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Topic: cos(1 deg) = irrational (Read 14962 times) 

Sir Col
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Re: cos(1 deg) = irrational
« Reply #1 on: Aug 22^{nd}, 2003, 5:43pm » 
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Nice problem... :: Let c=cos(x) and s=sin(x). So, e^{xi}=cos(x)+i sin(x)=c+si Therefore e^{5xi}=cos(5x)+i sin(5x)=(c+si)^{5}. (c+si)^{5}=c^{5}+5c^{4}si–10c^{3}s^{2}–10c^{2}s^{3}i+5cs ^{4}+ s^{5}. Taking real parts, cos(5x)=c^{5}–10c^{3}s^{2}+5cs^{4}. Using s^{2}=1–c^{2} throughout, cos(5x)=16c^{5}–20c^{3}+5c. We know that cos(45–30)=cos45cos30+sin45sin30=([sqrt]3+1)/(2[sqrt]2), and this is clearly irrational. Therefore, cos15=16cos3–20cos3+5cos3=([sqrt]3+1)/(2[sqrt]2). If cos3 was rational, 16cos3–20cos3+5cos3 would be rational, which is a contradiction, so we deduce that cos3 is irrational. Finally, using cos(3x)=4cos^{3}x–3cos(x), cos3=4cos^{3}1–3cos1. And by the same reaosning as above, we conclude that cos1 is irrational. quod erat demonstrandum. :: [e]Edited to add in new radical symbol. [/e]

« Last Edit: Aug 23^{rd}, 2003, 8:30am by Sir Col » 
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Icarus
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Re: cos(1 deg) = irrational
« Reply #2 on: Aug 23^{rd}, 2003, 7:31pm » 
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A somewhat simpler demonstration  though along the same lines: cos nx = 2cos (n1)x cos x  cos (n2)x Repeated application of this formula allows you to express cos n^{o} = P(cos 1^{o}) for some polynomial P with integer coefficients. Since the value of such a polynomial is rational if the argument is rational, if cos 1^{o} were rational, then so would be all values of cosine for integer degrees. Since we know that several integer degrees for which the cosine is irrational (for instance 45^{o}), cos 1^{o} must be irrational as well.


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Sir Col
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Re: cos(1 deg) = irrational
« Reply #3 on: Aug 24^{th}, 2003, 6:03am » 
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I'm really interested to know how you got this... cos nx = 2cos (n1)x cos x  cos (n2)x Could you please explain? However, I don't think that your method works, Icarus. Quote:Repeated application of this formula allows you to express cos n^{o} = P(cos 1^{o}) for some polynomial P with integer coefficients. 
 After each iterative step, the new formula will contain cos1 raised to increasing powers. For example, after the first stage, cosn = 2cos(n–1)cos1–cos(n–2) = 2[2cos(n–2)cos1–cos(n–3)]cos1–cos(n–2) = 4cos(n–2)cos^{2}1–2cos(n–3)cos1–cos(n–2).

« Last Edit: Aug 24^{th}, 2003, 6:05am by Sir Col » 
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towr
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Re: cos(1 deg) = irrational
« Reply #4 on: Aug 24^{th}, 2003, 8:01am » 
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on Aug 24^{th}, 2003, 6:03am, Sir Col wrote:After each iterative step, the new formula will contain cos1 raised to increasing powers. 
 If cos(1) were rational, cos^{n}(1) would still be rational. So the sum of any powers of cos(1) would still be rational.


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Sir Col
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Re: cos(1 deg) = irrational
« Reply #5 on: Aug 24^{th}, 2003, 8:21am » 
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Good point, towr. Do you know how he derived the iterative formula?


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Sir Col
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Re: cos(1 deg) = irrational
« Reply #7 on: Aug 24^{th}, 2003, 10:49am » 
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I really need to start making more use of that website. You won't believe how long it took me to get an expansion in terms of cos(x), for cos(5x). Before I thought of using Euler's formula – of which I was quite proud – I was battling with trying to simplify cos(2x+3x),but to no avail. I couldn't get is exclusively in terms of cosine. By the way, thanks very much for pointing out that formula, towr. When I get some time I may try to derive it for myself.


