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Calculating sin(π/5) Using a Regular Pentagon
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- Опубликовано: 3 апр 2024
- The exact value of sin(𝜋/5) is (sqrt(10-2sqrt(5))/4. 😎
We can find it using a regular pentagon, some similar triangles and the Pythagorean theorem...
Written Article:
www.cantorsparadise.com/the-e...
Interesting article about Gauss-Wantzel Theorem:
alephoneplex.com/2021/08/22/g...
![The Most Useful Curve in Mathematics [Logarithms]](http://i.ytimg.com/vi/OjIwCOevUew/mqdefault.jpg)
I know I'm being pedantic, but any well-defined function has an exact value at every point. The phrase you're looking for is "expressible in terms of radicals" (without involving complex numbers)
Being pedantic is a mathematician job
@@scottcarothers837a real number is algebraic if it is the solution of a nonzero polynomial with rational coefficients 👌
@@advaitkamath8442 I once got points deducted in a middle-school test on a question that asked to pick the false statement about the difference between two numbers. I took "The difference between any two numbers is not always positive" to be false because I thought of the difference as the distance between two numbers and didn't think of it as subtraction that can yield negative results.
How can a middle-school teacher ask students that and expect them to be that mathematically accurate if you haven't taught them in the same formality that characterizes high-level math?
I am not sure if the expected answer is even true or not.
It would have been more fair if they asked to explain your answer to see your thinking
@@scottcarothers837, pi and e, for example, are not algebraic. Numbers that aren’t algebraic are called transcendental, that is, they are not the roots of a non-zero polynomial with rational coefficients.
Sqrt(2) is algebraic because it is a root of/solution to the polynomial x^2 - 2 = 0. As are all rational numbers.
Proving that a number is transcendental is pretty difficult.
@@erezsolomon3838 Yes, you have reasonable logic. However, in doing math, you must always define everything rigorously. Subtraction has a great visual using distance, but it's not defined as distance. It had a more rigorous definition, while yes, your thinking was right. You didn't define it rigorously enough. The way you could have is saying the absolute value of the difference of 2 natural numbers is negative, which is false
Not only is sin(pi/7) algebraic (as others have pointed out) but it has an “exact value” (by which I mean you can write it in terms of whole numbers, addition, subtraction, multiplication, division and root extraction - the technical term for such a number is a “solvable number”).
It’s minimal polynomial has degree 6, but all the exponents are even, which means it’s a cubic polynomial in x^2. Cubic polynomials are all solvable (there’s a cubic formula), and so the six roots of the minimal polynomial are obtained by taking the (two) square roots of each of the three solutions of the cubic.
What is true about sin(pi/7) is that it is not constructible (which means given a line segment in the plane of length 1, you cannot construct a line segment of length sin(pi/7) with just a straightedge and compass). This is because constructible numbers are precisely the numbers that can be written as expressions involving whole numbers, +, -, *, /, and SQUARE roots (as opposed to arbitrary nth roots, which is what distinguishes the definition of a solvable number from a constructable number). Indeed, the expression for sin(pi/7) involves taking cube roots.
Thank you for the very informative comment!! (And correction)
Hello, I know I'm a bit late, but I'd like to contribute to the conversation.
It can be proven that the cosine of 2π/7 cannot be expressed in terms of radicals. As the previous comment mentioned, the expression for sine and cosine comes from the roots of a sixth-degree polynomial. The polynomial is as follows:
x^6 + x^5 + x^4 + x^3 + x^2 + x + 1
This is the minimal polynomial that expresses the seventh roots of unity (complex roots). These roots can be expressed as e^(i * 2kπ/7) with k belonging to the set {1, 2, 3, 4, 5, 6}. This polynomial can be "reduced" to the following expression:
x^3 + x^2 + x + 1 + (1/x) + (1/x)^2 + (1/x)^3
Now, a change of variable is applied where P = (x + 1/x), where x is the conjugate root of 1/x. This detail is important as this sum gives us a real value. This real value is equal to 2(cos(2π/7)) = P.
