A famous trig equation.

We look at a nice trigonometric identity.
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[@MizardXYT]

tan(2𝜋/3)+4sin(𝜋/3) = √3 tan(4𝜋/7)+4sin(𝜋/7) = √7 tan(4𝜋/11)+4sin(𝜋/11) = √11

[@goodplacetostop2973]

13:05 10+1 = 11 18:16

[@muckchorris9745]

Waiting for (michael penn)^300 xD

[@vedants.vispute77]

Video 3 mins ago and this one 5 mins ago. Explain.

[@goodplacetostop2973]

@@muckchorris9745 As a matter of fact, blackpenredpen suggested me a great idea for another video. I’m open to more suggestions 😀

[@goodplacetostop2973]

@@vedants.vispute77 Video goes up at 8AM EST. So I just wait for that exact time and posts the comment. Really that simple. But I’ve noticed the bell notification goes up 2-3 minutes after the actual upload of the video.

[@vedants.vispute77]

@@goodplacetostop2973 yeah ur right

[@skysong8140]

(A-B) can be solved by using the relation (A-B)^2 = (A+B)^2 - 4AB Would end up with two solutions but the negative one can be eliminated using the same logic as in the video.

[@mathsandsciencechannel]

Thank you sir. you have encouraged me to start my maths and science channel. lets learn more and sub to be notified. thanks

[@monikaherath7505]

How do people come up with identities like this? Do they just stumble upon it while solving a problem or just by messing around? It's an extremely unintuitive identity with an astoundingly elegant proof using complex roots of unity. Honestly my mind is completely blown knowing that you could have precisely calculated this identity hundreds of years ago using just a pencil and paper. Question: are the trigonometric components in this identity transcendental while its sum is algebraic?

[@kingkartabyo6206]

Regarding your last comment, no. Both terms are in fact algebraic. Say s=sin 2pi/11, then using formula for sin(11x) in terms of sin x, we get: 0=sin 2pi=a polynomial in s.

[@monikaherath7505]

@@kingkartabyo6206 Nice, thank you.

[@carlosgiovanardi8197]

If you are interested in these kind of expressions, see arxiv.org/pdf/0709.3755.pdf and math.stackexchange.com/questions/45144 Piece of cake!

[@karangupta1825]

It came on my KZbin home screen as a suggestion and couldn't resist 'not to click on it'

[@dhwyll]

It took me a while to figure out the factorization at 2:00. e^(i 20/11 pi) = e^(i pi (22/11 - 2/11)) = e^(i pi 22/11 - i pi 2/11) = e^(i pi 22/11) * e^(-i pi 2/11) = e^(i 2 pi) * e^(-i 2 pi/11)

[@victorclaytonbarnett2959]

Thank you

[@WestOfEarth]

Thanks! Stumped me.

[@hsjkdsgd]

I didn't know I had picked a tough problem to solve on my own.

[@manucitomx]

And I thought this was going to be a walk in the park. I’ve been had again.

[@laszloliptak611]

At 9:57 you want to refer to the geometric series summation formula: 1+x+x^2+...+x^n=(1-x^(n+1))/(1-x) Substituting x = -z^8 and n=9 we get 1-z^8+z^16- ... -z^72 = (1 -(-z^72)(-z^8))/(1-(-z^8))=(1-z^80)/(1+z^8). Same formula is used at 14:10.

[@Cypekeh]

your editing is subtly amazing

[@MathMan271]

wow great work Michael

[@n8cantor]

Is there any deeper reason as to why the cross terms in AB add up to 2(A+B)? I don't see any pattern in the exponents and the cross terms in other groupings don't seem to add up to anything special. For instance, in the 2nd tool equation, the z terms also come in inverse pairs (i.e. z^7 and z^4 are in separate groups and z^7*z^4 = 1) but if you multiply them out, the cross terms are a mess.

[@ssvemuri]

I am mesmerized by this result. Summing up two trig expressions can yield sqrt(11). VOW. just VOW. Using the same technique derive a closed form for (x-cube + x-squared - 2x-1) where x = 2cos(2pi/7) Hint: the answer is not an irrational number.

[@AlephThree]

Brilliant!

[@pattemsarvesh5002]

Great video 🙌

[@ThePhysicsMathsWizard]

wow, nice identity !!

