a golden value of cosine.

We find a closed form for cos pi/5.
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[@schweinmachtbree1013]

I feel like this could be part of the overkill playlist xD

[@dougr.2398]

Yes and no. It is a good demonstration of the advantage and simplicity of moving up one dimension by introducing complex exponentials, but just as a regular pentagon can be semi-simply constructed with a straight edge and compass, computing cosine and sine of pi divided by five is as easy as figuring the sine snd cosine of 18° ( π/5 = 72°; 72 + 18 = 90 ). There are rather simple and elegant geometric ways of doing this. However, the more ways you can find of demonstrating or proving a mathematical relationship (there are actually hundreds of ways of proving the Pythagorean Theorem) the more certain you may be of both your own skills and the validity of the found or discovered relationship

[@dougr.2398]

I may have confused this with a way of finding sin 15°.... I’ll have to check

[@bostonjuan9537]

Pro tip : watch series on Flixzone. Been using it for watching lots of of movies lately.

[@zairecallan83]

@Boston Juan Yup, been using Flixzone} for since november myself :D

[@cameronsolomon4249]

@Boston Juan definitely, been using flixzone} for months myself =)

[@josephsmith786]

He missed a golden opportunity for this video (which was originally posted on April 20th, 2022): writing cos(pi/5) as cos(pi*4/20)

[@PubicGore]

This isn't Flammable Maths.

[@MathNerdGamer]

@@PubicGore Now I want a Papa Flammy version of this video. :)

[@MohitRaj-1712]

I like the way he approaches a problem. FANTASTIC

[@Setiny]

An easier attempt: Notice that cos 3π/5 + cos 2π/5 = 0 Hence cos π/5 is a root of 4x^3 - 3x + 2x^2 - 1 = 0 It is easy to see that x = -1 is a root And we may obtain 4x^2 - 2x -1 = 0 By quadratic formula x = (1 +- sqrt 5) / 4 Clearly x > 0, and the answer is the positive one.

[@vinc17fr]

Few calculations, without using more than degree 2: Consider a = cos(2π/5) > 0 and b = cos(4π/5). One has 1 + 2a + 2b = 0 (regular pentagon) and b = 2a² − 1 (trigonometric formula). So 4a² + 2a − 1 = 0, whose positive root is a = (−1 + √5)/4. Then cos(π/5) = −cos(4π/5) = −b = 1/2 + a = (1 + √5)/4.

[@dugong369]

I feel like this solution deserves more recognition. I really like the derivation using similar triangles (blackpenredpen sin(18°) ) and I'd like to become more familiar with roots-of-unity methods, but your method is very simple and direct.

[@violetasuklevska9074]

Pi/5 is 36 degrees which is an angle often seen in normal pentagons and the golden ratio can also be found there. I'm too lazy to figure out the details, but it's sounds way simpler.

[@andreasproteus1465]

A golden triangle has angles 36°, 72°,72° (i.e. π/5, 2π/5, 2π/5) and sides ratio=φ. Hence cos(72°) = 1/2φ. From that you can easily find cos(π/5) = cos(36°) = φ/2, by using cos(2x) = 2 cos^2(x) -1 His method is too complicated.

[@violetasuklevska9074]

@@andreasproteus1465 Yeah, in hindsight a 36-72-72 triangle was enough.

[@vinc17fr]

@@andreasproteus1465 How do you prove that sides ratio=φ without using the cosine and/or sine values?

[@andreasproteus1465]

@@vinc17fr Golden triangles are formed by the sides and the diagonals of a pentagon. The ratio of the diagonal to side of a pentagon is φ because the construction of the pentagon is based on dividing a line segment to the "extreme and mean ratio" and using the small part for the side and the large part for the diagonal of the pentagon.

[@vinc17fr]

@@andreasproteus1465 Thanks. OK, the equation comes from the two homothetic triangles in the pentagon.

[@johnbailey8103]

Ah yes, the age old maths trick of multiplying by 1, why didn't I think of that? XD

[@1ucasvb]

Literally all of mathematics is basically "doing nothing" in creative ways: changing the perspective of the thing without changing the thing.

[@johnbailey8103]

@@1ucasvb That's why I have a very deep dislike for linear algebra

[@MuharremGorkem]

I always miss adding a zero. It seems harder than multiplying by 1 :)

[@nikitakipriyanov7260]

@@MuharremGorkem But that's the same way in logarithmic sense! Adding 0 to the logarithm is exactly the same as multiplying its argument by 1.

