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        riddles >> hard >> Trig identity involving pi/11
        
(Message started by: perash on Jan 1^st, 2007, 1:26pm)

Title: Trig identity involving pi/11
Post by perash on Jan 1^st, 2007, 1:26pm

prove that

tan(3pi\11)+4sin(2pi\11)=sqrt 11

Title: Re: prove
Post by towr on Jan 1^st, 2007, 1:52pm

Are the \ supposed to be / (divide)?

Title: Re: prove
Post by Michael_Dagg on Jan 1^st, 2007, 2:54pm

\ means / here.

This is well-known. Not too hard either. If you're not careful
you'll quickly get a messy expression.

Title: Re: prove
Post by JP05 on Jan 2^nd, 2007, 4:03pm

I working on this problem for a while today and came up with nothing in the way of showing the left side = the right side other than doing it numerically. Symbolically, Maple does not seem to be able to do much with the left side except make a humongous formula. It is a little smaller when you convert the trig to exp. But when I subtract the right side from the left side Maple does say it is = to 0 but only after I specify the digits of precision to be 300 or more. I know in come cases Maple does not know how to exactly deal with trig formulas if they do not fit into a certain identity scheme but most of the time I have alway seen it do very good with exp forms.

Title: Re: prove
Post by Michael_Dagg on Jan 2^nd, 2007, 4:30pm

Experimenting with this problem using Maple can be a fun thing
and if done right it will reveal some interesting relationships
and may give you a clue as how to do it analytically. But you
have to help it some in order to do this.

You are right that the complex exponential form is more compact
than the trig form.

What you need to do is define F = tan(3*x)+4*sin(2*x) where x = 2pi/22.
When you use the expand() command it will convert this to a rational
function in sin(x) and cos(x), and then Maples's convert(...,exp)
command will rewrite it as a rational function in exp(x*I).

You may then substitute the exp(x*I) factors with y, i.e. with
subs(exp(x*I)=y,%).

Now, y is an element of a cyclotomic field, that is, all the
computations are really taking place in Q[y] = Q[X]/phi(X) where
phi is the minimal polynomial of y, a primitive 22nd root of unity.

Thus phi is the 22nd cyclotomic polynomial, which you can get in
Maple by loading the number theory package with the command
numtheory[cyclotomic](22,X) . This will get you that polynomial.

So you can get your expression in the simplest possible way with
simplify(%, {phi}) . Unfortunately Maple is kind of stupid about
this: it tries to simplify numerator and denominator, but fails to
see that the whole rational expression can be written as a polynomial
in y.

You can help it out by using, say, the Euclidean algorithm to find
polynomials u(y) and v(y) so that u(y) phi(y) + v(y) d(y) = 1,
where d(y) is the denominator of the rational function; since phi
vanishes at y, this means 1/d(y) = v(y), a polynomial; then you
can use the simplify() command as above.

So with Maple you get you as far as expressing the left side as a
polynomial in y = exp(2 pi i / 22). It's a little bit of number theory
to know that within the cyclotomic field Q[y] you can also find the
real-quadratic field Q[sqrt(11)].

To know that your polynomial is exactly sqrt(11) is not the way to
proceed; instead check that it is equal to + or - sqrt(11) by showing
that its square equals 11. That, in turn, can be done with less
simplification than the previous paragraph suggested.

This should work:

F:=tan(3*x) + 4*sin(2*x);
convert(expand(F),exp);
subs(exp(x*I)=y,%);
numer(factor(%^2-11));
%/numtheory[cyclotomic](22,y);

The fact that the final answer has no denominator means the previous
one is a multiple of phi, and so vanishes when y is a primitive 22nd
root of unity. THerefore for this value of x, F^2 = 11, i.e. F is either
sqrt(11) or its negative.

But hey, to a number theorist, sqrt(11) just means a number whose square
is 11, without prejudicing one of these numbers of its Galois conjugates!

Title: Re: prove
Post by THUDandBLUNDER on Jan 2^nd, 2007, 4:41pm

on 01/02/07 at 16:03:34, JP05 wrote:

I working on this problem for a while today and came up with nothing in the way of showing the left side = the right side other than doing it numerically.