[@lovingphysics5865]

kzbin.info/www/bejne/gnXJemt_pcyJn8k

[@alaingoudemand9818]

@@lovingphysics5865 v

[@user-wp1uw8fv6y]

I tried for the answer by using trigonometric identities. First, define a constant "θ = π/11". Let A = tan(3π/11)+4sin(2π/11) = tan(3θ)+4sin(2θ). Multiply both sides by cos(3θ) ⇒ Acos(3θ) = sin(3θ)+4sin(2θ)cos(3θ) = sin(3θ)+2sin(5θ)-2sin(θ). Then square both sides ⇒ (Acos(3θ))^{2} = [sin(3θ)+2sin(5θ)-2sinθ]^{2} and expand right hand side ⇒ RHS = sin^{2}(3θ)+4sin^{2}(5θ)+4sin^{2}(θ) + 4sin(3θ)sin(5θ)-8sin(5θ)sin(θ)-4sin(θ)sin(3θ) Use two identities: "sin^{2}(α) = (1-cos(2α))/2" and "sin(α)sin(β) = (cos(α-β)-cos(α+β))/2". Then, RHS = (1-cos(6θ))/2+2(1-cos(10θ))+2(1-cos(2θ)) + 2(cos(2θ)-cos(8θ))+4(cos(6θ)-cos(4θ))+2(cos(4θ)-sin(2θ)) = 9/2 + (7/2)cos(6θ) -2(cos(2θ)+cos(4θ)+cos(8θ)+cos(10θ)) Besides, bring another identity: "Σ{0≤k≤n-1}cos(2πk/n)=0". Substitute "n=11" and "θ = π/11" ⇒ 1+cos(2θ)+cos(4θ)+cos(6θ)+…+cos(20θ)=0 And since "cos(α) = cos(2π-α) = cos(22θ-α)", it can be written as "1+2(cos(2θ)+cos(4θ)+cos(6θ)+cos(8θ)+cos(10θ))=0". Therefore, (Acos(3θ))^{2} = RHS = 11/2 + (11/2)cos(6θ) = 11cos^{2}(3θ), and A^{2}=11 Since A>0, the answer is A = tan(3π/11)+4sin(2π/11) = sqrt(11).

[@nawusayipsunam1643]

Yes. It is superb.

[@markregev1651]

That's the same idea as in IMO 1963? show that cos(pi/7)-cos(2pi/7)+cos(3pi/7)=1/2 Oh and another suggestion for a geometry video: Fuss' theorem about bicentric quadrilaterals

[@mathsandsciencechannel]

Thank you sir. you have encouraged me to start my maths and science channel. lets learn more and sub to be notified. thanks

[@elkincampos3804]

cos(π/7)=-cos(6*π/7) and cos(3*π/7)=-cos(4*π/7). Now, multiply the equation by 2. Take w=e^i(2*π/7). And use that 2*cos(x)=e^(i* x)+e^(-i*x). It follows the equality because w+w^2+w^3+w^4+w^5+w^6=-1.

[@vbcool83]

Didn't quite get the step at 10:02 - how does the 1-(z^8)^10 directly factor out 1+z^8? 1-(z^8)^10 = (1-((z^8)^5)^2) = (1-(z^8)^5)(1+(z^8)^5) = (1-z^8)(1+z^8+z^16+z^24+z^32)(1+z^8)(1-z^8+z^16-z^24+z^32) That's how 1+z^8 gets factored out.

[@mihaipuiu6231]

Very, very interesting equation. Good for you Penn. But very hard equation.

[@user-sp3pw6qs2r]

Also tan(π/11)+4*sin(3π/11)=√11

[@sundeep0207]

Great approach. Is there a method without using complex numbers?

[@get2113]

Very likely, but that route would entail searching through many tables of trig identities, some of might even have been proved via Euler. See your point but shortcuts are often needed to make progress.

[@sundeep0207]

@@get2113 Yes, an approach without using the complex numbers might be too complex 😁

[@dulcedeleche000]

Yes its possible you can find the solution of this question in IIT JEE Amit M Agrrwalas book. And when I did this question in 2008 publication KC Sinha trigonometry and there was no solution in this book. And yes the solution was purely trigonometric no complex numbers concept used.

[@georgeebberson6]

At 12:20, what happens to the 2 out front of the first "tool" expression? The i gets factored out, but rather than putting 2*z^10 inside the brackets (as well as 2*z) there's just z^10 and z respectively?

[@dimitrispapadakis2122]

You have 2*z^10 from the first tool, and - z^10 from the second tool, so you get z^10.