[@MuharremGorkem]

@@nikitakipriyanov7260 And now you suggest apply a function (log etc) to both sides? This does not make me any more comfortable :))) Instead of log, if we take sin of each side, this time you may add 2pi*n to both sides... So when to use which function? Furthermore, an added 0 (similarly a factor of 1) can be written infinitely many different ways. Math-ing is definitely an indefinitely creative activity :)

[@goodplacetostop2973]

9:40 Good Place To Stop Bonus homework for time travellers : Show that tan(3π/11) + 4sin(2π/11) = √11

[@goodplacetostop2973]

Wow, 2 months ago... I dont remember being so early in comments 😂

[@aahaanchawla5393]

@@goodplacetostop2973 how do you have access to the links before a public upload?

[@kostasbr51]

@@aahaanchawla5393 Perhaps he is the twin brother of Michael.

[@Noam_.Menashe]

How?

[@goodplacetostop2973]

@@aahaanchawla5393 Unlisted videos. They won’t appear in the Videos tab of the channel homepage but they can be seen if they are in public playlists.

[@MrRyanroberson1]

since zeta^5 = -1 by its definition, we know zeta^4 = -1/zeta, and zeta^6 = -zeta. we also know zeta^9 = 1/zeta, which means zeta^4 + zeta^6 = -(zeta + zeta^9) = -x. since we had -(zeta^4 + zeta^6) + 1, that means we get x+1. No messing around with the factorization of z^10-1

[@JobBouwman]

Using cos(A) = 1-2*sin^(A/2) we get: cos(pi/5)=1-2*sin^2(pi/10) Moreover, we have that sin(pi/10)=(1/2)/phi. This can be derived from the regular pentagon in which the ratio between its side and a diagonal equals phi = 1/2 + 1/2 sqrt(5)

[@michaelschmitt2427]

This is nice and short. But Michael's approach does not require any trig identity (including de Moivre's formula) and evokes some nice properties of the roots of unity, so it is more interesting I think. (On the other hand, Michael's method does require knowledge of formulas to factorize z^5-1.)

[@HoSza1]

@@michaelschmitt2427 Using trig identities is not a problem because those can be derived algebreaically. But he also uses geometry to calculate the ratio between side length and diagonal length of a pentagon, which is in contrast with Michael's purely algebraic solution.

[@anantprakashpandey2106]

Enlightening method of approach. Absolutely loved it.

[@manucitomx]

Thank you, Professor. That was truly inspiring.

[@AlephThree]

This is a similar calibre of teaching to what I received through the tutorial system at Oxford, over 20 years ago. Except it’s free and available to all. Incredible stuff.

[@behnamashjari3003]

Dr. Penn, In this solution, you confused the hell out of me, and I have a PhD in EE with minor in Math. You have skipped many intermediate steps. I'd rather use my calculator for cos(36°) than being so tortured! LOL. Your skipping major steps reminds me of a Charlie Chaplin hilarious silent movie in which he goes to a restaurant and orders a food which consists of a bowl of about 1000 green peas. He starts putting his fork in each pea, putting them one by one in his mouth. Someone next to him is watching and wonders this could take forever. After about 20 or so peas taken one by one by fork, Chaplin runs out of patience and puts his fork aside and takes the bowl into his mouth and gobbles up the whole thing in 5 seconds! Behnam Ashjari

[@MathElite]

I love golden numbers, they show up everywhere Well...I just like gold

[@gaiuslucidus276]

Spoiler alert please!

[@MathElite]

@@gaiuslucidus276 look at the title, I didn't spoil anything.

[@tomatrix7525]

@@gaiuslucidus276 since when is a spolier alert applicable to math! No hate though, I love the idea

[@SlidellRobotics]

It's rather easier to work out the formula for cos5x = 16cos⁵x -20cos³x + 5cosx, note that cos5x is -1, and arrange into the polynomial in c = cosx of: 16c⁵ -20c³ + 5c + 1 = 0. This factors as (c+1)(4c² - 2c - 1)². Then use the quadratic formula on the second order terms and it's just a matter of picking one of the positive roots. If you note that substituting f=2c in those quadratics, the defining equation for φ is easily recognizable. There's also a more geographic technique which involves an isosceles triangle with one angle of π/5 and two of 2π/5. Then bisect one of the large angles and do some similar triangle and isosceles triangle identities to get length ratios.

[@michaelz2270]

Dude, you just need to solve cos 2x + cos 3x = 0 for cos x, and use the formulas cos2x = 2 cos^2 - 1 and cos 3x = 4cos^3 x - 3 cos x. You get 4y^3 + 2y^2 - 3y - 1 = 0 where y = cos x. This factors into (y +1)(4y^2 - 2y - 1) = 0. Since y is not - 1, it's one of the roots of the quadratic (1 +- sqrt(5)) / 4. Since it's it quadrant 1 it's the positive root.