Mathworld has a few things to say about pi/11
http://mathworld.wolfram.com/TrigonometryAnglesPi11.html
There is a typo in Equation 9, the 6th typo I have spotted and submitted to Eric. 8)

Title: Re: prove
Post by Icarus on Jan 2^nd, 2007, 5:05pm

Would that be the use of "cot" instead of "tan"? I notice that there are two cotangent formulas, but none for tangent.

Title: Re: prove
Post by JP05 on Jan 2^nd, 2007, 5:43pm

on 01/02/07 at 17:05:01, Icarus wrote:

Would that be the use of "cot" instead of "tan"? I notice that there are two cotangent formulas, but none for tangent.

Based on the polynomial that the code above outputs that seems to be the case but what I don't understand is what do the subscripts mean on the (...)? First I thought it was the order of the polynomials but that can't be it unless 7, 8 and 9 are errors.

Title: Re: prove
Post by THUDandBLUNDER on Jan 2^nd, 2007, 6:27pm

on 01/02/07 at 17:05:01, Icarus wrote:

Would that be the use of "cot" instead of "tan"? I notice that there are two cotangent formulas, but none for tangent.

Well, they seem to be in alphabetical order, as are corresponding formulae on other pages.

on 01/02/07 at 17:43:08, JP05 wrote:

...but what I don't understand is what do the subscripts mean on the (...)? First I thought it was the order of the polynomials but that can't be it unless 7, 8 and 9 are errors.

From http://mathworld.wolfram.com/TrigonometryAnglesPi7.html
"Letting (P(x))_n denote the nth root of the polynomial P(x) using the ordering of Mathematica's Root function..."

Title: Re: prove
Post by Barukh on Jan 3^rd, 2007, 5:53am

on 01/02/07 at 16:30:42, Michael_Dagg wrote:

Is this the simplest way to solve this innocently looking identity?

::)

Title: Re: prove
Post by Michael_Dagg on Jan 4^th, 2007, 8:20pm

I think so. At least this is the way I approach these kind of things when
something ends up a mixed-trig relation with these kinds of angles.

You could try to use some of the same kind of mixed-trig identities
given in that link T&B provided to produce the desired result but then
you'd simply be using a variant of what you are trying to prove to begin
with.

Of course de Moivre theorem is key here. In this case it is figuring out
what n-th root of unity you want, which is not hard, and by the time you
compute one multiple of 11 you would have already reached for more paper.

Title: Re: prove
Post by JP05 on Jan 8^th, 2007, 7:30pm

Something that I thought would be informative to everyone who might have been interested in this problem is that before I was able to see what Michael Dagg has posted here I sent this problem to a mathematician at my university here. I won't say who and a I won't say where but he answered my email tonight and this is what he said

I gave your problem several hours of serious cerebration,but came up
empty.I've shown it to other colleagues as well. The appearance of the
naked number square-root of 11 is "unnatural" in a trigonometric iden-
tity (although,as you must have confirmed yourself,your proposed ident-
ity is numerically supported by a TI-83). What was your source for the
problem?
I have a hunch that Archimedes' method for approximating the circum-
ference of a circle via regular inscribed polygons might be relevant,
as that required him to compute (by geometry!) the sines and cosines
of pi/n for integers n,and your problem comes down to cos(pi/11) being
a root of a certain eighth-degree polynomial. That said,there may never-
theless be some insightful geometric approach with minimal computation--
"Tis a consummation devoutly to be wished for",as Shakespeare put it in
Hamlet's famous soliloquy.

Now the professor I asked is no newbie. I guess he is about 65 years old and does topology and is distinguished. I am guessing that his mention of cos(pi/11) being a root of a "certain 8-degree polynomial" is this exactly the same thing that Dagg said about cyclotomic polynomial of the full formula. I guess the point I am trying to make here is that how well known is the identity? Dagg said it is well known, apparently on Mathworld other people know about it, so how might a Ph.D. in topology not know this formula? Or should his canny Shakespeare quote just some way of tell me to "go fish"?

So, how many of your other people did not know this formula? I sure never saw it before.

