The function tangent and polynomials Andrzej Nowicki Version atg-01. May 15, 2015 , tan 5π , tan 7π It is known (see for example [1], [4]) that the numbers tan π8 , tan 3π 8 8 8 are distinct roots of the equation x4 − 6x2 + 1 = 0, and the numbers tan π7 , tan 2π , ..., 7 are distinct roots of the equation x6 − 21x4 + 35x2 − 7 = 0. tan 6π 7 This note contains some proofs of the above facts and their generalizations. Our note is based on [3], [2] and [5].
1
The polynomials fn(x) and gn(x)
If n is a nonnegative integer, then we denote by fn (x) and gn (x) the polynomials in one variable x defined by
(1 + ix)n = fn (x) + ign (x) . The polynomials fn (x) and gn (x) belong to the polynomial ring Z[x], that is, they are polynomials with integer coefficients. f0 (x) f1 (x) f2 (x) f3 (x) f4 (x) f5 (x) f6 (x) f7 (x) f8 (x)
= = = = = = = = =
1, 1, −x2 + 1 = −(x − 1)(x + 1), −3x2 + 1, x4 − 6x2 + 1 = (x2 + 2x − 1)(x2 − 2x − 1), 5x4 − 10x2 + 1, −x6 + 15x4 − 15x2 + 1 = (x2 + 2x − 1)(x2 − 2x − 1), −7x6 + 35x4 − 21x2 + 1, x8 − 28x6 + 70x4 − 28x2 + 1 = (x4 − 4x3 − 6x2 + 4x + 1)(x4 + 4x3 − 6x2 − 4x + 1), f9 (x) = 9x8 − 84x6 + 126x4 − 36x2 + 1 = (3x2 − 1)(3x6 − 27x4 + 33x2 − 1), f10 (x) = −x10 (x) + 45x8 − 210x6 + 210x4 − 45x2 + 1 = −(x − 1)(x + 1)(x3 − 4x3 − 14x2 − 4x + 1)(x4 + 4x3 − 14x2 + 4x + 1), f11 (x) = −11x10 (x) + 165x8 − 462x6 + 330x4 − 55x2 + 1,
g1 (x) g2 (x) g3 (x) g4 (x) g5 (x) g6 (x) g7 (x) g8 (x) g9 (x) g10 (x)
= = = = = = = = = =
x, 2x, −x3 + 3x = −x(x2 − 3), −4x3 + 4x = −4x(x − 1)(x + 1), x5 − 10x3 + 5x = x(x4 − 10x2 + 5), 6x5 − 20x3 + 6x = 2x(3x2 − 1)(x2 − 3), −x7 + 21x5 − 35x3 + 7x = −x(x6 − 21x4 + 35x2 − 7), −8x7 + 56x5 − 56x3 + 8x = −8x(x − 1)(x + 1)(x2 + 2x − 1)(x2 − 2x − 1), x9 − 36x7 + 126x5 − 84x3 + 9x = x(x2 − 3)(x6 − 33x4 + 27x2 − 3), 10x9 − 120x7 + 252x5 − 120x3 + 10x = 2x(5x4 − 10x2 + 1)(x4 − 10x2 + 5).
Each fn is an even polynomial, and each gn is an odd polynomial divisible by x. Thus we have the equalities gn (−x) = −gn (x).
fn (−x) = fn (x), Moreover, deg f2s = 2s, deg g2s = 2s − 1,
deg f2s+1 = 2s, deg g2s+1 = 2s + 1.
It is clear that we have the following equalities (1 − ix)n = fn (x) − ign (x), fn (x) =
(1+ix)n +(1−ix)n , 2
gn (x) =
(1+ix)n −(1−ix)n . 2i
If h is a nonzero polynomial in a variable x, then let us denote by w(h) the initial coefficient of h. For example, if h = 5x3 − 4x + 7, then w(h) = 5. It is easy to prove: Proposition 1.1. w (f2s ) = (−1)s , w (g2s ) = (−1)s+1 2s,
w (f2s+1 ) = (−1)s (2s + 1), w (g2s+1 ) = (−1)s .
Since fn (x) + ign (x) = (1 + ix)n , we have fn (x) + ign (x) = 1 + i
n 1
x1 −
n 2
x2 − i
n 3
x3 +
n 4
x4 + i
n 5
x5 − · · · .