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Icarus
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Re: cos(1 deg) = irrational
« Reply #8 on: Aug 24^{th}, 2003, 1:09pm » 
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Towr has it right. I looked it up in my copy of CRC Standard Mathematical Tables  something we preinternet types had to depend on for looking up all those formulas that you don't want to memorize (at least I'm young enough to have never used a slide rule! ). But I once worked it out myself as well. You can prove it and the corresponding formula for sine, sin nx = 2sin (n1)x cos x  sin (n2)x, together by induction. However, the page that towr linked has a nicer proof of the thing that I needed in my proof above: That cos nx is a polynomial with rational coefficients of cos x.


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SWF
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Re: cos(1 deg) = irrational
« Reply #9 on: Aug 24^{th}, 2003, 2:21pm » 
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This can be done without induction. Use the identity for cos(x+y) to find cos((m+1)x)=cos(mx+x) and cos((m1)x)=cos(mxx). Adding cancels out the sin() terms and leaves 2*cos(mx)*cos(x)=cos((m+1)x)+cos((m1)x). Let m=n1.

« Last Edit: Aug 24^{th}, 2003, 2:22pm by SWF » 
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Icarus
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Re: cos(1 deg) = irrational
« Reply #11 on: Aug 24^{th}, 2003, 6:20pm » 
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Related question: Show that cos x and sin x are algebraic if x is a rational multiple of [pi], and are transcendental if x is rational.

« Last Edit: Nov 8^{th}, 2005, 7:41pm by Icarus » 
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Sir Col
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Re: cos(1 deg) = irrational
« Reply #12 on: Aug 25^{th}, 2003, 4:55am » 
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Great puzzle, Icarus, but are you sure about the last part? cos(1/2) is transcendental? I'll assume that you meant algebraic for that too... It is sufficient to show that if cos(x) is algebraic, sin(x)=[sqrt](1–cos^{2}(x)) will also be algebraic. We have established that a polynomial for cos(qx), where q is integer, exists in terms of cos(x). Therefore a polynomial for cos(x) must exist in terms of cos(x/q). E.g. cos(3x)=4cos^{3}(x)–3cos(x) [smiley=bigto.gif] cos(x)=4cos^{3}(x/3)–3cos(x/3). So if cos(x) is algebraic, it follows that cos(x/q) is algebraic. If x=p[pi], cos(p[pi])=[smiley=pm.gif]1, which is algebraic, hence cos(p[pi]/q) will also be algebraic. If x=1, cos(1) is algebraic (see proofs above), therefore cos(p) is algebraic, and so too will be cos(p/q).

« Last Edit: Aug 25^{th}, 2003, 5:06am by Sir Col » 
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Icarus
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Re: cos(1 deg) = irrational
« Reply #13 on: Aug 25^{th}, 2003, 5:35pm » 
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Sorry, but you have misinterpreted something. cos(1) [ne] cos(1^{o}). cos(1^{o}) = cos([pi]/180), which is algebraic. When i refered to cos x. I meant exactly that, not "cosine of x degrees". However, your proof that cos r[pi] is algebraic [forall] r [in] [bbq] is good.


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Sir Col
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Re: cos(1 deg) = irrational
« Reply #14 on: Aug 25^{th}, 2003, 5:44pm » 
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Why, thank you. At the time I thought to myself, "Icarus just doesn't make mistakes, except for flying too close to the sun, that is." But I couldn't think where I was misunderstanding the problem. I should have realised that cos(1^{o})=cos([pi]/180). Duh! Anyway, I must get some shuteye now and work on it in the morning – well later this morning, as it's 1:45am. I just wish I hadn't peeked at the thread before going to bed; I'll spend most of the night awake now thinking about it. Wonders if he can do something with the power series expansion of cos(x)... zzz

« Last Edit: Aug 25^{th}, 2003, 5:47pm by Sir Col » 
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Icarus
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Re: cos(1 deg) = irrational
« Reply #15 on: Aug 25^{th}, 2003, 6:02pm » 
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on Aug 25^{th}, 2003, 5:44pm, Sir Col wrote:"Icarus just doesn't make mistakes, except for flying too close to the sun, that is." 
 Don't I wish!