Title: Re: prove
Post by balakrishnan on Jan 8^th, 2007, 7:40pm

first let us compute
Cos(0)+Cos(2pi/11)+Cos(4pi/11)+..Cos(10pi/11)
It is real of
Sum[exp(i*2*pi*k/11)]for k=0 to 5
which is real of [exp(12pii/11)-1]/[exp(2pi.i/11)-1]
which is real of
1/[1+exp(pi.i/11)]
which is
real of (1+exp(-i.pi/11))/(2+2Cos(pi/11))
which is
[1+cos(pi/11)]/[2(1+Cos(pi/11)]=1/2

Cos(0)+Cos(2pi/11)+Cos(4pi/11)+..Cos(10pi/11)=1/2
or
Cos(2pi/11)+Cos(4pi/11)+..Cos(10pi/11)=-1/2
I call pi/11 as z
So
2[Cos(4z)-Cos(8z)]=2[Cos(2z)+Cos(10z)]+1+2Cos(6z)+4Cos(4z)
which gives
4Sin(6z)Sin(2z)=4Cos(4z) Cos(6z)+1+2Cos(6z)+4Cos(4z)
or
3[1+Cos(6z)][1+Cos(4z)]=1+[1-Cos(4z)][1+Cos(6z)]+4Sin(6z).Sin(2z)
or
12(Cos(3z))^2.(Cos(2z))^2=1+4[Sin(2z)]^2.[Cos(3z)]^2+8.Sin(3z).Sin(2z).Cos(3z)
or
12.[Cos(3z)]^2=1+16[Sin(2z)]^2 [Cos(3z)]^2+8[Sin(3z)][Sin(2z)][Cos(3z)]
or
11.[Cos(3z)]^2=[Sin(3z)]^2+16[Sin(2z)]^2 [Cos(3z)]^2+8[Sin(3z)][Sin(2z)][Cos(3z)]
or
11.[Cos(3z)]^2=[Sin(3z)+4Sin(2z)Cos(3z)]^2
or
[Tan(3z)+4.Sin(2z)]^2=11
or
[Tan(3z)+4.Sin(2z)]=sqrt(11)..It is the positive sqroot because 3z<pi/2 and 2z<pi
So tan() and Sin() are positive

Title: Re: prove
Post by Icarus on Jan 8^th, 2007, 8:32pm

Yeesh. Where did you find that, balakrishnan?

Title: Re: prove
Post by balakrishnan on Jan 8^th, 2007, 8:48pm

on 01/08/07 at 20:32:57, Icarus wrote:

Yeesh. Where did you find that, balakrishnan?

In my head :D

Title: Re: prove
Post by Michael_Dagg on Jan 8^th, 2007, 9:01pm

Kudos!

That will do it. Those are the REFLECTIONS in this 22nd cyclotomic field!

Hence, (tan(3*x)+4sin(2*x))^2=11.

Title: Re: prove
Post by Barukh on Jan 9^th, 2007, 3:41am

That's much better, balakrishnan! :D :D

And IMHO doesn't require any recourse to cyclotomic fields. ;)

Title: Re: prove
Post by pex on Jan 9^th, 2007, 3:51am

Not being too much of a mathematical genius, I prefer a simple polynomial approach (thanks to Michael Dagg for pointing me in the right direction). I attach my solution, which I wrote a couple of days ago.

Title: Re: prove
Post by JP05 on Jan 9^th, 2007, 3:27pm

Those are both neat solutions. I guess the thing here is about the learning experience this forum has given us. I suppose that if some one just banged out the trig solution in this forum then we would not have learned as much. At least not for me. So, last night I sent out another email to a professor at Harvard and this his answer below. I am still finding this surprising but he dos give a solution.

"
I looked at the problem of verifying that identity and soon realized
that it becomes quite
horrendous if you express everything in terms of cos(pi/11) and
sin(pi/11) and then square
both sides. Since only even powers of these sines and cosines appear,
you can use the
Pythagorean Identity over and over again to finally come up with a
polynomial of degree
10 in cos(pi/11) with large, complicated coefficients. Verifying this
identity would be a
complicated matter but, in principle, it could be done.

A much easier way is to introduce the following complex number: z =
cos(pi/11) + i sin(pi/11),
which is a "primitive 22nd root of unity". Then sin(3pi/11) = (z^3 -
z^(-3))/2i , cos(3pi/11) =
(z^3 + z^(-3))/2, and sin(2pi/11) = (z^2-z^(-2))/2i. If you make these
substitutions, clear the
denominator by multiplying both sides by (z^3 + z^(-3)), then square
both sides and simplify
(this takes a bit of patience and perseverance), you will get:

