Download SOME TRIGONOMETRIC IDENTITIES RELATED TO POWERS OF

Folia Mathematica
Vol. 15, No. 1, pp. 41–62
Acta Universitatis Lodziensis
c 2008 for University of Lódź Press
SOME TRIGONOMETRIC IDENTITIES RELATED TO
POWERS OF COSINE AND SINE FUNCTIONS
ROMAN WITULA† , DAMIAN SLOTA††‡
Abstract. Some decomposition of certain basic symmetric functions of n-th
power of cosine and sine functions is presented here. Next the applications
of these decompositions to generating many new binomial and trigonometric
identities are discussed.
1. Introduction
In this paper we wish to investigate different trigonometric identities connected with the following trigonometric identities previously presented in [8]:
sin2n (x) + cos2n x − π6 + cos2n x + π6 + cos2n (x)+
+ cos2n x − π3 + cos2n x + π3 ≡ const ⇔ n = 1, 2, . . . , 5.
(1)
It is natural to relate these purpose to decompositions of the following four
symmetric functions of the n-th powers of cosine and sine functions:
(2)
Cn+ (x, ϕ) := cosn (x + ϕ) + cosn (x − ϕ),
(3)
Cn− (x, ϕ) := cosn (x − ϕ) − cosn (x + ϕ),
(4)
Sn+ (x, ϕ) := sinn (x + ϕ) + sinn (x − ϕ),
(5)
Sn− (x, ϕ) := sinn (x + ϕ) − sinn (x − ϕ).
for n ∈ N. The term ”decomposition” means here the trigonometric polynomial form of these functions. We note that some other decomposition of
the functions (2)–(5) using the Chebyshev polynomial of the second kind in
the separate paper [7] is discussed.
The basic decompositions of functions (2)–(5) are presented in Section 2
and in two next sections are applied to generating of trigonometric identities, starting from the simplest forms (see Section 3 and Lemma 8) to
‡
Institute of Mathematics, Silesian University of Technology, Kaszubska 23, Gliwice
44-100. Poland. E-mail: [email protected].
‡‡
Institute of Mathematics, Silesian University of Technology, Kaszubska 23, Gliwice
44-100. Poland. E-mail: [email protected].
AMS subject classifications: 26C05, 11B37, 11B83.
42
Roman Witula, Damian Slota
main identities shown in Lemma 10 and Corollaries 8, 10, 11, 12 and 14,
respectively.
Moreover at the end of the Section 2 an interesting application of these
decompositions to obtained some nontrivial binomial identities is also given.
2. The basic decomposition
Lemma 1 (see [3]). The following two classical identities hold
n−2
2
Cn+ (x, ϕ)
(6)
bn/2c n
=
cos (n − 2k)ϕ cos (n − 2k)x −
k
k=0
j k 1
n
n
n−1
−
−
2
2
2
b n2 c
X
and
2n−2 Cn− (x, ϕ) =
(7)
b(n−1)/2c X
k=0
n
sin (n − 2k)ϕ sin (n − 2k)x .
k
The proof of the above identities by induction follows and will be omitted
here.
The sequence of the following identities can easily be deduced from identities (6) and (7) (the formulas (12)–(14) are known see [1], the formula (15)
is probably new):
