a cosine. Sine squared plus cosine squared results in 1. The only algebra is the square of a difference.
Рік тому+4
@@MathVisualProofs I see. For me though, Sin square + Cos square equal 1 is more a Trig identity then just algebraic manipulation. Nonetheless, very nice visual proof. 🙂
@ Thanks! I do say "algebraic rules along with the pythagorean trigonometric identity" to indicate what needs to be done - so I agree that is more than algebraic manipulation.
This is a fact about the sine function, which measures the y-coordinate on the unit circle after rotating angle theta. So if you instead rotate pi-theta you have the same y-coordinate.
We are given the equation: c^2 = (a - b * cos(t))^2 + (b * sin(t))^2 We aim to show that this equation simplifies to: a^2 + b^2 = c^2 + 2 * a * b * cos(t) Expanding the terms First, expand the terms on the right-hand side of the equation: (a - b * cos(t))^2 = a^2 - 2 * a * b * cos(t) + b^2 * cos^2(t) (b * sin(t))^2 = b^2 * sin^2(t) Now substitute these expanded forms into the original equation: c^2 = (a^2 - 2 * a * b * cos(t) + b^2 * cos^2(t)) + b^2 * sin^2(t) Simplifying the equation Combine the terms: c^2 = a^2 - 2 * a * b * cos(t) + b^2 * (cos^2(t) + sin^2(t)) Recall the Pythagorean identity: cos^2(t) + sin^2(t) = 1 So, substitute this identity into the equation: c^2 = a^2 - 2 * a * b * cos(t) + b^2 Rearranging the terms Now, let's rewrite the equation to match the desired form. Rearranging the terms: c^2 = a^2 + b^2 - 2 * a * b * cos(t) Finally, add 2 * a * b * cos(t) to both sides to get: a^2 + b^2 = c^2 + 2 * a * b * cos(t) Conclusion Thus, we have shown that: c^2 = (a - b * cos(t))^2 + (b * sin(t))^2 leads to the equation: a^2 + b^2 = c^2 + 2 * a * b * cos(t)
And it is weird because in your second you added a right triangle and proved it like that i mean it does not make sense because you are solving for a different triangle now not an obtuse triangle so how both are related?
Beautiful. What do you mean when you say this is equivalent to Pythogarean theorem though ? If we didn't have it in the first place we wouldn't be able to prove the law of cosines, no ?
It is not called the law of cosines , it is called Alkashi's theory originally made by one of the greatest arabic mathematicians. I think we should give the honor to his name and not change his theory's name.
Do you know what the sine function does? It measures the y-coordinate on unit circle. If you rotate theta degrees you will be at the same y-coordinate as if your rotate theta degrees back from a pi rotation.
@@MathVisualProofs Ok i get it now, I think i got confused when trying to picture the obtuse triangle inside a circle and the epicentre (theta) not being in the middle of the imaginary circle 😅 👍
Dear Sir, I am truly impressed by your videos; they are truly remarkable. I am interested in sharing these videos on the "Chinese equivalent of UA-cam." I plan to include a link to your videos and want to emphasize that there will be no intention to generate any profit from them. My primary goal is to make your content accessible to a wider audience in China, especially those who do not have access to UA-cam. Thank you for your consideration.
انها نظرية الكاشي....العالم الرياضي .... المسلم الفارسي.... من كاشان...... وهي تعميم نظرية فيتاغورث.... ..... طريقة ممتازة...... لعرض البرهان.....نحن نتعلم...... من المهد...... الى اللحد
I mean, it is a published "proof without words." So in my mind, it's a pretty good visual proof. I added some commentary, but I think it stands alone too.
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](http://i.ytimg.com/vi/p6j2nZKwf20/mqdefault.jpg)
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Nice little visual proof, but I think that the "algebraic manipulation" that change a sin into a cos would have been worth explaining. In fact, it feels like the heavy lifting of this proof is in the algebraic manipulation.
The sine doesn’t become a cosine. Sine squared plus cosine squared results in 1. The only algebra is the square of a difference.
@@MathVisualProofs I see. For me though, Sin square + Cos square equal 1 is more a Trig identity then just algebraic manipulation.
Nonetheless, very nice visual proof. 🙂
@ Thanks! I do say "algebraic rules along with the pythagorean trigonometric identity" to indicate what needs to be done - so I agree that is more than algebraic manipulation.
Keep up these amazing videos!!
I’ll see what I can do. I’m over two years in and at some point I’ll run out of visual proofs I find interesting :)
I love binging your videos! Keep it up!
Excellent! Thanks for watching :)
Beautiful proof. It really makes the whole thing so intuitive
Thanks!
Great video man! Loving this channel!
Glad to hear it! Thanks for watching!
2:08 why does b*sin(pi-theta) = b*sin(theta) ?
This is a fact about the sine function, which measures the y-coordinate on the unit circle after rotating angle theta. So if you instead rotate pi-theta you have the same y-coordinate.
This was confusing since I thought of it like theta minus pi, and not like pi minus theta.