[@georgeebberson6]

@@dimitrispapadakis2122 ohhhhh I'm an idiot thanks 😂

[@mathsandsciencechannel]

Thank you sir. you have encouraged me to start my maths and science channel. lets learn more and sub to be notified. thanks

[@guill3978]

Is 1/ln(2) + 1/ln(3) a transcendental number?

[@nerdiconium1365]

From knowing e is transcendental, you can figure out each addend is at least irrational, but this feels like it might be in the same vein as “We don’t know if pi+e is rational, irrational, or transcendental” types of questions. Also, the reciprocals make it harder because it effectively switches the argument and the base of the logarithm.

[@jaa2174]

I get lost at 10:00. I believe you, but I would love to see where that result is coming from.

[@MathrillSohamJoshi]

I think he first split 1-z^80 as (1+z^40)(1-z^40) using the 'difference of squares' formula. Because(40=8(5)) we can factorise (1+z^40) such that one of its factors is the denominator, simplifying it to z^32-z^24+ z^16-z^8+1. Finally, multiplying the obtained expression by 1-z^40 gives us 1-z^8+z^16-z^24+z^32-z^40+z^48- z^56+z^64-z^72 Hope this helps :)

[@jaa2174]

@@MathrillSohamJoshi Thanks!

[@MathrillSohamJoshi]

@@jaa2174Welcome :)

[@l0new0lf000]

Nice hoodie

[@Zxv975]

I'm confused, what happened to the 1+z^8 denominator around 10:30?

[@zhaochengliu9970]

I think that's the geometric series summation, but in an inverse manner. The ratio of the series is z.

[@Zxv975]

@@zhaochengliu9970 ohhhhhh I see it now! It's an alternating geometric sum, got it. Thanks so much for that. I was thinking about it completely wrong because Michael mentioned about generalised difference of squares and I just couldn't see how that related. Also, did you mean to say the ratio is z^8 (or rather, I guess -z^8).

[@zhaochengliu9970]

@@Zxv975 Sure haha! I was struggling on that point too XD. It's kinda confusing. And yep, you are right, the ratio is actually -z^8.

[@claudioproy9489]

Can you generalize this for any prime in denominator?

[@BatmanPooping]

Where was it used that 11 is prime?

[@synaestheziac]

​@@BatmanPooping I think it has to do with the way in which certain exponents are inverses of each other mod 11 and the way in which the cross terms in the product of A and B reproduce the original exponents. I'm just learning about primitive roots mod p in Michael's number theory videos, so I'm not entirely sure what I'm talking about, but it seems like there are number-theoretic results hovering in the background herer.

[@BatmanPooping]

@@synaestheziac No. It has nothing to do with being prime. For any natural n, e^(2 pi i (n-1)/n)= e^(-2 pi i/n).

[@rbdgr8370]

What was the motivation behind using 11th root of unity for solving sin(2pi/11) and 22nd root of unity for solving tan(3pi/11)?Was it the step involved at 7:56 or there is a certain algorithm for such problems??

[@synaestheziac]

Seriously. This derivation had a whole bunch of steps I never would have thought of.

[@rbdgr8370]

@@synaestheziac after 7:56 it was comparatively easier as I have solved many ques on those concepts but before that I never would've guessed taking 22nd root of unity for solving tan(3π/11)

[@synaestheziac]

@@rbdgr8370 I got befuddled as soon as he constructed z to the 88th power, and then rewrote it as z to the 8th to the 10th - can you help me see how one would decide to make those moves?

[@ernest3109]

15:25 whaaat? why is the rest of the product equal to 2(a+b)?

[@skillerror951]

Фейспалм кирюш, все же очевидно

[@ernest3109]

@@skillerror951 мне не очевидно объясни

[@jounik]

It's just the pairwise sums of numbers 1-10 mod 11 where you pick one summand from A={10,8,7,6,2} and the other one from B={1,3,4,5,9}. Once you've set aside the 5 pairs that sum to 0, you get two each of the other values, i.e. two sets of A+B.

[@ernest3109]

@@jounik Still unclear. I guess I have to write down all remaining 20 parts to see the pattern. Probably I'm stupid.

[@jounik]

@@ernest3109 Not stupid and it's not exactly obvious, but it _is_ easy to verify which is all Michael said. I suppose there are some cyclic properties that would help see it directly but it's easier to just write them out. I know I did.

[@megacarlosloki]

Try OBMU second fase problem 5 :)

[@guibaroleo]

esse daí é brabo, né não

[@lawlietl1370]

De qual ano?

