WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER?

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1 WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I tought Topological Groups at the Göttige Georg August Uiversity. This was a very advaced lecture. I fact it was thought for third year studets. However, oe of the exercises was ot that advaced. Exercise. Show that SO ( ) is dese i SO ( ). Here, SO ( ) = {( ) cos ϕ si ϕ ϕ si ϕ cos ϕ } = {( ) a b a, b, a b a + b = 1} is the group of all rotatios of the plae aroud the origi ad SO ( subgroup of SO ( ) cosistig of all such matrices with a, b. The solutio we thought about goes basically as follows. ) is the Expected Solutio. Oe has si ϕ = ta ϕ 1+ta ϕ ad cos ϕ = 1 ta ϕ 1+ta ϕ. Thus we may put a := t ad b := 1 t for every t. Whe we let t ru through all the 1+t 1+t ratioal umbers this will yield a dese subset of the set of all rotatios. However, Mr. A. Scheider, oe of our studets, had a completely differet Idea. We kow =. Therefore, A := ( 3 is oe of the matrices i SO ( ). It is a rotatio by the agle ϕ := arccos 3. It would be clear that { A } is dese i SO ( ) if we kew arccos 3 is a irratioal agle.. The geeral questio What do we kow about the values of (co)sie? My pocket calculator shows arccos 3 3, Havig see that, I am immediately coviced that arccos 3 is a irratioal agle. But is it possible to give a precise reasoig for this? The truth is, I gave the exercise lessos. The lectures were give by Prof. H. S. Holdgrü. This fact is ituitively clear. For the iterested reader a proof is supplied i a appedix to this ote. 1 )

2 JÖRG JAHNEL Questios. Let α = m 360 be a ratioal agle. i) Whe is cos α equal to a ratioal umber? ii) Whe is cos α a algebraic umber? Oe might wat to make the secod questio more precise. ii.a) What are the ratioal agles whose cosies are algebraic umbers of low degree? For istace, whe is cos α equal to a quadratic irratioality? Whe is it a cubic irratioality? 3. Ratioal Numbers We kow that 1, 1, 0, 1, ad 1 are special values of the trigoometric fuctios at ratioal agles. Ideed, cos 180 = 1, cos 10 = 1, cos 90 = 0, cos 60 = 1, ad cos 0 = 1. It turs out that these are the oly ratioal umbers with this property. Eve more, there is a elemetary argumet for this based o the famous additio formula for cosie. Theorem. Let α be a ratioal agle. Assume that cos α is a ratioal umber. The cos α { 1, 1, 0, 1, 1}. Proof. The additio formula for cosie immediately implies cos α = cos α 1. For ease of computatio we will multiply both sides by ad work with cos α = ( cos α). Assume cos α = a is a ratioal umber. We may choose a, b b they do ot have ay commo factors. The formula above shows, b 0 such that cos α = a b = a b. b We claim that a b ad b agai have o commo factors. Ideed, assume p would be a prime umber dividig both. The, p b = p b ad p (a b ) = p a. This is a cotradictio. Therefore, if b ±1 the i cos α, cos α, cos 4α, cos 8α, cos 16α,... the deomiators get bigger ad bigger ad there is othig we ca do agaist that. O the other had, α = m 360 is assumed to be a ratioal agle. cos is periodic with period 360. Hece, the sequece ( cos k α) k may admit at most differet values. Thus, it will ru ito a cycle. This cotradicts the observatio above that its deomiators ecessarily ted to ifiity. By cosequece, the oly way out is that b = ±1. Oly 1, 1, 0, 1 ad 1 may be ratioal values of cos at ratioal agles. Of course, the same result is true for sie. si α = cos(90 α) ito accout. Oe just has to take the formula The Theorem shows, i particular, that Mr. Scheider is right. arccos 3 is ideed a irratioal agle.