@@TheEGod. theta-pi doesn’t give you the right identity.
@@MathVisualProofs Yeah thats why I said I was so confused. Since I was thinking of something inccorect.
@@TheEGod. Oh! I see. :)
We are given the equation:
c^2 = (a - b * cos(t))^2 + (b * sin(t))^2
We aim to show that this equation simplifies to:
a^2 + b^2 = c^2 + 2 * a * b * cos(t)
Expanding the terms
First, expand the terms on the right-hand side of the equation:
(a - b * cos(t))^2 = a^2 - 2 * a * b * cos(t) + b^2 * cos^2(t)
(b * sin(t))^2 = b^2 * sin^2(t)
Now substitute these expanded forms into the original equation:
c^2 = (a^2 - 2 * a * b * cos(t) + b^2 * cos^2(t)) + b^2 * sin^2(t)
Simplifying the equation
Combine the terms:
c^2 = a^2 - 2 * a * b * cos(t) + b^2 * (cos^2(t) + sin^2(t))
Recall the Pythagorean identity:
cos^2(t) + sin^2(t) = 1
So, substitute this identity into the equation:
c^2 = a^2 - 2 * a * b * cos(t) + b^2
Rearranging the terms
Now, let's rewrite the equation to match the desired form. Rearranging the terms:
c^2 = a^2 + b^2 - 2 * a * b * cos(t)
Finally, add 2 * a * b * cos(t) to both sides to get:
a^2 + b^2 = c^2 + 2 * a * b * cos(t)
Conclusion
Thus, we have shown that:
c^2 = (a - b * cos(t))^2 + (b * sin(t))^2
leads to the equation:
a^2 + b^2 = c^2 + 2 * a * b * cos(t)
Now this becomes hard to forget!
You should have pointed out the sin²+cos²=1. Many viewers may not know trig identities.
I do mention you need the "pythagorean trigonometric identity" so that people can search that if they aren't aware.
This is awesome
Glad you like it!
In the obtuse example was side a for all the base or soecific part of it ?
And it is weird because in your second you added a right triangle and proved it like that i mean it does not make sense because you are solving for a different triangle now not an obtuse triangle so how both are related?
I really liked your work and explanation. Can you help me with the name of the program you are using for the explanation
I use manim for this.
Beautiful. What do you mean when you say this is equivalent to Pythogarean theorem though ? If we didn't have it in the first place we wouldn't be able to prove the law of cosines, no ?
You can prove the law of cosines without the Pythagorean theorem.
I have doubt that if we have values of trigonometric ratios grater 90⁰, then why we can't use them directly?
Amazing❤
Thank you!
It’s not a fully “visual proof” since you included “algebraic manipulation”.
You must cote Al Kashi for this theorem as you cote Pythagore !
why is Pitagorean Theorem called by the mathematician's name and i almost never see this one called Al-Kashi's law of cosines?
Yes, PT should be called "Right triangle theorem". Unfortunately it is how it is.
Kind of speeded through. If you slow down a bit and added a few more algebraic steps, then the video would be great.
Noted. Thanks.
It is not called the law of cosines , it is called Alkashi's theory originally made by one of the greatest arabic mathematicians. I think we should give the honor to his name and not change his theory's name.
Irrelevant, and neither is a theory.
@@friedrichhayek4862 i meant Theorem but u got the point
very good....
Thanks!
One day the importance and need of VR and Augmented Reality will be realized in Schools and Universities
2:09 b*sin(pi-theta) = bsin(theta)
This makes absolutely no sense
Do you know what the sine function does? It measures the y-coordinate on unit circle. If you rotate theta degrees you will be at the same y-coordinate as if your rotate theta degrees back from a pi rotation.
@@MathVisualProofs Ok i get it now, I think i got confused when trying to picture the obtuse triangle inside a circle and the epicentre (theta) not being in the middle of the imaginary circle 😅 👍
Love it❤❤❤
Thanks!
우와..........
👍😀
Software name please?
Manim. It’s in the description of every video.
❤️
I understand some of these words.
Start with those and branch out to the others 👍😀
I still don't understand why it's Pi minus the angle alpha
That is the supplementary angle. Two angles that create a straight line must add to 180 degrees or pi rads
Dear Sir,
I am truly impressed by your videos; they are truly remarkable. I am interested in sharing these videos on the "Chinese equivalent of UA-cam." I plan to include a link to your videos and want to emphasize that there will be no intention to generate any profit from them. My primary goal is to make your content accessible to a wider audience in China, especially those who do not have access to UA-cam.
Thank you for your consideration.
انها نظرية الكاشي....العالم الرياضي .... المسلم الفارسي.... من كاشان...... وهي تعميم نظرية فيتاغورث.... ..... طريقة ممتازة...... لعرض البرهان.....نحن نتعلم...... من المهد...... الى اللحد
A little too rushed
👍
It's a nice video, but calling it a visual proof seems like a stretch.
I mean, it is a published "proof without words." So in my mind, it's a pretty good visual proof. I added some commentary, but I think it stands alone too.