+
22n−2 S2n
(x, ϕ) =
n
X
1 2n
2n
=
(−1)
cos 2(n − k)ϕ cos 2(n − k)x −
k
2 n
k=0
n−1
X
1 2n
2n
;
=
(−1)n−k
cos 2(n − k)ϕ cos 2(n − k)x +
2 n
k
(8)
n−k
k=0
+
22n−3 S2n−1
(x, ϕ) =
(9)
=
n−1
X
(−1)n−k−1
2n − 1
cos (2n − 2k − 1)ϕ sin (2n − 2k − 1)x ;
k
2n − 1
sin (2n − 2k − 1)ϕ cos (2n − 2k − 1)x ;
k
k=0
−
22n−3 S2n−1
(x, ϕ) =
(10)
=
n−1
X
k=0
n−k−1
(−1)
Some Trigonometric Identities Related to...
(11)
−
22n−2 S2n
(x, ϕ) =
n−1
X
(−1)n−k−1
k=0
(12)
22n−2 sin2n−1 (x) =
n−1
X
k=0
(13)
43
2n
sin 2(n − k)ϕ sin 2(n − k)x ;
k
(−1)n−k−1
2n − 1
sin (2n − 2k − 1)x ;
k
+
22n−1 sin2n (x) = 22n−2 S2n
(x, 0) =
n
X
1 2n
2n
;
=
(−1)n−k
cos 2(n − k)x −
2 n
k
k=0
(14)
2n−1 cosn (x) = 2n−2 Cn+ (x, 0) =
bn/2c X n
1 jnk
n−1
n
−
;
=
cos (n − 2k)x −
2
2
2
bn/2c
k
k=0
+
22n−2 C2n
x + π8 , π4 = cos2n x − π8 + cos2n x + 83 π =
n X
1 2n
2n
=
=
cos (n − k) π2 cos 2(n − k)x + (n − k) π4 −
2 n
k
k=0
2n
2n
1 2n
−
cos 4x + π2 +
cos 8x + π −
=
n−2
2 n
n−4
2n
2n
−
cos 12x + 23 π +
cos 16x + 2π − . . .
n−6
n−8
2n
2n
1 2n
+
sin(4x) −
cos(8x) +
=
2 n
n−2
n−4
2n
2n
+
sin(12x) +
cos(16x) +
n−6
n−8
2n
2n
+
sin(20x) −
cos(24x) +
n − 10
n − 12
2n
2n
(15)
+
sin(28x) +
cos(32x) + . . .
n − 16
n − 14
In the Tables 1 and 2 below the first seven decompositions of the type (6)–
(11) of every function Cn+ , Cn− , Sn+ and Sn− are presented.
44
Roman Witula, Damian Slota
Table. 1. The first seven decompositions of every function Cn+ and Cn−
n
+
(x, ϕ)
Cn
−
(x, ϕ)
Cn
1
2 cos(ϕ) cos(x)
2 sin(ϕ) sin(x)
2
1 + cos(2ϕ) cos(2x)
`
1
3 cos(ϕ) cos(x) + cos(3ϕ) cos(3x))
2
`
´
1
3 + 4 cos(2ϕ) cos(2x) + cos(4ϕ) cos(4x)
4
`
1
10 cos(ϕ) cos(x) + 5 cos(3ϕ) cos(3x)+
8
´
+ cos(5ϕ) cos(5x)
`
1
10 + 15 cos(2ϕ) cos(2x)+
16
´
+6 cos(4ϕ) cos(4x) + cos(6ϕ) cos(6x)
`
1
35 cos(ϕ) cos(x) + 21 cos(3ϕ) cos(3x)+
32
´
+7 cos(5ϕ) cos(5x) + cos(7ϕ) cos(7x)
sin(2ϕ) sin(2x)
`
1
3 sin(ϕ) sin(x) + sin(3ϕ) sin(3x))
2
`
´
1
4 sin(2ϕ) sin(2x) + sin(4ϕ) sin(4x)
4
`
1
10 sin(ϕ) sin(x) + 5 sin(3ϕ) sin(3x)+
8
´
+ sin(5ϕ) sin(5x)
`
1
15 sin(2ϕ) sin(2x)+
16
´
+6 sin(4ϕ) sin(4x) + sin(6ϕ) sin(6x)
`
1
35 sin(ϕ) sin(x) + 21 sin(3ϕ) sin(3x)+
32
´
+7 sin(5ϕ) sin(5x) + sin(7ϕ) sin(7x)
3
4
5
6
7
Table. 2. The first seven decompositions of every function Sn+ and Sn−
n
+
Sn
(x, ϕ)
−
Sn
(x, ϕ)
1
2 cos(ϕ) sin(x)
2 sin(ϕ) cos(x)
2
1 − cos(2ϕ) cos(2x)
`
1
3 cos(ϕ) sin(x) − cos(3ϕ) sin(3x))
2
`
´
1
3 − 4 cos(2ϕ) cos(2x) + cos(4ϕ) cos(4x)
4
`
1
10 cos(ϕ) sin(x) − 5 cos(3ϕ) sin(3x)+
8
´
+ cos(5ϕ) sin(5x)
`
1
10 − 15 cos(2ϕ) cos(2x)+
16
´
+6 cos(4ϕ) cos(4x) − cos(6ϕ) cos(6x)
`
1
35 cos(ϕ) sin(x) − 21 cos(3ϕ) sin(3x)+
32
´
+7 cos(5ϕ) sin(5x) − cos(7ϕ) sin(7x)
sin(2ϕ) sin(2x)
`
1
3 sin(ϕ) cos(x) − sin(3ϕ) cos(3x))
2
`
´
1
4 sin(2ϕ) sin(2x) − sin(4ϕ) sin(4x)
4
`
1
