705
Internat. J. Math. & Math. Sci.
Vol. 9 No. 4 (1986) 705-714
TRIGONOMETRIC IDENTITIES
MALVINA BAICA
Department of Mathematics and Computer Science
Whitewater
University of Wisconsin
Whitewater, Wisconsin 53190 U.S.A.
(Received November 20, 1985)
ABSTRACT.
In this paper the author obtains new trigonometric
)p-l-k
(p-1)(p,2)
pp-2
(1-cos
2
2
k=l
which are derived as a result of relations in a cyclotomic field
is a root of unity.
is the field of rationals and
where
3 and any
Those identities hold for every positive integer p
if not
difficult,
be
very
could
fields
cyclotomic
proof avoiding
identities of the form
(),
insoluble.
1
Two formulas
Z (-l)k(fk)
tanp-l-2k
and
k=l
p
(k
co
sP-2k
o
stated only by Gauss in a slightly different form without a proof, are
obtained and used in this paper in order to give some numeric applications of our new trigonometric identities.
KEY WORDS AND PHRASES. Trigonometric identities, cyclotomic field.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODE. 12A35, 12El2.
INTRODUCTION
The trigonometric identities which are being obtained in this
paper are a result of relations in a cyclotomic field ), where
The reader
is a root of unity.
is the field of rationals and
are the
which
will be familiar with the primitive p-th roots of unity
p-1 different roots of the irreducible polynomial
O.
xP-1 +xP-2+
which we shall denote by
cos+
.
+x+l
isin,
O,
p a prime
>
2
As is well known
_2k
p
k
1 2 9"’’ k-1
(0.2)
706
M. BAICA
Since the p-1 entities
(0.1) we
shall choose
cos+
, 2 ,...,
isin,
p-1 form all the different roots of
2
-,
(k:l).
(0.3)
The cyclotomic fields have been substantially investigated and the
author will make use of some comments. Yet there are still many open
problems in this domain. The full group of fundamental units has not
yet been revealed and this remains one of the most interesting
problems.
IDENTITIES
The following important formula has been proved by Pollard [3]
and many other authors and gives the discr+/-minant D() of the field
in the following explicit form
i.
()
(1.1)
We shall first investigate
T
T
l<i < j<_p-i
(i_j);
T2
D()
(
p-l)
We obtain from (1.2)
T
(
2)( 5)( 4)
(f2 31(2 41
(73
We shall calculate
We have
f41
(2
(3 p-l)
(-
-a
(1.2)
707
TRIGONOMETRIC IDENTITIES
2
i
6
p(p-)(p-,2)
2
o(p-)(p-2)
6
p(p-)(p-2)
5
- -
k=l
Now
cos 2 +i sin 2
(cos 2
p
-;
(z.4)
+ i sin
2 P
-)
I
cos 2% + i sin 2%
(p-1) (p-2)
p (p-i) (p-2)
6
6
(p)
We shall presume here p > 3 and shall return to this case later. (In
We then have
the case p
2, the primitive unity roots are + i.
or
that
either
0(6)
hence
(p-l)(p-2)
so
p-i
p,
0(6)
3
p-2
0(3) and p-i 0(3).
0(2). So in any case
Of course, since p is odd, p-I
61 (p-l)(p-2),
P
and since
p (p-1) (p- 2,)
I,
(p-I) (p-2)
(p-I) (p-2)
6
(p)
p-2
"’"
6
:
i, and from
(1.4)
(1.5)
k=l
We thus remain with
p-2
T
k=l
T
(1-
k) p-k-I
(1.6)
From (1.6) we obtain, since
?k
and from
T
cos 2k
--+i
sin
2
-, -:
2%k
,
2__k
p
k
(1.6)
.
[I
k)]P-l-k
(cos
k+i
sin
cos
k)-
i sin
k=l
p-2
[(1-
k] p-l-k
k=l
(2
k=l
sin2
-i2sin cos ) p-l-k
708
- M. BAICA
p-2
(-2i
sin
k=l
)P-l-k (cos
+isin
p-2
p-2
(-I) p-l-k
T
k=l
i p-l-k
k=l
(2
k=l
) p-l-k
sin )p-l-k
(1.7)
p-2
’
(cos
-1-k
+isin
k=l
We further have
-
(p-2) (-.i)
(-l)P-k-i
(-i)
(-I)p-k-i
(-i)’-
2
and since p is odd
(-i) p-2
-i,
k=l
hence
p-2
p-i
(i.8)
k=i
p-2
(p-I) (p-.2)
iP-l-k
i
k=l
p-2
(cos
k=l
=(e
+ i sin
k=l
6
ep
k=l
+ i sin
k=l
(p-l) (p-2)
With
-
k_ p-l-k
i__ p(p-i)(p-2)
p-2
=-e(p-l-k)k
(cos
i
)p-l-k
(-1)
iE
(p-l)(p-2)
6
=e
(p-l)(p-2,!6
(i. I0)
0(6).