So, our polynomial expressed in terms of P is as follows:
P^3 + P^2 - 2P - 1 = 0
Now comes the part I wanted to share. This polynomial is the minimal polynomial of 2(cos(2π/7)). By solving this cubic equation, we can find the exact value of P and consequently the values of cos(2π/7) and sin(2π/7). But it's not necessary to solve it to deduce that it is impossible to construct P with a compass and straightedge, because:
The polynomial is irreducible over the set of rational numbers.
The polynomial has no rational roots.
All its roots are real (since P is a real value).
Consequently, we are faced with a "casus irreducibilis", so P cannot be expressed in real radicals. Therefore, 2 * cos(2π/7) cannot be constructed with a compass and straightedge.
Knowing this, it can be asserted that both cos(2π/7) and sin(2π/7) cannot be constructed with a compass and straightedge. I know my writing is somewhat lengthy, but I wanted to share my discoveries (as I am self-taught and was just working on polygon constructions and thought I had something valuable to share with other math enthusiasts). I know I skipped details and rigor, but I believe my explanation is clear enough to be understood. Best regards, and I look forward to your comments.
I swear sinpi/7 is algebraic but non-constructable
Correct.
what does that mean
@deoxal7947 it can be expressed as the roots of a polynomial but cannot be constructed otherwise physically as (sqrt(2) can be constructed as the hypotenuse of a R.A.T with sides 1,1)
@@caracalthecat3040 It is constructable physically, I think you meant to say it can't be constructed geometrically
@@JacobM00se It is constructible with a compass, straightedge and parabola drawer.
sin(pi/7) is certainly algebraic. The degree of its minimal polynomial is 6, which makes it not constructible (because a constructible number would have minimal polynomial whose degree is a power of 2).
Thanks for the correction
6 is not a power of 2
sin(pi/7) is one of the roots of 64x^6-112x^4+56x^2-7
@Nice Math Problems are you still going to be falsely claiming that sin(π/7) is non-algebraic?
@@finnboltzC'mon it's about research
@@HarisRehmanGG Clarify please.
@@finnboltz He looked into π/7 differently, π/7 exists but he didn't mention it because he didn't look into it
@@HarisRehmanGG What do you mean by 'π/7 exists'?
Respectfully i feel like you're glossing over the most important point at 1:26, the equality of the values of the sides of the rightmost isosceles triangle to the radial segment is not obvious but is the key insight that let's you apply simple geometric ideas to find sin(π/5).
You make a fair point Brandon, thank you
It's not always clear what is obvious. It all depends on who your expected audience is. Which is hard to guess if you're uploading on RUclips with no major goal, and hard for us as audience to know if we haven't interacted wið ðe creator earlier
@@tomasbeltran04050 sure, but I made this comment because I think simpler/more intuitive concepts were given more time.
@@brandonklein1 That's right!!! I think that's really smart of you, in fact
@@tomasbeltran04050 I'm not trying to sound uppity / pretentious, in fact, if anything, I am doing the opposite. Of course what is easy/ obvious is subjective. I am not trying to say that any particular thing is easy or hard. What I'm saying is that in this video the point made at 1:26 was glossed over quickly while other points weren't.
But sin(pi/n) is algebraic.
Thank you for the correction! I have edited the video to remove the part where I said sin(pi/7) is not algebraic.
sin(pi/7) is Algebraic; it's the imaginary part of the first root of the polynomial x^6 + x^5 + x^4 + x^3 + x^2 + x + 1. But since it requires cube roots to express, it is not Constructible. Constructible numbers are numbers that can be written using only a finite number of addition, multiplication, or square root operations. Otherwise nice video and beautiful geometric demonstration for the Golden Triangle.
Thank you for the clarification PianoTriceratops 😁
The question that comes to mind is - is there an exact solution for ANY 'exact' value of sin, i.e. if the input can be written in some finite way, is there an output that can be written in some finite way (even if it involves millions of terms)
Yes good question. But pi/7 cannot, well it is not an algebraic number so it can't be expressed using sums, products, roots of integers
Maybe question should be asked as follows
Can be sin(π/5) expressed in terms of real radicals
@@nicemathproblems sin(π/7) can, in fact, be expressed with only field operations and roots of integers. You have been misinformed.