10 sin(ϕ) cos(x) − 5 sin(3ϕ) cos(3x)+
8
´
+ sin(5ϕ) cos(5x)
`
1
15 sin(2ϕ) sin(2x)−
16
´
−6 sin(4ϕ) sin(4x) + sin(6ϕ) sin(6x)
`
1
35 sin(ϕ) cos(x) − 21 sin(3ϕ) cos(3x)+
32
´
+7 sin(5ϕ) cos(5x) − sin(7ϕ) cos(7x)
3
4
5
6
7
At the end of this section it will be given the application some of above
decompositions to generating an interesting sets of binomial identities (see
also [2, 4]).
Corollary 1. Since for x, ϕ → 0 we have
n n
(x + ϕ)2
(x − ϕ)2
−
Cn (x, ϕ) = 1 −
+ ... − 1 −
+ ...
≈
2!
2!
n
≈
(x + ϕ)2 − (x − ϕ)2 = 2 n x ϕ,
2
Some Trigonometric Identities Related to...
45
so by (6) the following formula hold:
lim 2
x,ϕ→0
−
n−2 Cn (x, ϕ)
xϕ
b(n−1)/2c X
= lim
x,ϕ→0
k=0
n
k
sin (n − 2k) ϕ sin (n − 2k) x
,
ϕ
x
i.e.,
n−1
n2
b(n−1)/2c =
X
k=0
n
k
2
n − 2k .
Corollary 2. From (12) it can be easily derived the formula
2n+1
2n sin(x)
2
=
x
∞ X
n
2r−1
X
n−k+r 2 n + 1 (2 n − 2 k + 1)
=
(−1)
x2(r−n−1) ,
k
(2 r − 1)!
r=1 k=0
hence we get the sequence of the following binomial identities:
n
X
2r−1
k 2n + 1
(−1)
2n − 2k + 1
=0
k
k=0
for r = 1, 2, . . . , n,
n
X
2n+1
2n + 1
2n − 2k + 1
= 22n (2 n + 1)!,
(−1)k
k
k=0
n
X
2n+3 1
k 2n + 1
2n − 2k + 1
= (2 n + 1) 22n−1 (2 n + 3)!,
(−1)
3
k
k=0
n
X
2n+5 10 n + 3
k 2n + 1
(2 n + 1) 22n (2 n + 5)!,
(−1)
2n − 2k + 1
=
k
360
k=0
etc., since we have:
sin(x) n
n
n (5 n − 2) 4
(16)
= 1 − x2 +
x + ...
x
6
360
Similarly, from (13) we deduce the formula
2n
1 2 n −2n
2n−1 sin(x)
2
+
x
=
x
2 n
2r
∞ X
n
X
2 (n − k)
n−k+r 2 n
=
(−1)
x2(r−n) ,
k
(2 r)!
r=0 k=0
46
Roman Witula, Damian Slota
which by (16) implies the second sequence of the following binomial identities:
n
X
2n
1 2n
=
(−1)n−k
,
k
2 n
k=0
n−1
X
k 2n
(−1)
(n − k)2r = 0,
k
k=0
for r = 1, 2, . . . , n − 1,
n−1
X
2n 1
k 2n
(−1)
n−k
= (2 n)!,
2
k
k=0
n−1
X
2n+2
n
k 2n
(−1)
n−k
=
(2 n + 2)!,
k
24
k=0
n−1
X
2n+4 n (5 n − 1)
k 2n
(2 n + 4)!,
(−1)
n−k
=
2880
k
k=0
etc.
Corollary 3. From (14) we obtain
2
n−1
n 2 n (3 n − 2) 4
x + ... =
1− x +
2
24
X
∞ bn/2c
2r
X
1 j n − 1 k j n k
n
2r n (n − 2 k)
=
−
x2r ,
+
(−1)
2
2
2
(2
r)!
bn/2c
k
r=0
k=0
which implies:
bn/2c X
k=0
bn/2c X
k=0
n
k
n
k
n − 2k
2
= n 2n−1 ,
n − 2k
4
= n (3 n − 2) 2n−1 ,
etc.
Corollary 4. We have the formula (see [4, 6]):
bn/2c
Pn (x) = 2−n
X
k=0
(−1)k
n−k
k
2n − 2k
xn−2k ,
n−k
Some Trigonometric Identities Related to...
47
where Pn (x) denotes the n-th Legendre polynomial. Hence, by (14) for even
n ∈ N, we get
Z π
n
X
4n − 2k
2n − 2k
−4n
k 2n − k
P2n cos ϕ dϕ = π 2
(−4)
k
2n − k
n−k
0
k=0
n
X
4n − 2k !
= π 2−4n
(17)
(−4)k
.
2
k! (n − k)! (2 n − k)!
k=0
On the other hand, we have ([6]):
2 n−1
X
−4n 2 n
P2n cos ϕ = 2
+
ak,n cos 2 n − 2 k
n
k=0
for some ak,n ∈ Q, which by (17) implies the identity:
2 X
n
4n − 2k !
2n
k
=
(−4)
2
n
k! (n − k)! (2 n − k)!