In the case n = 4k we have f4k (x) = x4k − g4k (x) = −
4k 4k−1
4k 4k−2
x4k−2 +
x4k−1 +
4k 4k−3
2
4k 4k−4
x4k−4 − · · · −
x3 +
x4k−3 − · · · −
4k 3
4k 2
x2 + 1, 4k 1
x.
Hence: f4k (x) =
2k X
(−1)j
4k 2j
x2j ,
g4k (x) =
2k−1 X
(−1)j
j=0
4k 2j+1
x2j+1 .
j=0
By the same way we obtain the following 8 equalities: 2k P
f4k (x) =
(−1)j
4k 2j
j=0 2k P
f4k+1 (x) =
(−1)j
j=0 2k+1 P
(−1)j
f4k+2 (x) =
4k+1 2j
x2j ,
(−1)j
f4k+3 (x) =
j=0
2k−1 P
(−1)j
g4k (x) =
j=0
j=0 2k+1 P
x2j ,
2k P
g4k+1 (x) =
(−1)j
4k+1 2j+1
2k P
g4k+2 (x) =
(−1)j
j=0
4k+3 2j
x2j , x2j ,
2k+1 P
(−1)j
g4k+3 (x) =
j=0
x2j+1 ,
x2j+1 ,
4k+2 2j+1
x2j+1 ,
j=0
4k+2 2j
4k 2j+1
4k+3 2j+1
x2j+1 .
Thus, we have the following 4 equalities: s P
f2s (x) =
(−1)j
j=0 s P
2s 2j 2j x ,
g2s (x) =
2j (−1)j 2s+1 2j x , j=0
f2s+1 (x) =
g2s+1 (x) =
s−1 P
(−1)j
2s 2j+1 , 2j+1 x
(−1)j
2s+1 2j+1 , 2j+1 x
j=0 s P
j=0
The polynomials gn (x) are divisible by x. Let us introduce the notation: gn (x) , x for all n. Observe that if n is odd, then the polynomials fn (x) and hn (x) are of the same degree n − 1. hn (x) =
Proposition 1.2. For every integer k > 0 we have the following two equalities: x4k f4k+1
1 x
x4k+2 f4k+3
= h4k+1 (x),
1 x
= −h4k+3 (x).
Hence
x2s f2s+1
1 x
= (−1)s h2s+1 (x). .
Proof. Assume that n = 4k + 1. Then we have: x4k f4k+1
1 x
= x4k
2k P
(−1)j
j=0
=
2k P
(−1)j
j=0
=
1 x
2k P
2j 1
4k+1 2j
4k+1 2j
(−1)j
j=0
x
x4k−2j =
4k+1 2j+1
1 x
2k P
(−1)j
j=0
x2j+1 = x1 g4k+1 (x)
= h4k+1 (x). 3
4k+1 4k+1−2j
x4k+1−2j
.
Now let n = 4k + 3: x4k+2 f4k+3
1 x
= x4k+2
2k+1 P
x4k+2−2j =
j=0
=
2k+1 P
(−1)j
j=0
= − x1
2j
(−1)j
4k+3 2j
2k+1 P
(−1)j
j=0
4k+3 2j
4k+3 2j+1
1 x
1 x
2k+1 P
(−1)j
j=0
4k+3 4k+3−2j
x4k+3−2j
x2j+1 = − x1 g4k+3 (x)
= −h4k+3 (x). This completes the proof.
2
Divisibility properties of fn(x) and gn(x)
Proposition 2.1. For all nonnegative integers m, n we have the following two equalities:
fn+m (x) = fn (x)fm (x) − gn (x)gm (x) gn+m (x) = fn (x)gm (x) + fm (x)gn (x)
.
Proof. n+m
fn+m (x) + ign+m (x) =
1 + ix
=
n
= 1 + ix
1 + ix
fn (x) + ign (x)
fm (x) + igm (x)
=
m
fn (x)fm (x) − gn (x)gm (x) + i fn (x)gm (x) + fm (x)gn (x) .