[@Thankmel8r]

Perhaps you mean cos(2x+3x) = 0

[@michaelz2270]

@@Thankmel8r 2pi/5 and 3pi/5 are complementary angles, so their cosines are negative one another.

[@242math]

can see why they call it golden, you did a great job working this one

[@ARKGAMING]

I like how every second math video on KZbin has some gold joke in it cause of φ

[@Dionisi0]

Is this a JoJo reference?

[@hubb8049]

@@Dionisi0 no

[@jackhandma1011]

Nice to see an approach using roots of unity. I first learned about the value cos(pi/5) by using triangles.

[@michaelslack8900]

I feel like a few of those steps could have been sped up by writing - 1 in terms of zeta, for example

[@mikegallegos7]

Love the journey ...

[@ReCaptchaHeinz]

Okay this is incredible. One needs to be very into cosines to actually find cos(π/5) this way. Amazing

[@miro.s]

I know other more direct ways in fields of Chebyshev or inner product or complex geometry or algebraic geometry and never thought to do it in this way. Also interesting approach.

[@TwilightBrawl59]

If 1+sqrt(5) / 2 is the golden ratio, what are the silver and the bronze ratios? 🤔

[@vaxjoaberg]

kzbin.info/www/bejne/bZ21mJiKot2hosk

[@singularpoints3982]

This seems like a joke, but there is a silver and bronze ratio. In fact, there are an infinite number of "metallic ratios": en.wikipedia.org/wiki/Metallic_mean

[@robertveith6383]

That is *not* the Golden Ratio. [1 + sqrt(5)]/2 is.

[@SB_3.1415]

got out of hand pretty quick XD

[@harshparashar9210]

Great Sir👍👍 Keep it up🙏🙏

[@s1ng23m4n]

Amazing, thanx.

[@ameerunbegum7525]

That's Gold.

[@videolome]

Thank you. Nice problem. Here is a solution. Let z=exp(i pi/5). Then z^5+1=0. Clearly z is not -1. Dividing by (z+1) we get z^4-z^3+z^2-z+1=0. If we let w=z+1/z, then w^2-w-1=0. We conclude 2 cos(pi/5)=w= phi.

[@davidgillies620]

The proof using similar isosceles triangles with an apex angle of pi/5 is also quite instructive. Fewer cyclotomic polynomials though.

[@unonovezero]

And that's a good pl...

[@user-wp1uw8fv6y]

My trial: if θ = π/5, then 4θ = π - θ so that cos(4θ) = -cos(θ). Apply double angle formula twice: cos(4θ) = 2cos^{2}(2θ) - 1 = 2(2cos^{2}(θ) - 1)^{2} - 1 = 8cos^{4}(θ) - 8cos^{2}(θ) + 1 Let t = cos(π/5), then 8t^{4} - 8t^{2} + 1 = -t ⇔ 8t^{4} - 8t^{2} + t + 1 = 0. Factor LHS ⇒ (t + 1)(2t - 1)(4t^{2} - 2t - 1) = 0. Since cos(π/5) does not equal to -1 or 1/2, we get 4t^{2} - 2t - 1 = 0. Therefore, the positive value of t is (1+sqrt(5))/4.

[@pedroalonso7606]

An alternative would be using Moivre's formula for cos(5x)

[@malawigw]

Polynomials are such powerful!

[@SaveSoilSaveSoil]

Nice! But I don't readily see whether this method can be extended to other angles (pi/7, pi/9, etc) and how.

[@silvermica]

Oh, wow. That was cool.

[@luna9200]

Nice using complex numbers to show this value!

[@Bestape]

Popculture is into the 5th dimension these days. I like this definition of it. Thanks!

[@clintongryke6887]

...bringing the beauty of Mathematics to everyone.

[@elrichardo1337]

there's also the classic proof using the regular pentagon or 36-72-72 triangle

[@tomholroyd7519]

I love all the multiplying by 1 and adding of 0

[@donaldasayers]

There is an easy geometrical proof involving the isosceles triangle with angles pi/5, pi/5 and pi/10.

[@goodplacetostart9099]

Good Place To Start Trivial at 0:21

[@elaadt]

Crazy stuff

[@einsteingonzalez4336]

What about cos(π/18)? Also, for finding the closed-form expression for cos(2π/n), what if n is prime?

[@arkapratimdebnath]

This could also be done without such complication

[@kenbrady119]

I followed you through the introduction of zeta, but then Z came along and I couldn't figure out what you were taking about. Please slow down.

