SOME SPECIAL VALUES OF COSINE JAKE LEVINSON. Introduction We all learn a few specific values of cos(x) (and sin(x)) in high school such as those in the following table: x 0 6 π 4 π π π π cos(x) sin(x) 0 0 0 Surely there are other nice values of cos(mπ), where m? In fact, it turns out there are at least a couple that are more or less as nice as those in the above table, such as (.) cos( π 5 ) = 4 ( 5 ), cos( π ) = ( + 6), cos( π 8 ) = +. We ll find these and a few others, using basic Galois theory and algebraic number theory. In particular, we ll classify the values of m Z such that cos( π m ) is, respectively, rational, quadratic, biquadratic and quartic (i.e., a nested quadratic). Our goal is to proceed algebraically as much as possible, so we avoid embedding into C. In fact, the only truly analytic fact we need is (.) cos( π m ) 0 if m 4. We also observe that complex conjugation (always) restricts to the involution ζ ζ of Gal((ζ)/).. Setup (and rational values of cos( π m ) for m Z) Our approach is based on the following fact (pointed out by Adam Kaye): let ζ = ζ m = e πi/m C be a primitive m-th root of unity. Then (.) α = ζ + ζ = Re(ζ) = cos( π m ). In particular, α is real, and ζ satisfies a quadratic polynomial over (α), (.) αζ = ζ +. Assuming m,, so that [(ζ) : ] = φ(m), we therefore have the tower of fields (ζ) φ(m) (α) = (ζ) R φ(m) In particular, the degree of cos( π m ) over is φ(m)/. We ll consider the cases where it is degree 4, since those are, for example, root extensions of, hence reasonably tractable to work with.
.. Rational values. First of all, we see that α if and only if m =, or φ(m) =, which is to say m =,,, 4, 6. We note that when m is odd, the identities (.) ζ m = ζ (m+)/ m and ζ m = ζ m show that the cyclotomic extension is the same and that α m is a Galois conjugate of α m (the map ζ m ζ m (m+)/ is an automorphism of (ζ m )). In particular, we can skip m = 6: we ll get it easily once we have done m =. For the others, we have the cyclotomic polynomials (.4) f (x) = x, f (x) = x x = x +, f (x) = x x = x + x +, f 4 (x) = x4 x = x +. For the first two, we obtain ζ =,, respectively, and so (.5) α = cos( π m ) =, for m =,, as we learned in high school. For the other two, we observe that α = (ζ + ζ ) is the linear term of the minimal polynomial of ζ! We conclude that (.6) α =, 0 for m =, 4, respectively, as we had hoped. Finally, for m = 6, we compute using our result from m =, (.7) α 6 = ζ 6 + ζ 6 = ζ ζ = α, so α 6 =. We have shown: Corollary. The only values of m for which cos( π m ) are m =,,, 4, 6. We have, respectively, cos( π m ) =,,, 0,.. uadratic, Biquadratic and uartic values The quadratic values are straightforward, but will nicely illustrate the tools we ll use for the degree-4 case. We make use of the following standard facts: Fact. The cyclotomic field (e πi/p ), where p is an odd prime, contains p if p mod 4 and p if p mod 4. The field (e πi/8 ) contains. Fact. The Galois group of the m-th cyclotomic field is isomorphic to the unit group (Z/mZ). The unit u corresponds to the automorphism ζ ζ u. In particular, complex conjugation is always represented by the element Z/mZ. Also, given a field extension L/K, we denote by NK L and TrL K the norm and trace maps L K, respectively. Our approach is first to identify the extension (α) = R (ζ) as a root extension (i.e. a quadratic, biquadratic or nested quadratic), so that α can be expressed in a familiar basis. We then use the trace and norm of certain intermediate fields to finish the computations.