3 WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? 3 4. Algebraic Numbers There is the followig geeralizatio of the Theorem above from ratioal to algebraic umbers. Theorem. Let α be a ratioal agle. The i) cos α is automatically a algebraic umber. Eve more, cos α is a algebraic iteger. ii) All the cojugates of cos α are of absolute value. Proof. i) Let α = m 360. We use the well-kow formula of Moivre which is othig but the result of a iterated applicatio of the additio formula. 1 = cos α ( ) ( ) = cos α cos α si α + + ( 1) k cos k α si k α + k ( ) = ( 1) k cos k α (1 cos α) k k k=0 ( ) = ( 1) k cos k α k = k=0 m=0 m eve k= m k l=0 ( )( ( 1) m k m k ( ) k ( 1) l cos l α k l ) cos m α. The coefficiet of cos α is k=0 ( k) = 1 0. We foud a polyomial P [X] of degree such that cos α is a solutio of P (X) 1 = 0. I particular, cos α is a algebraic umber of degree. Algebraic umber theory shows that the rig of algebraic itegers i a algebraic umber field is a Dedekid rig, i.e. there is a uique decompositio ito prime ideals. The argumet from the proof above may be carried over. If ( cos α) = p e 1 1 per r is the decompositio ito prime ideals ad e i < 0 the ( cos α) = p e 1 i (powers of other primes). Ideed, cos α = ( cos α) ad cotais p i to a o-egative expoet. I the sequece ( cos k α) k the expoet of p i will ted to. As that sequece rus ito a cycle, this is a cotradictio. By cosequece, e 1,..., e r 0 ad cos α is a algebraic iteger. ii) We claim, every zero of the polyomial P (X) 1 obtaied i the proof of i) is real ad i [ 1, 1]. Ufortuately, the obvious idea to provide zeroes explicitly fails due to the fact that there may exist multiple zeroes. That is why istead of P (X) 1 we first cosider P (X) cos δ for some real δ 0. This meas, i the calculatio above we start with cos δ = cos α ad o more with 1 = cos α. There are obvious solutios, amely

4 4 JÖRG JAHNEL cos δ δ+360, cos,..., cos δ+( ) 360, ad δ+( 1) 360. For δ i a sufficietly small eighbourhood of zero these values are differet from each other. P (X) cos δ is the product of liear factors as follows, P (X) cos δ = 1 (X cos δ Goig over to the limit for δ 0 gives our claim. δ+360 )(X cos ) (X cos δ+( 1) 360 ). It is ot hard to see that for every ad every A there are oly fiitely may algebraic itegers of degree all the cojugates of which are of absolute value A.. Degrees two ad three It should be of iterest to fid all the algebraic umbers of low degree which occur as special values of (co)sie at ratioal agles. Observatio (Quadratic Irratioalities). i) Let x be a quadratic iteger such that x < ad x <. The, x = ±, x = ± 3, or x = ± 1 ± 1. ii) Amog the quadratic irratioalities, oly ± 1, ± 1 3, ad ± 1 ± may be values of cos at ratioal agles. Proof. Let x := a + b D where a, b ad D is square-free. x is a algebraic iteger for a, b. For D 1 (mod 4) it is also a algebraic iteger whe a ad b are both half-itegers ad a b. (Note e.g. that 1 ± 1 solve x x 1 = 0.) Assume a + b D < ad a b D <. Without restrictio we may suppose a 0 ad b > 0. If D 1 (mod 4) the D < ad D =, 3. If D 1 (mod 4) the D <, D < 3, ad thus D =. Ideed, oe has the well-kow formulae cos 4 = 1 ad cos 30 = 1 3. Correspodigly, cos 13 = 1 ad cos 10 = 1 3. Further, cos 36 = , cos 7 = , cos 108 = , ad cos 144 = The latter four values are closely related to the costructibility of the regular petago. So, virtually, they were kow i aciet Greece. Nevertheless, a formula like si 18 = cos 7 = does typically ot show up i today s school or Calculus books while the first four special values usually do. Cubic Irratioalities. We use the méthode brutale. If α 1, α, α 3 < the the polyomial x 3 + ax + bx + c = (x α 1 )(x α )(x α 3 ) fulfills a < 6, b < 1, ad c < 8. All these polyomials may rapidly be tested by a computer algebra system. The computatio shows there are exactly 6 cubic polyomials with iteger coefficiets ad three real zeroes i (, ). However, oly four of them are irreducible. These are the followig. i) x 3 x x + 1, zeroes: cos , cos , cos 7 180,