k=0
n
X
4n − 2k
2n − 2k
k 2n − k
=
(−4)
.
k
2n − k
n−k
k=0
3. Simple trigonometric identities
We mention only the simple form of trigonometric identities to discuss
here. Simultaneously this approach leads in Section 4 to direct our considerations to only symmetric trigonometric identities with respect to the
phase translations.
Lemma 2. Fix a, ϕ, ψ ∈ R. Then
sin2 x
fa,ϕ,ψ (x) := a
+ sin2 (x + ϕ) + sin2 (x + ψ) ≡ const
cos2 x
(fa,ϕ,ψ (x) is independent of x under this values a, ϕ, ψ) iff either:
ϕ − ψ = (2k + 1) π2
and
a=0
or
ϕ+ψ = kπ
and
a=
−2 cos(2ϕ)
2 cos(2ϕ)
.
Proof. First, we note that:
0
0
fa,ϕ,ψ (x) ≡ const =⇒ fa,ϕ,ψ
(x) ≡ 0 =⇒ fa,ϕ,ψ
(0) = 0.
48
Roman Witula, Damian Slota
Hence, we get:
sin(2ϕ) + sin(2ψ) = 0 ⇐⇒ sin(ϕ + ψ) cos(ϕ − ψ) = 0 ⇐⇒
⇐⇒ ϕ + ψ = k π
or ϕ − ψ = (2k + 1) π2
for some k ∈ Z.
If ϕ − ψ = (2k + 1) π2 then:
sin2 x
cos2 x
fa,ϕ,ψ (x) := a
+1
which implies a = 0. If ϕ + ψ = k π then, we have:
fa,ϕ,ψ (x) =
a sin2 x + S2+ (x, ϕ)
a cos2 x + S2+ (x, ϕ)
=
2 cos(2ϕ) + a sin2 x + 2 sin2 (ϕ)
a − 2 cos(2ϕ) cos2 x + 2 cos2 (ϕ)
and the proof is completed.
Corollary 5. We have:
sin2 x
a
+ cos2 (x + ϕ) + cos2 (x + ψ) ≡ const
cos2 x
iff either:
ϕ − ψ = (2k + 1) π2
and
a=0
or
ϕ+ψ = kπ
Proof. Set ϕ := ϕ +
π
2
and
and ψ := ψ +
a=
π
2
−2 cos(2ϕ)
2 cos(2ϕ)
.
in the Lemma 2.
Lemma 3. We have:
1) a cos4 (x) + S4+ (x, ϕ) ≡ const
iff either a = 1 and ϕ = ± π6 + k π, k ∈ Z
or a = −2 and ϕ = ± π2 + k π, k ∈ Z;
2) a sin4 (x) + S4+ (x, ϕ) ≡ const
iff either a = 1 and ϕ = ± π3 + k π,
or a = −2 and ϕ = k π, k ∈ Z.
k∈Z
Proof. The assertion from the following decomposition follows:
1)
a cos4 (x) + S4+ (x, ϕ) =
4 cos2 (ϕ) − 3 4 cos2 (ϕ) − 1 + a − 1 cos4 (x)−
− 4 cos2 (ϕ) 4 cos2 (ϕ) − 3 cos2 (x) + 2 cos4 (ϕ).
Some Trigonometric Identities Related to...
49
2)
a sin4 (x) + S4+ (x, ϕ) =
4 cos2 (ϕ) − 3 4 cos2 (ϕ) − 1 + a − 1 sin4 (x) +
+ 4 sin2 (ϕ) 4 cos2 (ϕ) − 1 sin2 (x) + 2 sin4 (ϕ).
Let us set:
ga,ϕ,ψ (x) := a cos3 (x) + cos3 (x + ϕ) + cos3 (x + ψ).
Lemma 4. We have:
1) ga,ϕ,ψ (x) ≡ const (is independent on x)
=⇒
either ϕ + ψ = 2 k π for some k ∈ Z
or a = 0 and ϕ − ψ = (2k + 1) π for some k ∈ Z.
=⇒
2) ga,ϕ,−ϕ (x) ≡ const
⇐⇒
⇐⇒
either a = 2 (−1)l+1 and ϕ = l π, l ∈ Z,
or a = 0 and ϕ = (2 l + 1) π2 , l ∈ Z.
Proof. 1) We have:
ga,ϕ,ψ π2 = ga,ϕ,ψ − π2 ⇔ sin3 (ϕ) + sin3 (ψ) = 0 ⇔
ϕ + ψ ϕ − ψ ⇔ sin(ϕ) + sin(ψ) = 0 ⇔ sin
cos
=0⇔
2
2
⇔ ϕ + ψ = 2 k π ∨ ϕ − ψ = (2 k + 1) π, k ∈ Z
and
ga,ϕ,ψ − ϕ = ga,ϕ,ψ − ψ ⇔ 2 a cos3 (ϕ) − cos3 (ψ) = 0 ⇔
⇔ a = 0 ∨ cos(ϕ) − cos(ψ) = 0 ⇔
ϕ + ψ ϕ − ψ ⇔ a = 0 ∨ sin
sin
=0 ⇔
2
2
⇔ a = 0 ∨ ϕ + ψ = 2 l π ∨ ϕ − ψ = 2 l π, l ∈ Z.
Hence, we conclude that either ϕ + ψ = 2 k π, k ∈ Z, or ϕ − ψ = (2 k + 1) π,
k ∈ Z and a = 0. In the second case ga,ϕ,ψ (x) ≡ 0.