.. uadratic values. For this case, we need φ(m) = 4, which yields m = 5, 8, 0,. As before, we save the case m = odd for the end. For m = 5, the cyclotomic polynomial is (.) f 5 (x) = + x + x + x + x 4. We have (Z/5Z) = Z/4Z, so there is exactly one quadratic subfield, which must be ( 5) by Fact. Hence (α) = ( 5), so α = a + b 5 for some rational a, b. We will compute the trace and norm of α (which are, respectively, a and a 5b ). The nontrivial automorphism σ Gal((α)/)must come from ζ ζ (or, equivalently, ζ ), since the other map restricts to the identity. Hence the Galois conjugate of α is ζ + ζ = ζ + ζ, so the trace and norm are (.) Tr (α) (α) = ζ + ζ + ζ + ζ 4 =, N (α) (α) = (ζ + ζ )(ζ + ζ ) = ζ + ζ 4 + ζ + ζ =. Hence a = and a 5b =, which gives (.) α 5 = cos( π 5 ) = ± 5. Finally, we know (analytically) that cos( π 4 ) should be positive, so we conclude (.4) cos( π 5 ) = 4 ( 5 ). We also get the m = 0 case as a Galois conjugate, namely (.5) α 0 = ζ 0 + ζ 0 = ζ 5 ζ 5 σ(α), the nontrivial Galois conjugate. We conclude (.6) cos( π 0 ) = 4 ( 5 + ). For m = 8,, we can proceed similarly. The cyclotomic polynomials are (.7) f 8 (x) = x 4 +, f (x) = x 4 x +. We observe that (ζ 8 ) contains (ζ 4 ) = (i), and by Fact, hence must be the biquadratic field (i, ). The real subfield is then ( ). Likewise, (ζ ) contains i and, so must be (i, i ), with real subfield ( ). As before, we compute the trace and norm, writing (.8) α 8 = a + b, α = c + d, a, b, c, d. Note that the traces are a, c and the norms are a b and c d. By abuse of notation, we can write the nontrivial automorphism in both cases as the map ζ ζ 5, so the remaining Galois conjugate is (.9) σ(α) = ζ 5 + ζ 5. Hence the trace and norm are (.0) Tr (α) (α) = ζ + ζ + ζ 5 + ζ 5. N (α) (α) = ζ 6 + ζ 4 + ζ 4 + ζ 6. For m = 8, we simplify using ζ 4 8 =, which gives Tr(α 8) = 0 and N(α 8 ) =. For m =, we instead simplify using ζ 6 =, to get Tr(α ) = 0, and ζ 4 = ζ, to get N(α ) =. After solving for a, b, c, d, we obtain: (.) α 8 = cos( π 8 ) = ±, α = cos( π ) = ±.
Finally, we know (analytically) that cos( π 4 ) and cos( π 6 ) should be positive. Putting these results together, we have the following: Corollary. The only m Z for which cos( π m ) is quadratic over are m = 5, 8, 0,. We have: m 5 8 0 cos( π m ) 4 ( 5 ) 4 ( 5 + )... The Only Biquadratic Value. It may come as a surprise that there is only one value of m making cos( π m ) biquadratic over, namely m = 4. We expected (correctly) that these would be the easiest non-quadratic values, but all the others are all nested quartics. The following facts explain why: Fact (Units mod m). The following facts determine the group structure of (Z/mZ) : (a) If p k is an odd prime power, then (Z/p k ) = Z/(p ) Z/p k. Moreover, any such isomorphism sends (Z/p k ) to the ordered pair ( p, 0) having order. (b) For any k, (Z/ k ) = Z/ Z/ k. Any such isomorphism sends (Z/ k Z) to an element of order. (c) If a, b are coprime, then (Z/ab) = (Z/a) (Z/b), and this isomorphism maps to (, ). Fact 4. Consider the group G = Z/ k Z/ kr and the element g = ( k,..., kr ), which has order two. Then the quotient G/ g has the same decomposition into cyclic groups, except with the smallest k i decremented by. In particular, if G/ g is isomorphic to (Z/) n, then every k i = and n = r. Thus Fact 4 shows that the only way for (α m ) to be biquadratic (with Galois group Z/ Z/) is if (ζ m ) is triquadratic, with Galois group (Z/mZ) = (Z/). By Fact, this can only occur for m = = 4. Note that the cyclotomic polynomial is then (.) f 4 (x) = x 8 x 4 +, and that ζ 4 also