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Sir Col
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Re: cos(1 deg) = irrational
« Reply #16 on: Aug 26^{th}, 2003, 6:02am » 
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I don't know if there's a more elementary proof for this, but I think the following method works; although it does make use of a theorem I researched and don't really know a whole lot about: Given x is some rational value, suppose that cos(x)=y. Clearly y [ne] 0, as x = [pi](4k[pm]1)/2 and we have already stated that x is rational. Using the identity, y=cos(x)=(e^{ix}+e^{–ix})/2, we get 2y=e^{ix}+e^{–ix}. Rearranging this and writing 2y=2y*e^{0}, e^{ix}+e^{–ix}–2y*e^{0}=0 (1) However, Linderman's Theorem states that A*e^{a}+B*e^{b}+C*e^{c}+... [ne] 0 if A,B,C,... and a,b,c,... are algebraic, each of A,B,C,... presented in the sum is nonzero, and a,b,c,... are distinct terms. As y is not zero and both x and y are algebraic numbers, equation (1) has no solution. Hence we prove that cos(x) is transcendental for all rational values of x. In fact, the proof is much stronger and holds for all algebraic values of x. I must say that I really enjoyed solving that problem. Thanks for asking the question, Icarus. Now we've proved that cos(x) is transcendental for all algebraic values of x and cos(x) is algebraic when x is a rational multiples of [pi]. I wonder was can be said, if anything, about algebraic multiples of [pi]... ?

« Last Edit: Aug 26^{th}, 2003, 6:57am by Sir Col » 
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Sir Col
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Re: cos(1 deg) = irrational
« Reply #17 on: Aug 26^{th}, 2003, 7:05am » 
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I've just been thinking about Linderman's Theorem. If we say that, A*e^{a}+B*e^{b}+C*e^{c}+...+ [alpha]*e^{0} [ne] 0; then A,B,C,...,[alpha] are nonzero algebraic numbers and a,b,c,... are also nonzero algebraic numbers – the nonzero clause is necessary now as the Theorem states that the exponents must be distinct and we have made the last term's exponent zero. However, we can write, A*e^{a}+B*e^{b}+C*e^{c}+... [ne] [alpha]*e^{0} = [alpha]. In other words, a slightly different, and perhaps more useful form, would be: "The sum of exponential terms with nonzero algebaric coefficients and nonzero algebraic exponents cannot be algebraic and must be transcendental."


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Icarus
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Re: cos(1 deg) = irrational
« Reply #18 on: Aug 26^{th}, 2003, 3:08pm » 
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Nice! I was thinking of the GelfondSchneider Theorem, which William quoted recently: GelfondSchneider Theorem: If [alpha] and [beta] are algebraic numbers with [alpha][ne]0, [alpha][ne]1, and [beta][notin][bbq], then [alpha][smiley=supbeta.gif] is transcendental. I believe you can get the result I actually stated by a clever choice of exponent. However, here Lindemann has trumped Gelfond. Well done.

« Last Edit: Aug 26^{th}, 2003, 3:13pm by Icarus » 
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Re: cos(1 deg) = irrational
« Reply #19 on: Aug 31^{st}, 2003, 9:12pm » 
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I'm a little rusty on this sort of thing, but I think I've got a simple proof. As we all know, every rational number is constructible. Since cos(n deg) can be expressed as a polynomial in cos(1 deg), it follows that, if cos (1 deg) is rational, the cosine of every angle is rational. But cos(20 deg) isn't constructible, which implies that it's not rational. QED?


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Sir Col
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Re: cos(1 deg) = irrational
« Reply #20 on: Sep 1^{st}, 2003, 9:59am » 
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Greetings, and welcome to the forum, ultrafiler! I'm afraid that your proof doesn't quite work, in that the assumption at the heart of your proof is, in fact, what we are trying to prove. To summarise your proof: There exists a polynomial for cos(n^{o}) in terms of cos(1^{o}). If cos(1^{o}) was rational, then cos(n^{o}) would be rational for all n. As cos(20^{o}) is irrational, cos(1^{o}) cannot be rational. What you have stated is the perfect introduction to the proof. However, my objection is that it is incomplete. That is, it does not demonstrate/prove that, (i) there exists a polynomial for cos(n^{o}) in terms of cos(1^{o}), and, (ii) cos(20^{o}) is irrational. Of course (ii) could quickly be resolved by picking cos(30^{o}) instead. However, (i) is neither trivial nor simple, and as I've already mentioned, is pretty much equivalent to the fact that cos(1^{o}) is irrational in the first place. Your post, however, does beg an interesting challenge... Prove that cos(20^{o}) is irrational.