This completes the proof. In particular, f2n (x) = fn (x)2 − gn (x)2 and
g2n (x) = 2fn (x)gn (x) . Proposition 2.2 ([5], [2]). The polynomials fn (x), gn (x) are relatively prime in Q[x]. Proof. Since Z[i][x] is a unique factorization domain, it is enough to prove that the polynomials fn (x) and gn (x) are relatively prime in the domain Q(i)[x]. Let n be a fixed positive integer and suppose that these polynomials are not relatively prime. Then fn (x) and gn (x) are divisible by an irreducible polynomial h(x) ∈ Q(i)[x]. Obviously deg h(x) > 1. Then h(x) divides the polynomials (1 + ix)n and (1 − ix)n . As h(x) is irreducible, we deduce that the polynomials (1 + ix) and (1 − ix) are divisible by h(x). This implies that h(x) divides the number 2. But it is a contradiction, because deg h(x) > 0.
4
Note next propositions concerning the divisibility of fn (x) and gn (x). Proposition 2.3. If m | n, then gm (x) | gn (x) (divisibility in Z[x]). Proof. We will prove by an induction that for every positive integer k, the polynomial gm divides the polynomial gkm . It is obvious in the case k = 1. Assume that k > 1 and gm divides gkm in Z[x]. Let gkm = u · gm with u ∈ Z[x]. Then by Proposition 2.1, we have g(k+1)m = gkm+m = fkm gm + fm gkm
= fkm gm + fm ugm = fkm + ufm gm and so, gm divides g(k+1)m in Z[x]. We may prove also: Proposition 2.4 ([2]). If m | n, then gn (x) = vm (x)gm (x), where vm (x) ∈ Z[x], and the polynomials vm (x) and gm (x) are relatively prime in Q[x]. Proof. Let n = dm, where d is a positive integer. If m = n, then vm (x) = 1 and we are done. Assume that m < n. Then d > 2 and we have fn + ign = (1 + ix)n = (1 + ix)dm = (fm + igm )d d 2
d−1 d gm − + idfm = fm
d−2 2 gm − i fm
d 3
d−3 3 gm + · · · , fm
2 d−1 , where u = u(x) is a polynomial belonging to Z[x]. gm + u · gm and hence gn = dfm Put d−1 vm (x) = dfm + ugm .
Then gn (x) = vm (x)gm (x). Since gcd(fm , gm ) = 1 (see Proposition 2.2), we have
d−1 d−1 gcd(gm , vm ) = gcd gm , dfm + ugm = gcd gm , dfm = 1.
Therefore, the polynomials gm (x) and vm (x) are relatively prime in Q[x]. Observe that the polynomials f3 = −3x2 + 1, f5 = 5x4 − 10x2 + 1 and f7 = −7x6 + 35x4 − 21x2 + 1 are irreducible in Q[x] (and also in Z[x]). Now we will prove that the same is true for all the polynomials fp , where p is an odd prime number. Let us recall that hn (x) = gnx(x) . Proposition 2.5. If p > 3 is a prime number, then the polynomials fp (x) and hp (x) are irreducible in Z[x]. Proof. fp (x) + igp (x) = (1 + ix)p = 1+
p 1
ix −
p 2
x2 − i
5
p 3
x3 + · · · ±
p p−1
xp−1 ∓ ixp .
p This implies that fp (x) = 1 − p2 x2 + p4 x4 + · · · ± p−1 xp−1 and gp (x) = x · hp (x), where p hp (x) = p1 − p3 x2 + · · · ∓ p−2 xp−3 ± xp−1 .
p As all the numbers p1 , p2 , . . . , p−1 are divisible by p, we deduce by Eisenstein’s theorem that the polynomials fp (x) and hp (x) are irreducible in Z[x].