2) By (6) we get:
ga,ϕ,−ϕ (x) = C3+ (x, ϕ) +
=
1
4
C3+ (x, 0) =
2 cos(3 ϕ) + a cos(3 x) +
a
2
3
4
2 cos(ϕ) + a cos(x),
50
Roman Witula, Damian Slota
which is independent on x iff
2 cos(3 ϕ) + a = 0
a = −2 cos(ϕ)
⇔
⇔
2 cos(ϕ) + a = 0
cos(3 ϕ) − cos(ϕ) = 0
a = −2 cos(ϕ)
a = −2 cos(ϕ)
⇔
⇔
⇔
sin(ϕ) sin(2 ϕ) = 0
sin(2 ϕ) = 0
a = −2 cos(ϕ)
either a = 2 (−1)l+1 and ϕ = l π, l ∈ Z,
⇔
⇔
π
ϕ = k 2 for some k ∈ Z
or a = 0 and ϕ = (2 l + 1) π2 , l ∈ Z.
Now, let us set:
ha,ϕ (x) := 2 a cos3 (x) + cos3 (x − ϕ) + cos3 (x + ϕ)
= a C3+ (x, 0) + C3+ (x, ϕ).
Lemma 5. We have:
ha,ϕ (x)
ha,ϕ (x)
1)
≡ const ⇔
≡ 6 sin2 (ϕ) cos(ϕ) ⇔
cos(x)
cos(x)
⇔ a + T3 cos(ϕ) = 0 ⇔ a = − cos(3ϕ);
q
q
p
p
3
3
2) 2 t(a) := a + a2 − 1 + a − a2 − 1 ⇒ T3 t(a) = a;
3) t(a) = cos(ϕ) for some a, ϕ ∈ R ⇔ |a| = 1 ⇔
a = (−1)k+1 and ϕ = k π, k ∈ Z,
where T3 (x) denotes the third Chebyshev polynomial of the first kind (see [5]).
Proof. 1) It follows from (6) for n = 3:
ha,ϕ (x)
=
cos(x)
1
2
⇔
cos(3 x) 3
cos(3 ϕ) + a
+ 2 cos(ϕ) + a ≡ const ⇔
cos(x)
ha,ϕ (x)
≡ 32 cos(ϕ) − cos(3 ϕ) ,
a = − cos(3 ϕ) and
cos(x)
which implies 1).
2) We have
i
h
2
3
a+ + a− − 3 =
2 T3 t(a) = a+ + a− − 3 a+ + a− = a+ + a−
h
2
2 i
3
3
= a+ + a−
a+ − 1 + a−
= a+ + a− = 2 a.
where a+ :=
p
3
a+
√
a2 − 1 and a− :=
p
3
a−
√
a2 − 1.
Some Trigonometric Identities Related to...
51
3) We note that t(a) is an odd function. Moreover we have:
q
q
p
p
3
3
1
t0 (a) = √
a + a2 − 1 − a − a2 − 1 .
6 a2 − 1
Hence, we deduce that t(a) is a decreasing function on (−∞, −1] and consequently an increasing one on [1, ∞). Should also be noticed, that t(1) = 1.
Lemma 6. We have:
C5+ (x, ϕ)
≡ const
cos(x)
⇐⇒
ϕ = (2 k + 1) π2 ,
k ∈ Z.
Proof. Since
C5+ (x, ϕ) = 2 cos(5ϕ) cos5 (x) + 5 sin(ϕ) sin(4ϕ) cos3 (x) +
+ 10 sin4 (ϕ) cos(ϕ) cos(x),
so we obtain:
C5+ (x, ϕ)
cos(5ϕ) = 0
≡ const ⇐⇒
sin(4ϕ) = 0
cos(x)
π
⇐⇒ ϕ = (2 k + 1) ,
2
k ∈ Z.
Now, let us set:
Fa,ϕ,ψ (x) := a cos6 (x) + cos6 (x + ϕ) + cos6 (x + ψ).
Lemma 7. We have:
Fa,ϕ,ψ (x) ≡ const ⇐⇒
⇐⇒ a = −2 and ϕ = k π and ψ = l π
for some k, l ∈ Z.
0
(x) ≡ 0, i.e.:
Proof. If Fa,ϕ,ψ (x) ≡ const then Fa,ϕ,ψ
a cos5 (x) sin(x) + cos5 (x + ϕ) sin(x + ϕ) + cos5 (x + ψ) sin(x + ψ) ≡ 0.
Hence, for x = 0 and x = π2 , respectively, we get:
(18)
cos5 (ϕ) sin(ϕ) + cos5 (ψ) sin(ψ) = 0,
(19)
sin5 (ϕ) cos(ϕ) + sin5 (ψ) cos(ψ) = 0.
Subtracting (19) from (18), we obtain:
sin(4ϕ) + sin(4ψ) = 0,
i.e.
sin 2(ϕ + ψ) cos 2(ϕ − ψ) = 0
which implies that either
(20)
ϕ+ψ =k
π
2
52
Roman Witula, Damian Slota
or
ϕ−ψ =
(21)
π
4
+k
π
2
for some k ∈ Z.
On the other hand, if we realize the operation: sin4 (ϕ)·(18)− cos4 (ϕ)·(19),
i.e.
sin4 (ϕ) cos5 (ψ) sin(ψ) − cos4 (ϕ) sin5 (ψ) cos(ψ) = 0,
(22) sin(2ψ) sin(ϕ+ψ) sin(ϕ−ψ) sin2 (ϕ) cos2 (ψ)+cos2 (ϕ) sin2 (ψ) = 0,