[@galveston8929]

Overkill. Start from sin(2pi/5)=sin(3pi/5) . Expand both sides, divide by sin(pi/5). You'll get the desiered quadratic eq.

[@aaryunik]

The thumbnail is literally gold!

[@vishalmishra3046]

If θ is 18° or -54°, then 2θ and 3θ add up to 5θ = 90° or -270°, therefore sin 2θ is equal to cos 3θ. 2 sinT cosT = sin 2T = cos 3T = 4 cos^3 T - 3 cosT = (1 - 4 sin^2 T) cosT. Therefore, 4 sin^2 T + 2 sinT - 1 = 0. So, sinT = sin(18° and -54°) = (-1 ± √5)/4. Therefore, cos(π/5 and 2π/5) = cos(36° and 72°) = sin(54° and 18°) = (√5 ± 1)/4. *So much simpler* . Right ?

[@harshparashar9210]

For IIT JEE we just remember value nothing else😉😉😉

[@stephanel2324]

Wow!

[@andreweberlein1509]

Now I'm curious if this approach can be applied to other rational multiples of π.....if so, then the trig functions of rational multiples of π would all be algebraic numbers, roots of some polynomial?

[@demenion3521]

that is true. take θ=2πm/n, then cos(θ)=Re(exp(iθ)). notice that exp(iθ)^n=1 and therefore exp(iθ) is a root of z^n-1, no matter the value of m. therefore exp(iθ) is an algebraic complex number and it is easy to show that those numbers can always be written in cartesian form with algebraic real and imaginary part. therefore cos(θ), sin(θ), tan(θ) etc are all algebraic

[@doontz111]

Can any value of cosine be calculated like this (like, giving an exact solution, not with a taylor series)?

[@miro.s]

Only cos kpi/n where n are special prime numbers, otherwise with coincidence to Galois Theory we are not able to find it in a symbolic way for n>5

[@vaibhavcm7503]

I just saw like 5 seconds of video and grabbed my calculator and put the value and got 0.809016...., I flashed to me it's somehow related to golden ratio, just multiplied it by 2 and bam!!!!!

[@micheltenvoorde]

Very nice! Maybe cos(pi/7) next? 😄

[@David-km2ie]

Haha I looked it up and claimed proven impossible to express using rational numbers + - * / and root extractions

[@DavidSavinainen]

@@David-km2ie But you can express it with a closed form, it’s (Spoiler alert if ever Michael makes a video on this) (1+(98/(-1+3i√3))^(1/3)+(7(-1+3i√3)/2)^(1/3))/6

[@synaestheziac]

There’s gotta be a way to prove this from a regular pentagon, right?

[@egillandersson1780]

Haha ! The geometric way is easier, but I like this walk by the complex world !

[@manuela1803]

A good rational aproximation is: cos(π/5)^2+sin(π/5)^2=1 sin(x)~x (when x~0) sin(π/5)~π/5 cos(π/5)^2~(25-π^2)/25 π~3 cos(π/5)^2~16/25 cos(π/5)~4/5 cos(π/5) is positive: 0

[@mcwulf25]

Surely it's easier to use double angle rules to calculate cos(2pi/5) and cos(3pi/5), which is just -cos(2pi/5). I worked this out while trying to get to sleep one night!!!!!!!?

[@txikitofandango]

I tried to do cos(π/5) = cos(π/6 + π/36 + π/216 ...) Use the cosine addition formula and others ??? Profit: golden ratio

[@AzureSky6612]

If you look at a golden triangle the angle is found in it.

[@mastertoru5531]

Is it really that because z^10 is even therefore the negative version (conjugate) are roots as well? Isn't because the polynomial has real coefficients, therefore the conjugates are also roots of the equation?

[@demenion3521]

both is true

[@singularpoints3982]

He doesn't use conjugates, he finds that zeta^4 = -(zeta^9), which comes from evenness of the polynomial. The real coefficients tell you that zeta^4 is the complex conjugate of zeta^6 (and thus when you add them they give a real number, same with the pair zeta and zeta^9).

[@MrWarlls]

For the first step, you just need to say that you use the Euler's formula for cos.

[@schweinmachtbree1013]

yeah but it was cool though

[@MrWarlls]

@@schweinmachtbree1013 , I absolutly agree with you.

[@louisdu54]

I think more teachers should do this, it reminds you how to find it by yourself and helps understand IMO

[@br00zer31]

Tis a gold place to stop.

[@seroujghazarian6343]

sin(3pi/10)=phi/4 as well

[@replicaacliper]

How'd you obtain that factorization at 5:50?