Proposition 2.6 ([5]). If n, k are nonnegative integers, then: (1) fn | g2kn ; (2) fn | f(2k+1)n ; (3) the polynomials fkn and gn are relatively prime; (4) gcd (gkn+r , gn ) = gcd (gr , gn ) for integers r > 0. Proof. (1). Since g2n = 2fn gn , we have fn | g2n . But g2n | g2kn (see Proposition 2.3). Hence, fn | g2kn . (2). It is obvious for k = 0. Let k > 0 and assume that fn | f(2k+1)n . Since f(2k+3)n = f(2k+1)n+2n = f(2k+1)n f2n − g(2k+1)n g2n and fn | g2n , we have fn | f(2k+3)n . (3). It is obvious for k = 0, because f0 = 1. If k = 1, then it followsfrom Proposi tion 2.2. Let k > 1 and assume that gcd (fkn , gn ) = 1. Then we have: gcd f(k+1)n , gn = gcd (fkn+n , gn ) = gcd (fkn fn − gkn gn , gn ) = gcd (fkn fn , gn ) = gcd (fkn , gn ) = 1. (4). We use the previous properties: gcd (gkn+r , gn ) = gcd (fkn gr + gkn fr , gn ) = gcd (fkn gr , gn ) . But gcd (fkn , gn ) = 1, so gcd (gkn+r , gn ) = gcd (gr , gn ). Now, by Proposition 2.6 and the Euclid algorithm, we have: Theorem 2.7. If n, m are positive integers, then
gcd (gm , gn ) = ggcd(m,n) .
3
Zeros of the polynomials fn(x) and gn(x)
Let n > 1 be a fixed integer. Consider the numbers of the form
jπ tj = tan n
!
,
where {0, 1, 2, . . . , n − 1} and n 6= 2j. If n jest odd, then we have n distinct real numbers t0 = 0, t1 , . . . , tn−1 . If n = 2m is even, then we have n − 1 distinct real numbers t0 = 0, t1 , . . . , tm−2 , tm−1 , tm+1 , tm+2 , . . . , tn−1 . 6
Theorem 3.1. All zeros of each polynomial gn (x) are real numbers. More precisely, if n is odd, then all the real numbers t0 , t1 , . . . , tn−1 are distinct zeros of gn . If n = 2m is even then all the real numbers t0 , t1 , . . . , tm−1 , tm+1 , tm+2 , . . . , tn−1 are distinct zeros of gn . Proof. Assume that j ∈ {0, 1, . . . , n − 1}, n 6= 2j. Then we have:
n
fn (tj ) + ign (tj ) = (1 + itj ) = 1 + =
cos jπ n
−n
n i tan jπ n
cos jπ + i sin jπ n n
= (−1)j cos jπ n
sin
= 1 + i cos
jπ n jπ n
n
n
−n
and this implies that gn (tj ) = 0. Theorem 3.2. All zeros of each polynomial fn (x) are real numbers. Proof. It follows from Theorem 3.1 and the equality g2n = 2fn gn . As a consequence of the above theorems we obtain Proposition 3.3. For every n > 1, the polynomials fn (x) and gn (x) are square-free. The next proposition is also a consequence of the above theorems. Proposition 3.4. If q 6=
1 2
is a rational number, then the number tan (qπ) is algebraic.
Proof. Put r = tan(qπ). Then ±r = tan nj π , where j, n are nonnegative integers with n > 1 and j < n, and of course n 6= 2j. We know, by Theorem 3.1, that ±r is a zero of the polynomial gn (x). All the coefficients of gn (x) are integers, so ±r is algebraic, and so r is algebraic. Theorem 3.5. If q is a rational number, then all the numbers sin(qπ), cos(qπ), tan (qπ), cot(qπ) (except when undefined) are algebraic. Proof. It is a consequence of Proposition 3.4 and the following known equalities: sin α =
2t , 1 + t2
cos α =
1 − t2 , 1 + t2
cot α =
1 , tan α
where t = tan α2 . Theorem 3.6. If q is a rational number, then all the numbers sin(q o ),
cos(q o ),
tan (q o ),
cot(q o )
are algebraic. Proof. Since the angle 180o (180 degrees) equals π radians, we have q o = Therefore, this theorem is a consequence of the previous theorem.
7
q π. 180
4
Equalities with tangents
Look again at Theorems 3.1, 3.2 and the initial coefficients described in Proposition 1.1. Observe that as a consequence of the above facts we obtain the following equalities. Proposition 4.1.
f2m (x) = (−1)
2m m Q
x−
k=1
f2m+1 (x) = (−1)m (2m + 1)
tan (2k−1)π 4m Q k∈A
,
, x − tan (2k+1)π 4m+2
where m > 1 and A = {0, 1, 2, . . . , 2m} r {m}. Proposition 4.2. 2m Q
g2m+1 (x) = (−1)m x
k=1 m+1
g2m (x) = (−1)
kπ x − tan 2m+1 ,
2mx
Q k∈A
kπ x − tan 2m π ,
where m > 1 and A = {0, 1, 2, . . . , 2m − 1} r {m}. Since tan (π − α) = −tan (α), we may rewrite the above two propositions in the following way. Proposition 4.3.