(1.8), (1.9), (i. I0), (1.7)takes the form
(p-2) (p-l)
’2
T: (-i) 2 i
(p-i) (p-2)
6
(-i)
p-2
II (2sin) p-i-k.
(i. il)
k:i
But from (1.2)
T 2, so with (p-l)(p-2)
D()
(p-2) (p-l)
D()= (-I)P-I (i 2)
D()= (-i)
2p-l-k
k=i
(p-l) (p-R)
6
((-1) 2)
2
(2
k=i
sin
m()
2
2
2
)
p-l-k
(i cos k) p-l-k
k=l
-2
2P-l-k(2 sin2 )P-l-k
k=l
(p-2)(p-1)
(-I)
0(6)
(-i) pp-2
709
TRIGONOMETRIC IDENTITIES
from (i.
i).
Hence we obtain our identity
pp-2
’-jP-l-k
(i- cos
2
(I.12)
k=l
or
(i
2
pp-2.
cos
(1.13)
k=l
It is interesting that the identity (I.13) is valid also for p
(p-l) (p-2)
2
=21=2
We have p
I, 2
3, k
1
2 1
,, tl-cos --)
2(1 +
cos 120;
2(1
k=l
But the proof is not valid here, since 31P.
cos 120 + i sin 120 from (i.I)
2
D(?)
since
D()
Also
9
? 2)2 2(12 2 + 2)
I,
(-I)
[(cos
-
(2i sin
(?-i)(
3 3-2
+ i
+
I)
-4
(cos
We find directly, with
(72 2?3
+
/
O, and for
+ i
sin)]
-3-
Now consider the Vandermonde determinant:
1
1
1
1
2
1
2
4
p-I
2(p-l)
1
p-i
2(p-l)
(p-l)2
1
T.
oi < j<p-i
j=l
T as in (1.6).
Multiplying this determinant by itself we obtain
0
p
0
0
0
0
0
p
V2=
0
0
p
0
0
p
0
0
33-2
3
-3.
sin)-
)2
+
lz)
3.
(-i)
pp.
1
710
-
M. BAICA
In view of
{
(k+l)j
j
kj.
?
:o
:o
Since
2
0)
(x-
0 if
p
k+l
p i
p-l,
for x
i +x+x +... +x
i we obtain
0=l
j=l
(1-J)
Consequently V=pT.
p.
From the definition of T we obtain
II- k12= 2(1- cos
2k
p ), and then
Hence
=-
IT21
IT21
TPl =p-elVPl pp-2.
(ll-kl2) p-k-1.
k=l
k=l
(2(I- cos
2k
p
But
))p-k-l.
Hence (1.13) follows.
Here we did not use the assumption that p is a prime number.
(1.13) is
Therefore (1.13) holds for every positive integer p >_ 3.
an interesting identity and any proof avoiding cyclotomic fields could
be very difficult, if not insoluble.
NUMERIC APPLICATION
We shall give a few examples for the identity (1.13).
purpose, we need a short review of known formulas.
Let again be
cos 2k + i sin 2k
k
O, 1 ,.. p-1
2.
(cos
-2
Choosing again k
(cos
cosP
For this
--,
2 k k
0,1,...,p-1
+ i sin
we have
l, 2%
,
-.)
+ i sin
)P
(i tan
+
0
i
l) p
1
0
I) p
1
O,
(2.1)
cosp
and for the imaginary part of (2.1) we obtain, after cancelling by i;
(i tan
+
tanP: ()tanP-2+ ()tanP-4+---+(-I)
Excluding
tion for
tan=
tan
0, we can divide (2.2) by
tan
ptan=0. (2.2)
and obtain the equa-
1
(-1)k () tanP-l-2k
O.
This formula was also stated by Gauss [i].
the cos # can be obtained from (2.3) since
2
1
tan2=
We obtain from (2.3)
cos
cos2 =
=
(2.3)
We shall still show how
_
711
TRIGONOMETRIC IDENTITIES
2
(_)k() (-cos2p-1-2k
(2.4)
O.
2
(cos2)
Setting
cos2
we obtain from
x
(2.4)
p-l-2k
(_l)k(k)
2
(l-x)
p,-1- 2k’
(2.6)
0
2
x
and in expanded form
-+ ...+
x
(-)
p
o
x
x
or
p-l-2k
(_l)k(l_x)
2
xk(A) O.
for cos2.
(2.7)
But there is another formula
This equation is of degree
which is obtained in the following way. We obtain for the
for
cos
real part of
(cos+
-1
isin)8
I
0
+Z (-1)k() cosP-2ksin2k=
0
or
-i+
(-l)k(2Pk) cosP-2k (1-cos2) k
(2.8)
0
which is easily transformed into
+
2k+2i
co
O.