[@juliang8676]

test its true

[@replicaacliper]

@@juliang8676 well I'd like to be able to rederive it myself, especially that last factor

[@juliang8676]

@@replicaacliper i think its a standard result, easiest way I can think of is splitting it into (x^5+1)(x^5-1) and obviously -1 is a factor of the left bracket and 1 is a factor of the right bracket, then you just have two quartic's to solve which shouldnt be too difficult

[@replicaacliper]

@@juliang8676 ahh okay that actually makes sense! Yeah I could see why the x^4+x^3+x^2+x+1 was a factor as those are the fifth roots of unity other than 1, but I didn't see that the other terms came from the x^5+1 term, nice

[@singularpoints3982]

@@replicaacliper Another way of seeing why the alternating factor shows up is from the evenness of z^10-1. You know that you get all the 5th roots of unity, and by even symmetry, you also get their negatives. So if you can get to (z+1)*(z-1)*(z^4+z^3+z^2+z+1)*(something), you can take (z^4+z^3+z^2+z+1) and send z -> -z, giving a polynomial whose roots are the negative of the fifth roots of unity (besides -1 which is already accounted for): z^4-z^3+z^2-z+1.

[@robertponga4653]

what is cos(pi/5)? Me: grabs calculator = 0.809 Michael Penn: Let's go on an adventure!

[@vishalmishra3046]

*Challenge Question* - Calculate sin or cos (2π/4294967295). *Hint* 4,294,967,295 = 2^32 - 1.

[@Etothe2iPi]

Why not just draw a regular pentagon?

[@BBQsquirrel]

I find the solution presented in this video a little bit too........ ......complex to understand

[@soranuareane]

How in the world do people figure out the factorization of z^10-1 in terms of reals.

[@kalolord]

cos(π/5) = sin(π/2-π/5) approximate sinx = x cos(π/5) = 3/10π approximate π = 3 final answer: 0.9

[@natepolidoro4565]

*C* joins the battle

[@felipelopes3171]

Old news. Much better is cosine of 2pi/17 :)

[@synaestheziac]

Aww, what happened to “C has entered the battle”?

[@CM63_France]

Hi, For fun: 1 "so let's may be go ahead and", 1 "so let's go ahead and do that".

[@808spvrk]

دمت گرم

[@JamesLee-pk8oj]

just use a triagle with angle 36, 72, 72 and Angle bisector theorem。

[@jxp3543]

anyone have absolutely no clue what is going on but thinks maths looks pretty so kept watching? 😅😅😅

[@denis0dns]

As a Math-Student I didn't understand anything xD

[@ReCaptchaHeinz]

Nobody is talking about the screams in 2.25? :(

[@ReCaptchaHeinz]

2:25

[@haierpad5669]

Thanks god we have calculators (yes, i'm engineer XD)

[@CTJ2619]

the background noises are a bit of a distraction

[@rudychan2003]

i don't like that euler's formula; complex and confusing and complicated! 😕 cos (π/5) is cos 36° = sin 54° !! = (√5 + 1) ÷ 4. i already know this is 'golden ratio' ÷ 2! From trigonometry equation: sin 3x - sin x = 2.cos2x.sin x = 2.cos2x.(sin2x/2.cos x) = sin 4x / 2. cos x Much-much Easier, much easy to understand than euler's imaginary root: √-1 ??? (didnot exist!) 😜

[@robertveith6383]

You need some missing grouping symbols. For instance, for the last fraction, it needs to be sin(4x)/(2cos(x)) or sin(4x)/[2cos(x)] because of the Order of Operations.

[@JW-ss8es]

This is overdone and does not give insight why golden ratio appears. considering an isosceles triangle with top angle pi/5, make an angle bisector line of one bottom angle, you got a similar triangle with the original triangle. the length of bottom line equals the length of angle bisector line. using similar ratio, we immediately got the golden ratio equation equation.

[@oliverhees4076]

blackpenredpen has a geometric way of finding the value if you want to see the geometric way of doing this; however, I think there's a certain beauty in seeing things both geometrically and algebraically. It's nice to see two completely different approaches to the same problem.

[@jonathanbonicel1654]

Or you can just use a calculator and get the result in seconds

[@RainBoxRed]

Kids upstairs?

[@kujmous]

/swear words/

[@phyarth8082]

Dah, we know we watch Da Vinci Code Iliuminati Galileo society, and satanistic Pentagram star in reality is not satanistic is just usage of golden ratio usage in bad hands becomes pentagram satanistic. Golden ratio is a bit overrated math constant. Phi can not compete with pi() or prime numbers, but I think it is to much overrated.

[@ZainAlAazizi]

AlBeiruni proved this geometrically.