5 WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? ii) x 3 + x x 1, zeroes: cos 7 180, cos , cos , iii) x 3 3x + 1, zeroes: cos 40, cos 80, cos 160, iv) x 3 3x 1, zeroes: cos 0, cos 100, cos 140. The zeroes foud are ideed special values of (co)sie at ratioal agles. They are related to the regular 7-, 9-, (14-, ad 18-)gos. 6. A outlook to the case of arbitrary degree At this poit it should be said that, as i real life, whe someoe is willig to ivest more the she/he has the chace to ear more. For the story discussed above it turs out it is helpful to ivest complex umbers ad some abstract algebra. For example, oe has cos α = e αi + e αi = e m πi + e m πi = ζ m + ζ m showig immediately that cos α is a sum of two roots of uity. I particular, it is a algebraic iteger. It is also possible to aswer the geeral questio from the itroductio for algebraic umbers of higher degree. Theorem. Let α = m 360 be a ratioal agle. Assume that m, ot have ay commo factors. The, 0 do i) cos α is a ratioal umber if ad oly if ϕ(), i.e. for = 1,, 3, 4, ad 6. ii) cos α is a algebraic umber of degree d > 1 if ad oly if ϕ() = d. Here, ϕ meas Euler s ϕ-fuctio. Proof. Note that ϕ() is always eve except for ϕ(1) = ϕ() = 1. The well-kow formula cos α = ζm +ζ m implies that ζ m solves the quadratic equatio X (cos α)x +1 = 0 over (cos α). Further, ζ m geerates (ζ ) as m ad are relatively prime. Thus, [ (ζ ) : (cos α)] = 1 or. As (cos α) that degree ca be equal to 1 oly if ζ, i.e. oly for = 1,. Otherwise, [ (cos α) : ] = [ (ζ): ] = ϕ(). It is ow easily possible to list all quartic ad quitic irratioalities that occur as special values of the trigoometric fuctios si ad cos. ϕ() = 8 happes for = 1, 16, 0, 4, ad 30. Hece, cos α is a quartic irratioality for α = 4, 48 ; 1, 67 1 ; 18, 4 ; 1, 7 ; 1, ad 84. These are the oly ratioal agles with that property i the rage from 0 to 90. ϕ() = 10 happes oly for = 11 ad =. cos ,..., cos are quitic irratioalities. These are the oly oes occurrig as special values of cos at ratioal agles betwee 0 ad 90.

6 6 JÖRG JAHNEL Appedix Let us fially explai the correctess of the desity argumet from the itroductio. Why do the multiples of a irratioal agle fill the circle desely? Fact. Let ϕ be a irratioal agle. The { ϕ } is a dese subset of the set [0, 360 ) of all agles. This meas, for every ϱ [0, 360 ) ad every N there exists some such that ϕ ϱ < 360. N Proof. Cosider the N + 1 agles ϕ, ϕ,..., (N + 1)ϕ. As they are all irratioal, each of them is located i oe of the N segmets (0, 1 N 360 ), ( 1 N 360, N 360 ),..., ( N N 360, N 1 N 360 ), ad ( N 1 N 360, 360 ). By Dirichlet s box priciple, there are two agles aϕ, bϕ (a < b) withi the same box. It follows that 0 < sϕ := (b a)ϕ < 1 N 360. Put M := 360. Clearly, M > N. sϕ Now, let ϱ (0, 360 ) be a arbitrary agle. We put R := max { r {0, 1,..., M} rsϕ ϱ }. The Rsϕ ϱ < (R + 1)sϕ. which implies ϱ (R + 1)sϕ sϕ < 1 N 360. As N may still be chose freely we see that there are multiples of ϕ arbitrarily close to ϱ. Refereces [1] Gradstei, I. S.; Ryshik, I. M.: Tables of series, products ad itegrals, Vol. 1, Germa ad Eglish dual laguage editio, Verlag Harri Deutsch, Thu 198 [] Hardy, G. H.; Wright, E. M.: A itroductio to the theory of umbers, Fifth editio, The Claredo Press, Oxford Uiversity Press, New York 1979 [3] Lag, S.: Algebraic umber theory, Secod editio, Graduate Texts i Mathematics 110, Spriger-Verlag, New York 1994

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