4(z^10 + z^(-10)) + 4(z^8 + z^( -8 )) + 4(z^6 + z^(-6)) + 4(z^4 + z^(-4))
+ 4(z^2 + z^(-2)) + 4
= 0, which is valid because w = z^2 is a "primitive 11th root of
unity", and any primitive 11th
root of unity w satisfies:

w^5 + w^(-5) + w^4 + w^(-4) + w^3 + w^(-3) + w^2 + w^(-2) + w + w^(-1)
+ 1 = 0

As you can see, the polar form of complex numbers and DeMoivre's Theorem are crucial in this verification.
"

Title: Re: prove
Post by THUDandBLUNDER on Jan 9^th, 2007, 4:02pm

From Mathworld (http://mathworld.wolfram.com/TrigonometryAnglesPi23.html)
"Trigonometric functions of pi/p for p prime have an especially complicated Galois-minimal representation. In particular, the case cos(pi/23) requires approximately 500 MB of space using the Mathematica command FunctionExpand[Cos[Pi/23]."

:o

Title: Re: prove
Post by Icarus on Jan 9^th, 2007, 4:37pm

on 01/08/07 at 20:48:44, balakrishnan wrote:

In my head :D

Good! That means you can answer my question: How did you figure it out? Obviously, you didn't start with the idea "lets add together the cosine of the even multiples, and see if it can't be shown out of that". There are many choices you've made in the direction of the calculation that seem unmotivated at the time. Of course, you must have started at the desired formula and worked the other way, but even then I'm curious at the reasons you did some things which at that point in the calculation seem arbitrary - though in the end they give the desired result.

Title: Re: prove
Post by JP05 on Jan 9^th, 2007, 4:42pm

THUDandBLUNDER this is very interesting. That is a large big gob of memory. I guess this might have to do with what your link said about a complicated Galois something. Anyone know how to explain this Galois? It turns out that this problem boils down to knowing that cos(pi/11) is the root of 8th degree polynomial. But it is not obvious because you miss it since the polynomials that are discussed have degrees that are higher. I do notice that there is a mix of [] and () in the trig solution. Mathematica requires [] and Maple (). So, I am wondering if there was a little help from Mathematica producing that trig solution. If so I would be intersted to know how you did it.

I am also guessing that a reflection must mean something like since sin(5pi/11) = sin(6pi/11) or something like this because this happens across the y axis of the circle at those two values and maybe more subtle ones that are not that obvious.

Title: Re: prove
Post by balakrishnan on Jan 9^th, 2007, 5:49pm

I backtracked.
You can see that the steps backwards are more obvious than the forward steps ;)
I continuosly aimed at reducing the squares.
Then it is a matter of identifying what
Cos(0)+Cos(2pi/11)+Cos(4pi/11)+..Cos(10pi/11)
is which is -1/2..So this becomes your first step ;D

Title: Re: prove
Post by Icarus on Jan 9^th, 2007, 6:11pm

Please let me state that I am not accusing balakrishnan of plagiarism. I am merely curious what motivated him to take certain steps. My first suspicion was that he was reproducing something he found elsewhere. I have no problem with that - a significant number of posts in the hard forum fall into that category. I have made a number myself that have more to do with memory than original thinking. But I have no reason for, or any intention of doubting his statement otherwise.

Both of my earlier posts are simply because I am curious about how this solution was arrived at. Since happily it is balakrishnan's own, he can answer my questions!:D My questions do not in any way cast doubt on the authorship, as the solution obviously was derived by somebody.

But the thing I was curious about was what guided you to get the Cos(0)+Cos(2pi/11)+Cos(4pi/11)+...+Cos(10pi/11) form. You are in there adding and subtracting terms with no apparent immediate reason, and suddenly out of this horrendous grabbag of trig expressions pops this nice neat progression. When I look at it, I see other routes you could go at the time that seem as likely as the way you did - but those just lead to nastier and nastier formulas.

Actually, JP05's quoted work lends some explanation of why this is the way to go. Note that in both it and in pex's solution, this same cosine progression is present (though in disguise).

[ edit ] I see that the post that suggested balakrishnan might have plagiarized has been wisely removed since I started this reply. [ /edit ]

Title: Re: prove
Post by Michael_Dagg on Jan 14^th, 2007, 8:46pm

GOODNESS! A few days gone and I miss a lot!

KUDOS - Pex your solution is most desirable. Nice work (I bet that it got a few Huhs?, WHEW)!!!

So, there you have it -- cyclotomic polynomials in a trig problem, image that.