We obtain for the first element under the sigma sign with k
(2.9)
0
2k
Equation (2.9) is of degree p; one of its roots is,
1.
k O, cos O= l, so the polynomial (2.9) is divisible by
2 2.2=
quotient polynomial is of degree p-l, its p-1 roots -A-,
9-
for
from--=--0
cos- The
712
-
M. BAICA
,
are pairwise equal
p
viz. cos
2%k
cos
hence this quotient polynomial is the perfect
2%k)
---.,
square of
(2
k
1 ,...,p- 1
a polynomial
k=l,...,.
of degree
the roots of which are cos k,
Formula (2.9) is also stated by Gauss Ill in a slightly different
form.
Another way to obtain cos
directly is obtained in the following
method investigated also by Perron for symmetric polynomial. We have
the irreducible equation (O. 1)
x-I
p-2 +
+ x
Dividing by x
0.
+ x+l
we obtain
+
+
+
+
+
(2.10)
x2+
x+
+
V
+i=0.
We obtain easily, setting
1
x+:y,
x
/
x
y2
1
x3
y
+ .p_l
/
1
y3
i
sin
+
X
But x +
of
1
(2.10)
cos+
i
cos.
is 2
sin
+
cos-
TWO EXAMPLES
We shall first investigate
.C..ase A p 5.
3.
Equation
(2.3)
yields
tan4 (52 tn2 + (54):
Setting tan2
x, we obtain
2-
O;
l Ox + 5--0
x
tan2
x
1
co’s2
1 +
5 +
2
tan2#
cs2 6+2
cos#= / 4-,1
6 + 2V,
16
2
cos=
y, so that a root
_+
Equation
hence
TRIGONOMETRIC IDENTITIES
(2.9)
24
713
yields
(_)k
5
2k+2i
-
)(k
=
cos 5-
O,
5- 20 cos3+ 5 cos#- 1 0
dividing (3.2) by cos- l, we obtain
16 cos 5#- 20 cos3#+ 5 cos#- i
cos4 +
cos
1
cos
oose
(4
cos2 +
4
+
2
4
ooscos- 1
e
)= o.
0 and
Finally we shall use formula
cos+
cos=
(2.10)
y2 y-
O
-
shall now verify deny
cos=
We have
I-
(/-l)
O,
(1.1)
1
4
y
for p
-
5 and prove, wh
(6-2’)(’-1)
8/- 16 8(V-2)
(-.2)
8(y-2) 5
(1-cos 2) 2
(2 sin2) 2
64
256
8
4
16
(1 _6-216 4 )2 .(10j5 4)2
8
2
+
4
26(1- cos4) 5(1- cos2)2(1-cosS)
5
1
cos=
5
26
i
4
(1-cos 4) 3
Hence
4
and obtain
Hence
y2_ 2+y+l
(3.2)
(-2)
8
which proves
5(+5)
8
+1)
4
4
26 55
8
8
4
4
53
(5.4).
C,as.e, #
p
7.
2
Here it will be extremely difficult to state explicitly
and the more so to verify our identity (1.13).
For we obtain
cos,
from
(2.3)
tan6 ()tan4 + ()tan2 + ()= O,
cos--=
M. BAICA
714
tan6-
tan4 + 35 tan2-
21
with
tan2
y3
21
y2
O, and
7
y,
(3.5)
O.
+ 35 y- 7
From (2.9) we obtain
-I +
64
Dividing
64
26 cosT#- F(!)+
+
0
cos7#-112 cos5#+56 cos3#-7
oosS#
+
cos#-l=
(3.6)
O.
(3.7)
(3.7) by cos#- 1 yields
.0os3#-7. cos#- 1 64 0os6#+64 cos5#
cos7#-ll2 0os5#+56
cos # 1
cos4#- os3# + 8 os2# + 8 oos # + (8 co3# + 4 co2#- 4 os #- )2
o. 8 os3# + cos2#- o #- o, ettng cos#
y3
+
y2
o.
ey
We finally get from (2. lO)
x3 +
+
+
2
x+w
cos#=
y,
(3.8)
+
x +
+ 1
O;
(3.9)
cos#=
(3.9) yields
y3
y3
y2
2 + y + 1
O,
O, which is equation
2y- 1
+
We leave it to the reader to calculate cos# from
and then to verify our identity (1.13).
3y +
y2
ACKNOWLEDGEMENT.
(3.8).
(3.5) or (3.8),
This paper is dedicated to Professor Dr. L. Carlitz.
BIBLIOGRAPHY
Ill
[2]
[3]
GAUSS, CARL FRIEDRICH, Untersuch_u._gen inter hbhere Mathematik,
Chelsea Publ. Co., Bronx, New York (905), p. 399.
PERRON, OSKAR,_AIEebra, Walter de Gruyter and Co. (1927),
vol. 2, p. 82, 83.
POLLARD, HARRY, The theory of Algebraic Numbers, The Carus
Monogr. A.M.M. 63 p 56, 5.