No. sin(3°) has a "nice" expression, as it does for all multiples of 3°, but sin(1°), for example, does not
@@hamsterdam1942 But what is a 'nice' expression? You haven't defined that clearly.
Shouldn't sin(π/7) be the imaginary portion of one of the roots of x⁷+1=0? Given that it factors into (x+1)(x⁶-x⁵+x⁴-x³+x²-x+1)=0, it should be in the factors of that hexic factor. The fundamental theorem of algebra states that any polynomial with real coefficients can be factored into linear or quadratic terms with real coefficients, and since we know that the only real solution is x=-1, this hexic should be a triplet of quadratic factors. I guess my question is how an algebraic number can have a non-algebraic component.
Thank you for the correction! I have edited the video to remove the part where I said sin(pi/7) is not algebraic.
@@nicemathproblems given that the definition of algebraic is any number that is the root of an integral polynomial, we should know that the whole complex number is algebraic by virtue of being a root of x⁷+1, but the real question is how it can be algebraic and yet have transcendental components, especially since it should in theory be extractible using a system of polynomial equations.
@@ProactiveYellow It doesn't have transcendental components. sin(π/7) is a perfectly good algebraic number, and in fact it can also be expressed using radicals.
you can add non algebraic numbers together and have it make an algebraic number, like if you add e to 1-e you get 1
@@ultrapea I never said otherwise. But here, sin(π/7) is most definitely algebraic. I think whoever made this video is confusing algebraic with constructible.
Is it just me, or is 7 is just such a pain in the neck in everything ever? I hate 7. I even remember thinking about this as a child, just looking at a foam cutout of that wretched integer and thinking "man, this thing sucks".
I don't care if it's mathematically invalid. 7 = 0/10.
Hahaha 😂. Some ancient people love 7... 7 wonders of the world etc.
I think it's because 7 is coprime with 10, and the only number smaller than 7 that's also coprime with 10 is 3. But 3 is not coprime with 10-1 so it's still easy to work with.
Basically I'm saying it's the smallest number that's annoying to do math with in base 10. So your frustration is valid!
@@davishallwe should switch bases to base 6 so that we can avoid that god forsaken number.
360 is highly divisible. It can be divided evenly by every number under 10… except 7 >:(
You should start using 7 as base for the number system. Hahahahahahahaha. 7 is absolutely awesome. Look:
1/7 = 0.142857...
2/7 = 0.285714...
3/7 = 0.428571...
4/7 = 0.571428...
5/7 = 0.714285...
6/7 = 0.857142...
How special is that??? I found that when I was a kid. The decimal representation of 7 contains
142857
as period, a cyclic number!!! Maaan, if that's not cool, I don't what it is.
Sin(pi/7) is one root of
64x^6-112x^4+56x^2-7=0
Take x² = y
64y^3-112y^2+56y-7=0
So sin(pi/7) is a root of a polynomial equation as well as it can be represented in terms of radicals as we have the formula for a cubic. I got the above equation by expanding sin(7x) in terms of sin(x).
Yes I see that now
Hello, could you tell me how you arrived at the formula for that polynomial? Can you find the formulas for the cosine?
@@EvidLekan
sin (7x) = sin (4x + 3x)
= sin(4x)cos(3x) + sin(3x)cos(4x)
= 2sin(2x)cos(2x)(4 cos³x - 3 cos x) + (3 sin x - 4 sin³x)(1- 2sin²(2x))
= 4sin(x)cos²(x)(1 - 2sin²(x))(4cos²x - 3) + (3 sin(x) - 4sin³(x))(1- 8sin²x(1-sin²x))
= 4sinx(1-sin²x)(1-2sin²x)(1-4sin²x) + (3sinx -4sin³x)(1-8sin²x(1-sin²x))
Put x = pi/7 you will get a equation in terms of sin(pi/7)
@@EvidLekan you can do the same thing for cosine.