It turns out that this not anything new but the trig solution is new, at least to me, as I have not
seen it before now, but just because I haven't seen it means "just that." However, it's both logical
and clever: logical because it removes the nasty trig and clever because it pulls in a geometric series
that is not obvious!

JP05 resources are interesting -- don't try that with a homework problem!

I used the term "reflections" a little losely in this number field but, yes, JP05 your equality is correct,
in this same way. In fact, if you are up to making some discoveries in these kinds of number fields you
might also find tan(4 pi/11) + 4 sin(pi/11) = sqrt(11) for these same reasons.

Title: Re: prove
Post by SWF on Mar 6^th, 2007, 8:34pm

This riddle is on the brink of slipping into oblivion due to its non-descriptive title. How about changing the title from "prove" to something that might help in finding this again- maybe "Trig identity involving pi/11".

It wasn't easy, but I found a way do it using basic trig identities:

(1) 2*sin(A)*cos(B) = sin(A+B) + sin(A-B)
(2) sin(2A)=2*sin(A)*cos(B) and cos(2A)=1-2*sin(A)^2 = 2*cos(A)^2-1
(3) sin(3A)=sin(A)*(3-4*sin(A)^2) and cos(3A)=cos(A)*(1-4*sin(A)^2)
(4) sin(A)^2 + cos(A)^2 = 1

Start with 2*sin(11x)*cos(x):

2*sin(11x)*cos(x) = sin(12x) + sin(10x) (from (1) with A=11x, B=x)

= sin(12x) + 2*sin(6x)*cos(4x) - sin(2x) (from a rearrangment of (1) with A=6x, B=4x)

= 2*sin(6x)*cos(6x) + 2*sin(6x)*(1 - 2*sin(2x)^2) - sin(2x) (from (2) )

= 2*sin(6x)*(2*cos(3x)^2 - 1 + 1) - 4*sin(6x)*sin(2x)^2 -sin(2x) (from (2) )

= sin(2x)*[2*(3-4*sin(2x)^2)*2*cos(3x)^2 - 4*sin(6x)*sin(2x) - 1 ] (from (3) and factoring out sin(2x)

= sin(2x)*[12*cos(3x)^2 - 16*sin(2x)^2*cos(3x)^2 - 8*sin(3x)*cos(3x)*sin(2x) - cos(3x)^2 -sin(3x)^2] (from (2), (4) and expanding)

= 2*sin(x)*cos(x)*[11*cos(3x)^2 - 16*sin(2x)^2*cos(3x)^2 - 8*sin(3x)*cos(3x)*sin(2x) -sin(3x)^2] (from (2) )

Factoring and dividing through by 2*cos(x) leaves:

sin(11x) = sin(x)*[ 11*cos(3x)^2 - (4*sin(2x)*cos(3x) + sin(3x) )^2 ]

If sin(11x) is zero and sin(x) is not zero (i.e. x=N*pi/11, N not a multiple of 11), the term in brackets is zero and dividing by cos(3x)^2 leaves:

11 = ( 4*sin(2x) + tan(3x) ) ^2

Title: Re: Trig identity involving pi/11
Post by Michael_Dagg on Mar 7^th, 2007, 4:20pm

KUDOS -- that's great SWF!

Title: Re: Trig identity involving pi/11
Post by JiNbOtAk on Mar 8^th, 2007, 4:58pm

I can't believe I'm saying this, but I could actually understand SWF's approach ! :P

Attempting it is a different story though, nice job SWF.

Title: Re: Trig identity involving pi/11
Post by Michael_Dagg on Mar 9^th, 2007, 9:21pm

Of course, I hope others see this too as it is elegant, clear,
concise and constructive proof. There are some cagey
chess moves there but those are SWF secrets -- Gauss once
said, responding to admiration of the ingenuity of his proofs,
"that he'd simply removed the scaffolding (as is done with
impressive buildings) which had let him do it all."

(Try using SWF result to get also tan(4 pi/11) + 4 sin(pi/11) = sqrt(11) ).

Title: Re: Trig identity involving pi/11
Post by Michael Dagg on Nov 7^th, 2008, 4:07pm

FYI there is recent paper by V.H. Moll in the
College Mathematics Journal, Vol. 39 No. 5.,
pp. 395-399 regarding trig identties involving
sqrt 11 that is neat.