and operation: sin4 (ψ)·(18) − cos4 (ψ)·(19), i.e.
sin4 (ψ) cos5 (ϕ) sin(ϕ) − cos4 (ψ) sin5 (ϕ) cos(ϕ) = 0,
(23) sin(2ϕ) sin(ϕ + ψ) sin(ϕ − ψ) sin2 (ϕ) cos2 (ψ) + cos2 (ϕ) sin2 (ψ) = 0
and we assume that (21) holds, then from (22) and (23) we get sin(ϕ + ψ) =
0, i.e.:
ϕ + ψ = l π,
ϕ − ψ = π4 + k π2
for some k, l ∈ Z, so:
ϕ=
π
8
+ l π2 + k
π
4
and ψ = l π2 − k
π
4
− π8 .
Hence, we obtain:
Fa,ϕ,ψ (x) = a cos6 (x) + C6+ x + l π2 , k π4 + π8 =
1 6 6
6
−5
=2 a
+
cos(2x) +
cos(4x) + cos(6x) +
2 3
2
1
−4
10 + 15 cos π4 + k π2 cos 2 x + l π +
+2
+ cos 43 π + 32 k π cos 6 x + 3 l π =
5(a + 2)
−5
l −4
π
π
=
+
15
2
a
+
(−1)
2
cos
+
k
cos(2x)+
4
2
24
3
+ 4 a cos(4x) + a 2−5 + (−1)l cos 43 π + 32 k π cos(6x)
2
which, from the linear independence of trigonometric system, implies a =
0 and, in consequence, Fa,ϕ,ψ (x) is not const, contrary to our assumptions.
If we suppose now that (20) holds, i.e. ϕ + ψ = π2 + l π, l ∈ Z, then:
Fa,ϕ,ψ (x) := a cos6 (x) + cos6 (x + ϕ) + sin6 (x − ϕ)
and
Fa,ϕ,ψ (ϕ) = Fa,ϕ,ψ (−ϕ)
Some Trigonometric Identities Related to...
53
i.e.:
cos6 (2ϕ) = 1 + sin6 (2ϕ) =⇒ sin(2ϕ) = 0 ⇐⇒ ϕ = k π2 , k ∈ Z.
Hence:
Fa,ϕ,ψ (x) = (a + 1) cos6 (x) + sin6 (x) 6≡ const .
Next let us assume that ϕ + ψ = l π, l ∈ Z, then:
Fa,ϕ,ψ (x) = a cos6 (x) + C6+ (x, ϕ) =
= 2−5 a + 2 cos(6ϕ) cos(6x) + 3 · 2−5 a + 2 cos(4ϕ) cos(4x) +
+ 15 · 2−5 a + 2 cos(2ϕ) cos(2x) + . . .
hence
Fa,ϕ,ψ (x) ≡ const
(24)
a = −2 cos(2ϕ),
cos(6ϕ) = cos(4ϕ) = cos(2ϕ).
2 (cos(t) − 1) (cos(t) + 21 ) = 0,
cos(t) (cos(t) − 1) (cos(t) + 1) = 0,
⇐⇒
We note that
cos(t) = cos(2t) = cos(3t)
⇐⇒
which implies cos(t) = 1. Thus, from (24) it follows that:
a = −2 ∧ cos(2ϕ) = 1
⇐⇒
a = −2 ∧ ϕ = k π,
k ∈ Z.
Now, let us set:
fn (x) := sin2n (x) + cos2n x −
gn (x) := cos2n (x) + cos2n x −
2n
π
6 + cos
2n
π
3 + cos
π
6 ,
π
3 .
x+
x+
We have the following identities:
n
1
2
3
4
5
6
fn (x)
3
2
9
8
`
´
3
10 − cos(6x)
32
`
´
3
35 − 8 cos(6x)
128
`
´
27
14 − 5 cos(6x)
512
`
3
462 − 220 cos(6x)
2048
gn (x)
+ cos(12x)
´
3
2
9
8
`
´
3
10 + cos(6x)
32
`
´
3
35 + 8 cos(6x)
128
`
´
27
14 + 5 cos(6x)
512
`
3
462 + 220 cos(6x)
2048
+ cos(12x)
´
54
Roman Witula, Damian Slota
n
fn (x) + gn (x)
n
1
3
7
2
9
4
15
8
105
64
189
128
`
3
462
1024
8
3
4
5
6
9
10
11
+ cos(12x)
fn (x) + gn (x)
3
858 + 7 cos(12x)
2048
`
´
45
429 + 8 cos(12x)
16384
`
´
51
715 + 24 cos(12x)
32768
`
´
969
286 + 15 cos(12x)
262144
`
´
1197
442 + 33 cos(12x)
524288
`
´
´
The form of fn (x) + gn (x) for n ≥ 12 is more complicated (there are at
least three terms in the respective decomposition), for example we have:
3
f12 (x) + g12 (x) = 4194304
6118 221 + 22 cos(12 x) + cos(24 x) .
Remark 1. We have also
n
n
C2+ x, π6
+ C2+ x, π3
−
n
− (−1)n + 1 21 cos(2 x) + n (−1)n − 1