« Last Edit: Sep 1^{st}, 2003, 10:03am by Sir Col » 
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ultrafilter
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Re: cos(1 deg) = irrational
« Reply #21 on: Sep 1^{st}, 2003, 10:04am » 
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on Sep 1^{st}, 2003, 9:59am, Sir Col wrote:Greetings, and welcome to the forum, ultrafiler! I'm afraid that your proof doesn't quite work, in that the assumption at the heart of your proof is, in fact, what we are trying to prove. To summarise your proof: There exists a polynomial for cos(n^{o}) in terms of cos(1^{o}). If cos(1^{o}) was rational, then cos(n^{o}) would be rational for all n. As cos(20^{o}) is irrational, cos(1^{o}) cannot be rational. My objection is two fold. It does not demonstrate/prove that, (i) there exists a polynomial for cos(n^{o}) in terms of cos(1^{o}), and, (ii) cos(20^{o}) is irrational. Of course (ii) could quickly be resolved by picking cos(30^{o}) instead. However, (i) is neither trivial nor simple, and as I've already mentioned, is pretty much equivalent to the fact that cos(1^{o}) is irrational in the first place. In other words, it is one that needs to be proved to show that cos(1^{o}) is irrational. 
 Quite possibly. Like I said, I'm more than a little rusty. But I thought that had been shown earlier in the thread.


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Icarus
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Re: cos(1 deg) = irrational
« Reply #22 on: Sep 1^{st}, 2003, 1:37pm » 
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I think that Sir Col was assuming you were trying to prove it from scratch, since in my post showing that cos n^{o} is a polynomial of cos 1^{o}, it is only a very short way from there to the full theorem. I assume that you proved this independently, and only bothered to give here the difference between your proof and mine. And Sir Col is wrong about your showing that cos 20^{o} is irrational. cos 20^{o} is known not to be constructable (it is the key to the proof that it is impossible to trisect angles in contruction). And since as you said, all rationals are constructable, cos 20^{o} must therefore be irrational. But that seems a very hard way to go about it. All you need is a single integer degree angle whose cosine is irrational. And the values cos 30^{o} = [sqrt]3/2 and cos 45^{o} = [sqrt]2/2 are both much better known as irrational numbers.


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Sir Col
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Re: cos(1 deg) = irrational
« Reply #23 on: Sep 1^{st}, 2003, 2:09pm » 
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Isn't that what I said? The point I was trying to make was that it is no more obvious that cos(20^{o}) is irrational than it is that cos(1^{o}) is irrational. The proof for its irrationality is by no means trivial, which was why I presented the challenge... "Prove that cos(20^{o}) is irrational." The problem with elementary proofs, like this, is that they are easy/hard depending on the results that are assumed to be true, and upon which you present the proof. The question is always, how much do I need to prove. For example, I used Euler's Formula in my proof, but perhaps I should have proved that. Then, I might have to show that ([sqrt]3+1)/(2[sqrt]2) is irrational, and so on. Where does one stop in proving something from 'first principles'. At the end of the day, the result that cos(1^{o}) is irrational, is itself fundamental and might be used as an assumption in a more difficult problem.

« Last Edit: Sep 1^{st}, 2003, 2:09pm by Sir Col » 
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Icarus
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Re: cos(1 deg) = irrational
« Reply #24 on: Sep 1^{st}, 2003, 2:21pm » 
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It sounded to me like you were saying he had not shown cos 20^{o} is irrational. But he did, starting from a couple of results from construction theory. Admittedly, cos 20^{o} is not constructable is a considerably more advanced result than the one he derived from it: cos 1^{o} is irrational. But (to the best of my recollection), the irrationality of cos 1^{o} is not used in showing cos 20^{o} is not constructable. Which means that making use of this result is valid. I.e. Ultrafilter's proof works. It's just not a good way to go about it.


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