cool video but these sound effects are *really* irritating at times
Agreed, maybe turn them a little down
I didn't even notice ðem
agreed
sin(pi/9) also has an expression with roots but it has to use complex numbers
*Simplest derivation as a positive root of a quadratic equation*
5 x 18° = 90°, therefore 2 sin 18° cos 18° = sin (2x18° = 36°) = cos (3x18° = 54°) = 4 cos^3 18° - 3 cos 18° = cos 18° (4 cos^2 18° - 3 = 1 - 4 sin^2 18°)
Canceling cos 18° from both sides, we get the quadratic 4 x^2 + 2 x - 1 = 0, where x = sin 18° (positive root of the quadratic) and - sin 54° (negative root of the quadratic, since 5 x -54° = -270° same as +90°)
Therefore, cos 72° = sin 18° = (√5 - 1) / 4 and cos 36° = sin 54° = (√5 + 1) / 4
Hence, sin (π/5) = sin (180°/5) = sin 36° = √(1 - cos^2 36°) = (10 - √20) / 4.
Therefore, sin (18° N) and cos (18° N) are computable for any integer N, positive, zero or negative N.
very nice🥰
Very nice derivation. Well done!
2:52 yes it is. It's a root of 64x⁶-112x⁴+56x²-7.
What it isn't is constructible. It cannot be expressed using only integers, addition, subtraction, multiplication, division, and square roots.
Also, adding cube roots into the mix doesn't fix this problem, unless you introduce complex numbers. Technically, though, there aren't ANY trig functions of rational multiples of π which arent constructible but are representable using cube roots of reals (though I'd love to be proven wrong). Same goes for fifth roots, sixth roots, seventh roots, etc. Only square roots and 2^n-th roots.
Thank you for the correction! I have edited the video to remove the part where I said sin(pi/7) is not algebraic.
This derivation depends entirely on the assumption (at 1:29) that the triangle can be divided to become golden with chord 'a' equal to the short legs on the left. Without clear reason to suppose that geometry holds, the derivation is circular - generating an answer that is consistent with the assumption made. One way to prove the triangle at 1:29 is golden, would be to work backwards from the value of sin pi/5 (hence circularity). It is not enough to claim the triangle to be "a known property of a regular pentagon", because the same argument can be made for any features (such as a value for sin pi/5).
Nice video i guess but it's hard to tell whether you're an AI or not, and you have some serious audio consistency issues to sort out.
Haha I'm not an AI as far as I know, but yes you're right I recorded some of the audio at different times and didn't take the time to iron it out smoothly 😅
Bro i was watching the first 20 seconden and was really confused and then i got what u were saying😂
I think why π/5 is largely skipped in secondary and university level math courses is basically due to utility. Much of the practical world uses triangles, squares and hexagons in some form or another. I wonder if easy tesselation of the plane partially explains that bias. Try finding a pentagon in everyday reality and it’s rather difficult. The most obvious example is the Pentagon outside Washington DC but there’s not much beyond that. You could make the argument that five-pointed stars are just stellated pentagons but the outlay of the shape is missing and doesn’t lead to an easily visible analysis on site. I doubt anyone looks at stars on most flags and say “there’s a 36° degree angle” or “there’s a pentagon hidden in plain sight”.
Yes there is definitely something to what you're saying
cos(pi/5) is phi/2. Doing a math test i found this out by making the arccosine of phi/2 and it turned out to be precisely pi/5. So i proved it with a pentagon. Now i can also say that sin(2pi/5) is also phi/2. Then i found out someone else also proved it in a similar way on the internet.
That's awesome 🥰
Numberphile has a great video on the constructability of heptadecagons
Love numberphile!
But (since this is Nice Math Problems) wouldn't it actually be nicer to write √(5/8 - √5/8) instead of the √(10 - 2√5)/4 that you use?
(I'm undecided, Wolfram math engine gives the former, which also has the "rhyme" of the 5 and 8).