0,
for n = 1,



2,
for n = 2, 3,
=
2 (2 x),
2
+
3
cos
for n = 4,



2 + 5 cos2 (2 x),
for n = 5.
1
2
n−1
cos(2 x)
=
The next result indicates the direction in which attempts at generalizing
certain results from Section 3 should follow.
Lemma 8. Let ϕ, ψ ∈ R and
Θn (x) = Cn+ (x, 0) + 2 Cn+ (x, ϕ) + 2 Cn+ (x, ψ),
n ∈ N, x ∈ R.
If Θn (x) ≡ const for some two different values of n ∈ N and at least for one
odd value of n ∈ N, then
Θn (x) = Cn+ (x, 0) + 2 Cn+ (x, 52 π) + 2 Cn+ (x, 54 π),
n ∈ N, x ∈ R.
If Θn (x) ≡ const, for two different even values of n ∈ N, then
+
+
+
(x, 25 π) + 2 C2n
(x, 45 π),
(x, 0) + 2 C2n
Θ2n (x) = C2n
n ∈ N, x ∈ R.
Proof. Directly from decomposition (6) it follows, that if
Θk (x) ≡ const
and
Θl (x) ≡ const
for k, l ∈ N, k < l, so, in view of the linear independence of the trigonometric
system one of the following three conditions holds:
(
cos(ϕ) + cos(ψ) = − 21 ,
(25)
cos(2ϕ) + cos(2ψ) = − 12 ,
Some Trigonometric Identities Related to...
55
whenever (−1)k + (−1)l = 0; or
(
cos(ϕ) + cos(ψ) = − 21 ,
(26)
cos(3ϕ) + cos(3ψ) = − 12 ,
whenever (−1)k + (−1)l = −2; or
(
cos(2ϕ) + cos(2ψ) = − 21 ,
(27)
cos(4ϕ) + cos(4ψ) = − 12 ,
whenever (−1)k + (−1)l = 2.
Ad (25) The given system implies:
(
cos2 (ϕ) + 2 cos(ϕ) cos(ψ) + cos2 (ψ) = 41 ,
cos2 (ϕ) + cos2 (ψ) = 34 ,
i.e.
cos(ϕ) cos(ψ) = − 41 .
So, system (25) is equivalent to the following system:
(
cos(ϕ) + cos(ψ) = − 12 ,
(28)
cos(ϕ) cos(ψ) = − 41 .
The solutions cos ϕ and cos ψ of (28) form the set of the roots of polynomial
x2 + 12 x − 14 ,
i.e. the set
√ (29)
cos ϕ, cos ψ = 14 (−1 ± 5) = cos( 25 π), cos( 54 π) .
Ad (26) We have:
(
i.e.
cos(ϕ) + cos(ψ) = − 21 ,
T3 cos(ϕ) + T3 cos(ψ) = − 12 ,
(
cos(ϕ) + cos(ψ) = − 12 ,
4 cos3 (ϕ)) + cos3 (ψ) = −2.
By transforming the second equation we obtain, respectively:
2
4 cos(ϕ) + cos(ψ)
cos(ϕ) + cos(ψ) − 3 cos(ϕ) cos(ψ) = −2,
4 − 21 41 − 3 cos(ϕ) cos(ψ) = −2,
cos(ϕ) cos(ψ) = − 14 ,
so the system (28) holds and the equalities (29) are satisfied.
56
Roman Witula, Damian Slota
Ad (27) Condition (27) from condition (25) for ϕ := 2ϕ and ψ := 2ψ
follows. Hence, we get:
(30)
cos(2ϕ), cos(2ψ) = cos( 25 π), cos( 45 π) .
Now, the assertions of Lemma 8 from (29) and (30) follows.
4. Some generalizations
Let us now step down to present the announced generalized trigonometric
identities of the same nature as identity (1). Each one should be preceded
by essential technical lemmas describing the values of some trigonometric
sums.
Lemma 9. Let n, r ∈ N, (2n − 1) 6 | r. Then the following equality holds:
(31)
n−1
X
2kr
σn (r) :=
exp i
π =
2n − 1
k=0