satisfies ζ 4 =. For this case, we see that (ζ 4 ) contains (ζ ) and (ζ 8 ), hence is precisely (i,, ). The real subfield is thus (α 4 ) = (, ), so we can write (.) α 4 = ζ 4 + ζ 4 = a + b + c + d 6. A direct calculation shows that the Galois conjugates of α 4 are (.4) β = ζ 5 4 + ζ 9 4, γ = ζ 7 4 + ζ 7 4, δ = ζ 4 + ζ 4. We observe that α 4 = δ. The corresponding automorphism ζ 4 ζ 4 sends ζ ζ, hence fixes α =. Thus it must be the map that negates and 6. We equate coefficients, concluding that (.5) a = c = 0. By similar reasoning, the map ζ 4 ζ 7 4 fixes α 8 =. Thus it negates 6, so (.6) (α + γ) = (b ) = 8b 4
and, using ζ 4 =, (.7) (ζ 4 + ζ 4 + ζ7 4 + ζ 7 4) = ζ 8 4 ζ 4 4 + 4 =, where the last equality is from the cyclotomic polynomial. A similar computation with α + β shows that d = ±, and so (.8) α 4 = ± ( ± 6). Both signs must be the same (and both give positive real numbers), but we have chosen our ζ s so that α 4 = α =, which forces the inner sign to be +. Corollary. The only m Z for which cos( π m ) is biquadratic is m = 4, and the resulting value is cos( π 4 ) = 4 ( + 6)... uartic Values. Lastly, we consider the other values of m for which φ(m) = 8, namely m = 5, 6, 0, 0. As usual, we will get m = 0 for free as a Galois conjugate. In all three of these cases, the Galois groups are (.9) Gal((ζ)/) = Z/ Z/4, Gal((α)/) = Z/4. In particular, (α) contains a unique quadratic subfield. In each case, we will identify the subfield as ( r) for some r, then express (α) as ( r, a + b r). In other words, we will ultimately write (.0) cos( π m ) = c + c r + c a + b r + c 4 ar + br r. We will work through the case m = 0; the cases m = 5, 6 are similar. We make the following observations: (ζ 5 ), ( 5), (i) are all subfields of (ζ 0 ); Since Gal((ζ 0 )/) = Z/4 Z/, there is a unique biquadratic subfield, which must therefore be (i, 5). The other two degree-4 subfields have Galois groups isomorphic to Z/4; these must be (α) and (ζ 5 ) (to distinguish them, note that (α) is real and (ζ 5 ) is not.) Putting these thoughts together, we see that the lattice of subfields of (ζ 0 ) is the following: (ζ 0 ) (i, 5) (ζ 5 ) (α) = (ζ) R ( 5) (i) ( 5) In particular, we conclude that the quadratic subfield of (α 0 ) is ( 5). By a similar analysis (identifying the biquadratic subfield and its real quadratic degree- subfield) we determine that the quadratic subfield of (α 5 ) is ( 5) and the quadratic subfield of (α 6 ) is ( ). 5
Now we proceed as follows: we know that α 0 is quadratic over ( 5). We will compute its (relative) norm and trace, (.) Tr (α) ( 5) (α) = T = a + b 5, N (α) ( 5) (α) = N = c + d 5. These determine the minimal polynomial of α over ( 5), namely (.) α T α + N = 0, so we can solve for α using the quadratic formula. The Galois conjugates of α over are (.) β = ζ 0 + ζ 7 0, γ = ζ 7 0 + ζ 0, δ = ζ 9 0 + ζ 0. We note that ζ ζ 9 fixes ζ 4 0 + ζ 4 0 = α 5. We proved earlier that (α 5 ) = ( 5), so in fact this is the identity automorphism of ( 5). Thus the nontrivial conjugate of α over ( 5) is δ. Consequently, (.4) (.5) T = α + δ = ζ 0 + ζ 9 0 + ζ 0 + ζ 9 0 = 0, where we have used ζ0 0 =. In particular, N = α δ = ζ 0 0 + ζ 0 + ζ 8 0 + ζ 0 0 = (ζ 0 + ζ 0 ), (.6) α 0 = ± 5 + 5. As always, we take the positive root. We apply this same method to the other values m = 5, 6, 0. Our results are summarized as follows: Corollary 4. The only m Z for which cos( π m ) is a nested quartic are m = 5, 6, 0, 0. We have: m 5 6 0 0 cos( π m ) 8 ( + 5 + 0 6 5 ) + 4 Note: the field α 5 and α 0 generate the same field extension over. 0 + 5 8 ( + 5 + 0 + 6 5 ). 6