[@lawlietl1370]

@@guibaroleo nada melhor do que mais um BR por aqui

[@shailendragupta7517]

This can be done without using complex number

[@amberdeshbhratar9342]

How to prove 1/cos(π/7) + 2cos(π/7)/cos(2π/7) =4

[@sundeep0207]

Was able to transform this problem to the below one. Proving one is equivalent to proving the other. Show that Cos(pi/7)+Cos(3pi/7)+cos(5pi/7)=0.5 Got stuck here. Any thoughts on this?

[@sundeep0207]

Was able to arrive at the proof. Please find it below. Part 1 of the proof Show that the given problem is equivalent to proving Cos(π/7) + Cos(3π/7) + Cos(5π/7) = 0.5 Consider X = 1/Cos(π/7) + 2Cos(π/7)/Cos(2π/7) - 4. Our original problem is proving that X=0. Multiplying both sides by Cos(π/7)*Cos(2π/7), we get Cos(π/7)*Cos(2π/7)*X = Cos(2π/7) + 2*Cos(π/7)*Cos(π/7) - 4*Cos(π/7)*Cos(2π/7) For the purpose of brevity, I will replace the LHS of the equation with Y. We can see that X=0 if and only if Y=0 . Hence, the given problem is same as showing Y=0. Y = Cos(2π/7) + 2*Cos(π/7)*Cos(π/7) - 4*Cos(π/7)*Cos(2π/7) Y = Cos(2π/7) + 2*Cos(π/7)*Cos(π/7) - 2*(2*Cos(π/7)*Cos(2π/7)) Y = Cos(2π/7) + 2*Cos(π/7)*Cos(π/7) - 2*(Cos(3π/7)+Cos(π/7)) Y = Cos(2π/7) + Cos(2π/7) + 1 - 2*(Cos(3π/7)+Cos(π/7)) Y = 2*Cos(2π/7) + 1 - 2*(Cos(3π/7)+Cos(π/7)) Y = -2*Cos(5π/7) + 1 - 2*(Cos(3π/7)+Cos(π/7)) Therefore, Y = 1 - 2*(Cos(π/7) + Cos(3π/7) + Cos(5π/7)) From this we can observe that Showing X=0 is same as showing Cos(π/7) + Cos(3π/7) + Cos(5π/7) = 0.5 Part 2 of the proof Prove that Cos(π/7) + Cos(3π/7) + Cos(5π/7) = 0.5 Consider the equation Z^7 + 1 = 0 The roots of this equation are β, β^3, β^5, β^7, β^9, β^11 and β^13 , where β = e^(iπ)/7 We can see that the sum of the roots of this equation is 0. i.e., β + β^3 + β^5 + β^7 + β^9 + β^11 + β^13 = 0 Considering the real parts on both sides, we get Cos(π/7) + Cos(3π/7) + Cos(5π/7) + Cos(7π/7) + Cos(9π/7) + Cos(11π/7) + Cos(13π/7) = 0 Cos(π/7) + Cos(3π/7) + Cos(5π/7) + (-1) + Cos(9π/7) + Cos(11π/7) + Cos(13π/7) = 0 Cos(π/7) + Cos(3π/7) + Cos(5π/7) + Cos(9π/7) + Cos(11π/7) + Cos(13π/7) = 1 Cos(π/7) + Cos(3π/7) + Cos(5π/7) + Cos(5π/7) + Cos(3π/7) + Cos(π/7) = 1 2*(Cos(π/7) + Cos(3π/7) + Cos(5π/7)) = 1 Therefore, Cos(π/7) + Cos(3π/7) + Cos(5π/7) = 0.5 Please let me know if something is not clear.

[@amberdeshbhratar9342]

thank you 😁

[@yashagrahari]

It would seem more magical if you don't see down to the floor.

[@TrueBagPipeRock]

k

[@eusterich3035]

Yeah kinda like memorise shit

[@eusterich3035]

All iit jee aspirant just rote learn it 😂😂😂

[@mathsandsciencechannel]

Thank you sir. you have encouraged me to start my maths and science channel. lets learn more and sub to be notified. thanks

[@vedants.vispute77]

All people here listen, Prove that the probability of getting a like on this video is zero. Disproove this statement.

[@Thaplayer1209]

There are 2 dislikes now what have you done

[@goodplacetostop2973]

Asking people on the internet to not do something is the best way to provoke that thing 😛

[@vedants.vispute77]

@@drsonaligupta75 I saved myself.