f2m (x) = (−1)m
m Q 2 k=1
x − tan2 ( (2k−1)π 4m )
m
f2m+1 (x) = (−1) (2m + 1) g2m (x) = (−1)
m+1
g2m+1 (x) = (−1)m x
2mx
m−1 Q 2
x −
k=0 m−1 Q
2
x −
k=1
2 (2k+1)π tan ( 4m+2 )
kπ tan2 ( 2m )
m Q 2 k=1
.
kπ x − tan2 ( 2m+1 )
Let us recall (see Proposition 1.2) that x2m f2m+1
1 = (−1)m h2m+1 (x). x
(x) where h2m+1 (x) = g2m+1 . The zeros of the polynomial h2m+1 (x) are real numbers of x kπ the form tan 2m+1 , where k = 1, . . . , 2m. Thus, we have:
8
Proposition 4.4. The zeros of the polynomial f2m+1 (x) are real numbers of the form cot
kπ , 2m + 1
k = 1, 2, . . . , 2m.
Thus, we have the equalities
f2m+1 (x) = um
2m Q k=1
kπ x − cot 2m+1 = um
m Q 2
kπ , x − cot2 2m+1
k=1
where um = (−1)m (2m + 1). Recall that hn (x) =
gn (x) . x
h4k+1 (x) =
Look again at the polynomial hn (x) in the case n = 4k+1:
2k X
(−1)j
4k+1 2j+1
2k Y
jπ ) . x2 − tan2 ( 4k+1
x2j =
j=0
j=1
Comparing the constant terms, we obtain the equality
paring the coefficients of x4k−2 , we obtain the equality
4k+1 1
2k Q
=
j=1
4k+1 2
=
2k P j=1
jπ ). Comtan2 ( 4k+1
jπ tan2 ( 4k+1 ). Hence,
we have the following two identities: √ jπ tan = 4k + 1, 4k + 1 j=1 2k Y
2k X
2
tan
j=1
jπ = 2k(4k + 1). 4k + 1
Using the same in the case n = 4k + 3, we obtain: 2k+1 Y
tan
j=1
√ jπ = 4k + 3, 4k + 3
2k+1 X
tan2
j=1
jπ = (2k + 1)(4k + 3). 4k + 3
Assume now that n is even. For n = 4k we have: h4k (x) = −4k
2k−1 Y
x2 − tan2 ( jπ ) = 4k
j=1
2k−1 X
(−1)j
4k 2j+1
x2j ,
j=0
4k 1
Comparing the constant terms, we obtain the equality paring the coefficients of x4k−4 , we obtain the equality
4k 3
= 4k
= 4k
2k−1 Q
j=1 2k−1 P j=1
tan2 ( jπ ). Com4k
tan2 ( jπ ). Hence, 4k
we have the following two identities: 2k−1 Y j=1
jπ tan 4k
= 1,
2k−1 X
2
tan
j=1
jπ 4k
1 = (2k − 1)(4k − 1). 3
Using the same in the case n = 4k + 2, we obtain: 2k Y j=1
tan
jπ = 1, 4k + 2
2k X
tan2
j=1
jπ 1 = 2k(4k + 1). 4k + 2 3
Therefore, we proved the following proposition. 9
Proposition 4.5. m−1 Q j=1 m Q j=1
tan
tan
jπ 2m
jπ 2m+1
m−1 P
= 1,
=
√
j=1 m P
2m + 1,
j=1
tan2
jπ 2m
=
1 3 (m
− 1)(2m − 1), .
2
tan
jπ 2m+1
= m(2m + 1).
By the same method, using Proposition 4.4, we obtain the following equality with cotangents. Proposition 4.6 ([1] 123).
m X
jπ m(2m − 1) = . 2m + 1 3
cot2
j=1
5
Examples and applications (1)
tan π5
(2)
tan2
cot2
(3)
π 5
= 36o . √ · tan 2π = 5. 5
5.1. Multiples of π 5
π 5
+ tan2 + cot2
q
= 10.
= 2.
2π 5
2π 5
q √ √ 2π 5 − 2 5, tan 5 = 5 + 2 5.
