Irish Math. Soc. Bulletin
Number 73, Summer 2014, 21–28
ISSN 0791-5578
TWO TRIGONOMETRIC IDENTITIES
HORST ALZER AND WENCHANG CHU
Abstract. We show that the trigonometric identities
n−1
Yn
2kπ o(n−k)`+m
1 − cos
= 2(1−n)(`n/2+m) n`n+2m
n
k=1
and
n−1
Yn
cos(2θ) − cos
k=1
n sin(nθ) o`n+2m
2kπ o(n−k)`+m
= 2(1−n)(`n/2+m)
n
sin θ
are valid for all `, m ∈ Z and 2 ≤ n ∈ N. They extend the results
due to Baica and Gregorac, who proved the identities for the special
case ` = 1, m = −1. Moreover, we determine all `, m, n such that
the first trigonometric product just displayed is an integer.
In 1986, Baica [1] applied methods from cyclotomic fields to provide
a rather long and complicated proof for the following interesting
trigonometric identity:
n−1
Yn
2kπ on−k−1
1 − cos
= 2(1−n)(n/2−1) nn−2
(1)
n
k=1
where n = 2, 3, 4, · · · . Baica also remarked that “any proof avoiding
cyclotomic fields could be very difficult, if not insoluble” [1, P. 705].
In 1989, Gregorac [3] used properties of Chebyshev polynomials to
present a new proof of (1). Actually, he proved the identity
n−1
Yn
k=1
n
on−2
2kπ on−k−1
(1−n)(n/2−1) sin(nθ)
=2
cos(2θ) − cos
n
sin θ
(2)
for n = 2, 3, 4, · · · , which, letting θ tend to 0, leads to (1).
2010 Mathematics Subject Classification. 26A09, 33B10.
Key words and phrases. Trigonometric identity, finite product, sharp bound,
divisibility.
Received on 9-9-2013; revised 16-3-2014.
c
2014
Irish Mathematical Society
21
22
HORST ALZER AND WENCHANG CHU
Here, we extend (1) and (2). First, we offer an elementary short
and simple proof of a generalization of Baica’s identity. In order to
verify our result we only make use of three well-known properties of
sine and cosine,
1 − cos(2θ) = 2 sin2 θ,
sin(π − θ) = sin θ,
n−1
Y
kπ
sin
= 21−n n.
n
(3)
(4)
(5)
k=1
Formula (5) as well as many related formulas involving trigonometric
functions can be found in [2, Eq. 4.14].
We have the following extension of identity (1).
Theorem 1. Let `, m be integers and let n ≥ 2 be a natural number.
Then,
n−1
Yn
k=1
2kπ o(n−k)`+m
1 − cos
= 2(1−n)(`n/2+m) n`n+2m .
n
(6)
Proof. Applying (3) yields
n−1
n−1
Yn
Yn
2kπ o(n−k)`+m
kπ o2[(n−k)`+m]
(n−1)(`n/2+m)
=2
.
1 − cos
sin
n
n
k=1
k=1
(7)
From (4) we conclude that
n−1
Yn
k=1
n−1
n−1
kπ o(n−k)`+m Y n
(n − k)π ok`+m Y n
kπ ok`+m
=
=
.
sin
sin
sin
n
n
n
k=1
k=1
(8)
TWO TRIGONOMETRIC IDENTITIES
23
Using (8) and (5) gives
n−1
Yn
k=1
n−1
n−1
k=1
n−1
Yn
k=1
n−1
Yn
kπ o2[(n−k)`+m] Y n
kπ o(n−k)`+m Y n
kπ o(n−k)`+m
sin
sin
sin
=
n
n
n
=
=
k=1
n−1
Yn
k=1
kπ o(n−k)`+m
sin
n
k=1
kπ ok`+m
sin
n
`n+2m
kπ o`n+2m
sin
= 21−n n
.
n
Combining (7) and (9) leads to (6).