Hmm yes there is something nice about the 5 and 8 in there 😊
Indeed sin1 in degrees has exact value. It can be derived from cubic equation involving sin3. And sin 3 = sin (18-15). sin 18 is derived from half angle formula of sin 36, and sin15 is half angle formula for 30
Other values find kar sakte hai like 18
I would have thought sin(rπ) is algebraic for every rational r. My brain isn't currently functioning, but I'm sure it's provable with e^iθ.
No you're right, others have pointed out that it's algebraic but non constructible
What software was used to create the graphics?
Manim (Python library created by 3Blue1Brown)
Or rather "sin(𝜋/5) and (sqrt(10-2sqrt(5))/4 have exactly the same value", regarding some strange comments on "exact values". Each of the expressions describes an exact value, and it happens to be the same, which is nicely shown.
Thanks S-tefan 😊
How do you know sin(π/7) is non-algebraic?
I think bcs π/7 rad isn't a integer(in degrees)
@@Mediterranean81 π/8 radians isn't an integer number of degrees either, but one can quite easily show that sin(π/8) is algebraic and even derive a radical expression for it.
@@finnboltz Pi/8 have a constant deg value
@@Mediterranean81 So does π/7.
It is algebraic: The root of 64 x^6 - 112 x^4 + 56 x^2 - 7 near x = 0.433884. It's simply not constructible using a compass and a straightedge. The video creator made a minor error.
Also sin(pi/5) = 1/2 * square root(3 - golden ratio)
Cool
what are these sounds bro
😅 the sounds of an amateur video creator
Australian narrator detected.
😂
Wtf What do you mean the regular heptagon isn’t constructible?? How did I never know this
😂
*REAL CHALLENGE - Large denominators*
Calculating sin(π/D) is super easy, when denominator D = 5 (small). What if D = 2^32 - 1 (large) ? No, it is *NOT a very long expression* , unless you're using a computer which obviously cannot simplify irrational expressions with the creativity of human mind.
I'll wait and then post the simple expression and see if you're surprised. The simplicity of the solution may be a little un-intuitive.
well, exact value is kindof a wierd statement, is the square root of 2 ever exact?
square root of 2 is an exact value, any approximation of it is not
Good point! I prefer the term "closed form solution".
7 😊
7^{-1} 👹
Smart ❤😂🎉😍🤩
That voice feels very AI generated.
No it's just amateur audio recording. Maybe I'll turn on the webcam next time to convince you I'm a real person 😅.
Pi is an irrational number. Dividing it by any number other than a multiple of itself yields another irrational number.
Yet taking the sin of pi radians gives out 0, a decidedly rational number. Just goes to show that sin is not a rational function.
@@lumipakkanen3510 No, what it shows is trigonometric functions are derived from pi.
@@davidhess6593 Trigonometric functions can be defined in terms of power series with no mention of pi whatsoever.
Eh, no... Dividing an irrational number by any rational multiple of itself of yields a rational number.
Simple Proves:
sin 54° - sin 18° = 1/2 &
sin 54°× sin 18° = 1/4
You will get:
sin 18° = (√5 - 1) / 4 &
sin 54° = (√5 +1) / 4
sin 36° = 2•sin 18°•cos 18° = 🤓
......
I already know this, since 2000
(24 years ago) 😎
I have given a THUMBS-DOWN👎to this video because:
1) @1:57 you have NOT shown the rigorous proof that both triangles are similar. In case you say that we could read Wikipedia to get the proof, I would answer that using the same logic, this video has no meaning of existing since we could also read Wikipedia to know that sin(pi/5) has an exact value.
2) Many people have already pointed out that sin(pi/7) IS algebraic.
But RUclips no longer shows the number of thumbs-down. Yeah, I know.
If you could upload a better video to replace this one, I would give a thumbs-up.
Thanks for taking the time to provide your honest feedback. You make 2 very valid points.
Please leave the annoying sound effects at home. Does not improve the video.
Thanks for the feedback. Will do 👍
You don't say
I look opposite of 𝜋 divide by 6 in mean points!