(4)
tan π5
(5)
The numbers tan π5 , tan 2π , tan 3π , tan 4π are zeros of the polynomial 5 5 5
=
h5 (x) = x4 − 10x2 + 5. (6) The numbers cot π5 , cot 2π , cot 3π , cot 4π są are zeros of the polynomial 5 5 5 f5 (x) = x4 − 10x2 + 1. 5.2. Multiplies of π7 . (1) (2) (3)
tan π7 · tan 2π · tan 3π = 7 7 tan2
cot2
π 7
π 7
+ tan2 + cot2
(4) The numbers tan
2π 7
2π 7 π , 7
√
7.
+ tan2 + cot2
3π 7
3π 7
= 21.
= 5.
tan 2π , . . . , tan 6π are zeros of the polynomial 7 7 h7 (x) = x6 − 21x4 + 35x2 − 7.
(5) The numbers cot π7 , cot 2π , . . . , cot 6π are zeros of the polynomial 7 7 f7 (x) = −7x6 + 35x4 − 21x2 + 1. 10
5.3. Multiplies of π8 = 22, 5o . (1) tan π8 · tan 2π · tan 3π = 1. 8 8
π 8
+ tan2 2π + tan2 3π = 7. 8 8 q q √ √ 3π = 1, tan = (3) tan π8 = 3 − 2 2, tan 2π 3 + 2 2. 8 8 (4) The numbers tan π8 , tan 3π , tan 5π , tan 7π are zeros of the polynomial 8 8 8
(2)
tan2
f4 (x) = x4 − 6x2 + 1. 5.4. Multiplies of π9 = 20o . · tan 3π · tan 4π = 3. (1) tan π9 · tan 2π 9 9 9 (2) (3)
tan2
cot2
π 9
π 9
+ tan2 + cot2
(4) The numbers tan
2π 9
2π 9 π , 9
+ tan2 + cot2
3π 9
3π 9
+ tan2
+ cot2
4π 9
4π 9
= 36.
=
28 . 3
tan 2π , . . . , tan 8π are zeros of the polynomial 9 9
h9 (x) = x8 − 36x6 + 126x4 − 84x2 + 9 = (x2 − 3)(x6 − 33x4 + 27x2 − 3) , . . . , cot 8π are zeros of the polynomial (5) The numbers cot π9 , cot 2π 9 9 f9 (x) = 9x8 − 84x6 + 126x4 − 36x2 + 1 = (3x2 − 1)(3x6 − 27x4 + 33x2 − 1). π 5.5. Multiplies of 10 = 18o . π (1) tan 10 · tan 2π · tan 3π · tan 4π = 1. 10 10 10
(2) (3)
tan2
π 10
π tan 10 =
+ tan2
q
√ 5−2 5 , 5
2π 10
+ tan2
tan 3π = 10
3π 10
q
√ 5+2 5 . 5
+ tan2
4π 10
= 12.
, where j ∈ {1, 2, 3, 4, 6, 7, 8, 9}, are zeros of the (4) All numbers of the form tan jπ 10 polynomial h10 (x) = 10x8 − 120x6 + 252x4 − 120x2 + 10 = 2(5x4 − 10x2 + 1)(x4 − 10x2 + 5). π (5) The numbers tan 10 , tan 3π , tan 7π , tan 9π are zeros of the polynomial 10 10 10
f5 (x) = 5x4 − 10x2 + 1.
11
Literatura [1] T. Andreescu, Z. Feng, 103 Trigonometry Problems. From the training of the USA IMO team, Birkh¨auser, Boston - Basel - Berlin, 2005. [2] X. Lin, Infinitely many primes in the arithmetic progression kn − 1, The American Mathematical Monthly, 122(1)(2015), 48-50. [3] T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964. [4] A. Nowicki, Real Numbers and Functions (in Polish), Podróże po Imperium Liczb, part 10, Second Edition, OWSIiZ, Toruń, Olsztyn, 2013. [5] A. Nowicki, S. Spodzieja, Polynomial imaginary decomposition for finite extensions of fields of characteristic zero, Bull. Pol. Sci. Acad., 51(2)(2003), 157-168. Nicolaus Copernicus University, Faculty of Mathematics and Computer Science, 87-100 Toruń, Poland, (e-mail:
[email protected]).
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