(9)
Next, we extend Gregorac’s identity (2). We need the following
formulas:
π
− θ = cos θ,
(10)
sin
2
sin(2θ) = 2 sin θ cos θ,
(11)
x+y
x−y
sin
,
2
2
(12)
Yn
kπ o
sin(nθ)
n−1
cos θ − cos
=2
.
sin θ
n
(13)
cos y − cos x = 2 sin
n−1
k=1
Identity (13) is the well-known product representation for the Chebyshev polynomials of the second kind.
Theorem 2. Let `, m be integers and let n ≥ 2 be a natural number.
Then, for θ ∈ R,
n−1
Yn
k=1
n
o`n+2m
2kπ o(n−k)`+m
(1−n)(`n/2+m) sin(nθ)
cos(2θ) − cos
=2
.
n
sin θ
(14)
Proof. Using (10) gives
n−1
Y
k=1
n−1
n−1
k=1
k=1
(n − k)π θ Y
kπ θ θ Y
sin
−
=
sin
−
=
cos
+ . (15)
2n 2
2n
2
2n 2
kπ
24
HORST ALZER AND WENCHANG CHU
Now, we apply (12), (15) and (11). Then we have
n−1
Yn
k=1
n−1 kπ θ kπ θ kπ o Y
cos θ − cos
2 sin
=
+
sin
−
n
2n 2
2n 2
k=1
n−1
kπ θ kπ θ Y
2 sin
=
+
cos
+
2n 2
2n 2
=
k=1
n−1
Y
sin
kπ
+θ .
n
k=1
(16)
From (4) and (12) we obtain
n−1
Y
k=1
sin2
kπ
n
+θ =
=
n−1
Y
k=1
n−1
Y
sin
sin
kπ
=2
kπ
+ θ sin
n
n−1
Yn
k=1
+ θ sin
n
k=1
1−n
(n − k)π
n
kπ
n
−θ
−θ
2kπ o
cos(2θ) − cos
.
n
(17)
Applying (13), (16) and (17) yields
n sin(nθ) o2m
sin θ
2m(n−1)
n−1
Yn
2m(n−1)
k=1
n−1
Y
=2
=2
m(n−1)
=2
2m
sin
k=1
n−1
Yn
k=1
kπ o2m
cos θ − cos
n
kπ
n
+θ
2kπ om
cos(2θ) − cos
.
n
(18)
TWO TRIGONOMETRIC IDENTITIES
25
From (4) and (12) we get
n−1
Y
sinn
k=1
kπ
n
+θ =
=
=
n−1
Y
k=1
n−1
Y
k=1
n−1
Y
sinn−k
k
sin
+ θ sink
(n − k)π
kπ
n
k=1
=2
n
n
sink
(1−n)n/2
kπ
n−1
Yn
k=1
kπ
n
k
+ θ sin
− θ sink
kπ
n
+θ
kπ
n
+θ
+θ
2kπ ok
cos(2θ) − cos
.
n
(19)
Combining (13), (16) and (19) gives
n sin(nθ) on
sin θ
(n−1)n/2
=2
n−1
Yn
k=1
2kπ ok
cos(2θ) − cos
.
n
(20)
Finally, (18) and (20) lead to
n sin(nθ) o2m+`n
sin θ
= 2
= 2
(n−1)(`n/2+m)
n−1
Yn
(n−1)(`n/2+m)
k=1
n−1
Yn
k=1
2kπ om+k`
cos(2θ) − cos
n
2kπ o(n−k)`+m
cos(2θ) − cos
.
n
This is equivalent to (14).
Remark 1. Setting ` = 1 and m = −1 in (6) and (14), respectively,
gives (1) and (2).
Remark 2. Let `, m ∈ Z and 2 ≤ n ∈ N with `n + 2m > 0.
Applying (14) and the well-known inequality
sin(nθ) ≤ 1 (n = 1, 2, 3, ...)
n sin θ
we obtain for all θ ∈ R:
n−1
Yn
k=1
2kπ o(n−k)`+m
cos(2θ) − cos
≤ 2(1−n)(`n/2+m) n`n+2m .
n
(21)
26
HORST ALZER AND WENCHANG CHU
Setting θ = 0 we conclude from (6) that the given upper bound is
sharp. If `n + 2m < 0, then (21) holds with “≥” instead of “≤”.
