We claim that a 2 2b 2 and b 2 again have no common factors. Indeed, assume p would be a prime number dividing both. This is a contradiction. Hence, the sequence 2 cos 2 k k may admit at most n dierent values. Thus, it will run into a cycle.
This contradicts the observation above that its denominators necessarily tend to innity. Only 1, 1 2 , 0, 1 2 and 1 may be rational values of cos at rational angles. Of course, the same result is true for sine. The Theorem shows, in particular, that Mr. Schneider is right. Algebraic Numbers There is the following generalization of the Theorem above from rational to algebraic numbers.
Then i cos is automatically an algebraic number. Even more, 2 cos is an algebraic integer. We use the well-known formula of Moivre which is nothing but the result of an iterated application of the addition formula. The coecient of cos n is. In particular, cos is an algebraic number of degree n. Algebraic number theory shows that the ring of algebraic integers in an algebraic number eld is a Dedekind ring, i. The argument from the proof above may be carried over.
In the sequence 2 cos 2 k k the exponent of p i will tend to. As that sequence runs into a cycle, this is a contradiction. By consequence, e 1 ,. Unfortunately, the obvious idea to provide n zeroes explicitly fails due to the fact that there may exist multiple zeroes. There are n obvious solutions, namely 4 J. For in a suciently small neighbourhood of zero these values are dierent from each other. Going over to the limit for 0 gives our claim.
It is not hard to see that for every n and every A there are only nitely many algebraic integers of degree n all the conjugates of which are of absolute value A.
Degrees two and three It should be of interest to nd all the algebraic numbers of low degree which occur as special values of co sine at rational angles. Observation Quadratic Irrationalities. D where a, b and D is square-free. For D 1 mod 4 it is also an algebraic integer when a and b are both half-integers and a b. Note e. If D 1 mod 4 then. The latter four values are closely related to the constructibility of the regular pen- tagon. So, virtually, they were known in ancient Greece.
Nevertheless, a formula like sin Cubic Irrationalities. We use the methode brutale. All these polynomials may rapidly be tested by a computer algebra system. The computation shows there are exactly 26 cubic polynomials with integer coe- cients and three real zeroes in 2, 2. However, only four of them are irreducible. These are the following. The zeroes found are indeed special values of co sine at rational angles. They are related to the regular 7-, 9-, , and gons.
For the story discussed above it turns out it is helpful to invest complex numbers and some abstract algebra. In particular, it is an algebraic integer. It is also possible to answer the general question from the introduction for algebraic numbers of higher degree. Then i cos is a rational number if and only if n 2, i.
Here, means Eulers -function. Further, m n generates n as m and n are relatively prime. As cos that degree can be equal to 1 only if n , i. It is now easily possible to list all quartic and quintic irrationalities that occur as special values of the trigonometric functions sin and cos. These are the only rational angles with that property in the range from 0. These are the only ones occurring as special values of cos at rational angles between 0.
Why do the multiples of an irrational angle ll the circle densely? Let be an irrational angle. As they are all irrational, each of them is located in one of the N segments 0.
It follows that 0. As N may still be chosen freely we see that there are multiples of arbitrarily close to. References [1] Gradstein, I. Open navigation menu. Close suggestions Search Search. User Settings. Skip carousel.
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Is this content inappropriate? Report this Document. Flag for inappropriate content. Download now. You might be looking for Niven's theorem :. I'm not sure right now how to prove that. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams?
Learn more. Ask Question. Asked 9 years, 11 months ago. Active 1 year, 11 months ago. Viewed 14k times. Bart Michels I doubt there is a simple characterization when it is rational. Show 3 more comments. Active Oldest Votes. Add a comment. Michael Hardy Michael Hardy 1. I learned this trick from Robert Israel back on sci. Martin Sleziak Sign up or log in Sign up using Google.
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