The representation (6) reveals that if `, m ∈ Z, 2 ≤ n ∈ N, then the
product
n−1
Yn
2kπ o(n−k)`+m
1 − cos
Pn (`, m) =
n
k=1
is a rational number. In view of this result it is natural to ask
whether there exist numbers `, m, n such that Pn (`, m) is an integer.
The next theorem answers this question.
Theorem 3. Let `, m be integers and n ≥ 2 a natural number. The
product Pn (`, m) is an integer if and only if `n + 2m = 0
or
or
`n + 2m > 0 with n = 2r
`n + 2m < 0 with n = 2r
(r = 1, 2);
(3 ≤ r ∈ N).
(22)
(23)
Proof. Using (6) we obtain:
if `n + 2m = 0, then Pn (`, m) = 1;
if n = 2r (r = 1, 2) and `n + 2m > 0, then
Pn (`, m) = 2(`n+2m)/2 ∈ Z;
if n = 2r (r ≥ 3) and `n + 2m < 0, then
Pn (`, m) = 2−(2
r
−2r−1)(`n+2m)/2
∈ Z.
Now, let Pn (`, m) ∈ Z. We asssume (for a contradiction) that none
of (22), (23) and `n + 2m = 0 is satisfied. We have
P2 (`, m) = 2`+m ,
n 3 o3`+2m
,
P3 (`, m) =
2
P4 (`, m) = 22`+m ,
n 5 o5`+2m
P5 (`, m) =
.
4
Case 1. `n + 2m > 0.
Then, P3 (`, m) ∈
/ Z and P5 (`, m) ∈
/ Z. Let n ≥ 6. From
2(n−1)(`n+2m)/2 · K = n`n+2m
(K ∈ N)
(24)
TWO TRIGONOMETRIC IDENTITIES
27
we conclude that 2 divides n`n+2m . This implies that n is even. Let
n = 2r q, where r ≥ 1 and q is odd. Then, (24) leads to
2((n−1)/2−r)(`n+2m) · K = q `n+2m .
Since q is odd, we obtain
n−1
− r ≤ 0.
2
(25)
Hence,
2r ≤ 2r q = n ≤ 2r + 1.
If follows that r = 1 or r = 2. However, this contradicts (25), since
n ≥ 6.
Case 2. `n + 2m < 0.
Then, Pn (`, m) ∈
/ Z for n = 2, 3, 4, 5. Let n ≥ 6. From (24) we
obtain
n−(`n+2m) · K = 2−(n−1)(`n+2m)/2 .
This yields n = 2r with r ≥ 3. A contradiction. The proof is
complete.
Acknowledgement. We thank the referee for a careful reading of
the manuscript.
References
[1] M. Baica, Trigonometric identities, Intern. J. Math. Math. Sci. 9 (1986),
705–714.
[2] H.W. Gould, Combinatorial identities: Table III: Binomial identities derived
from trigonometric and exponential series, www.math.wvu.edu/∼gould.
[3] R.J. Gregorac, On Baica’s trigonometric identity, Intern. J. Math. Math.
Sci. 12 (1989), 119–122.
Horst Alzer received his doctoral degree in mathematics from the University of
Bonn and his habilitation in mathematics from the University of urzburg. His
main interest of research are inequalities and special functions.
Wenchang Chu received, in July 1987, his PhD in Combinatorial mathematics
under the supervision of Prof. L. C. Hsu at Dalian University of Technology
(P.R.China). Since then he has been employed in Academia Sinica (Beijing),
Università di Roma La Sapienza, Freie Universität Berlin, INRIA (Paris), and
28
HORST ALZER AND WENCHANG CHU
currently in Salento University (Italy). His research interests cover broadly the
interaction among classical analysis, combinatorics and special functions.
(Alzer) Morsbacher Str. 10, 51545 Waldbröl, Germany
E-mail address, Alzer: H.Alzer@gmx.de
(Chu) Dipartimento di Matematica e Fisica “Ennio De Giorgi”, Università del Salento, Lecce-Arnesano, P. O. Box 193, 73100 Lecce,
Italy
E-mail address, Chu: chu.wenchang@unisalento.it