rπ

i
1

,
 2 − 2 tan
2(2n − 1)
=
rπ


,
 12 + 2i cot
2(2n − 1)
whenever r ∈ 2N,
whenever r ∈ 2N − 1,
where 2N (2N − 1) denotes the set of even (odd) positive integers.
Proof. We perform the following transformations:
−1
2nrπ
2rπ
σn (r) = 1 − exp i
1 − exp i
=
2n − 1
2n − 1
−1
2rπ
irπ
1 − exp i
=
= 1 − exp (i r π) exp
2n − 1
2n − 1
−1
2rπ
irπ
1 − exp i
=
= 1 − (−1)r exp
2n − 1
2n − 1
−1
irπ
= 1 − (−1)r−1 exp
=
2n − 1

rπ
rπ

1
−1

cos
,
for r ∈ 2N,
 2 exp −i
2(2n − 1) 2(2n − 1)
=
rπ
rπ


sin−1
,
for r ∈ 2N − 1,
 2i exp −i
2(2n − 1)
2(2n − 1)
which implies the desired identity.
Some Trigonometric Identities Related to...
57
Corollary 6. Let n, r ∈ N. Then:
(32)
1+2
n−1
X
k=1
cos
2kr
π
2n − 1
=
whenever (2n − 1) 6 | r,
whenever (2n − 1) | r,
0,
2n − 1,
and
(33) 2
n−1
X
k=1
sin
2kr
π
2n − 1
=

0,






rπ
,
− tan
=
1)

2(2n − 

rπ


,
 cot
2(2n − 1)
whenever(2n − 1) | r,
whenever (2n − 1) 6 | r ∧ r ∈ 2N,
whenever (2n − 1) 6 | r ∧ r ∈ 2N − 1.
In the next Lemma, identity (1) shall be generalized. The Lemma is
derived on the grounds of (6), (7) and Corollary 6.
Lemma 10. Let us set (for n, r ∈ N):
(34)
Φ+
r,n (x)
:=
Cr+ (x, 0)
+2
n−1
X
k=1
Cr+
2kπ
x,
2n − 1
and
(35)
Φ−
r,n (x)
:=
n−1
X
k=1
Cr−
2kπ
x,
2n − 1
.
Then we have for r ∈ 2N − 1:

0,
r < 2n − 1,






r


(2n − 1) r−2n+1 22−r cos (2n − 1)x , 2n − 1 ≤ r < 3(2n − 1),





2
+
r
Φr,n (x) =
2−r
(2n
−
1)
×

r−2n+1 2


2



× cos


(2n −1)x +


r

2−r

2
cos
3(2n
−
1)x
, 3(2n − 1) ≤ r < 5(2n − 1),
+

r−6n+3
2
58
Roman Witula, Damian Slota
and for r ∈ 2N:
` r ´ 1−r
8
(2n − 1) r/2
2
,
>
>
>
“` ´
>
>
>
r
>
(2n − 1) r/2
21−r +
>
>
!
>
>
>
`
´”
>
r
>
>
+ r−2(2n−1) 22−r cos 2(2n − 1)x ,
>
>
>
>
2
<
“` ´
Φ+
(x)
=
r
r,n
(2n − 1) r/2
21−r +
>
>
>
!
>
>
>
`
´
>
r
>
>
+ r−4n+2 22−r cos 2(2n − 1)x +
>
>
>
2
>
!
!
>
>
>
`
´
>
r
2−r
>
>
cos 4(2n − 1)x ,
+ r−8n+4 2
:
r < 2(2n − 1),
2(2n − 1) ≤ r < 4(2n − 1),
4(2n − 1) ≤ r < 6(2n − 1).
2
We have for r ∈ 2N − 1 and r ≤ 2 n − 1:
Φ−
r,n (x)
(36)
=2
1−r
(r−1)/2 X
k=0
r
k
(r − 2 k)π
2 (2 n − 1)
cot
sin (r − 2 k) x
and for r ∈ 2N and r ≤ 2(2 n − 1):
Φ−
r,n (x)
(37)
= −2
1−r
r/2 X
r
k=0
k
tan
(r − 2 k)π
2 (2 n − 1)
sin (r − 2 k) x .
Lemma 11. The following identity hold:
n−1
X
k=1
(2 k − 1) r π
(−1) exp i
2n − 1
k
=
1 − (−1)n+r−1 e−i
2rπ
1 + ei 2 n−1
rπ
2 n−1
=
1 − (−1)n−1 ei
i
1+e
2 (n−1) r π
2 n−1
2rπ
2 n−1
rπ
(−1) ei 2 n−1 =
rπ
rπ
(−1) ei 2 n−1 = −
ei 2 n−1 − (−1)n+r−1
2rπ
1 + ei 2 n−1
=
(38)
=

rπ
π

i 2 rn−1
−i
sin

rπ
2 (2n−1)
e
−1


e−i 2 (2 n−1) =
− i 2rπ
=


r
π

2
n−1
e
+1

cos 2 n−1



i


1
rπ
rπ

for (n + r) ∈ 2N − 1,
 = 2 1 − sec 2 n−1 − 2 tan 2 n−1 ,


rπ
π

i 2 rn−1
cos

rπ

2 (2 n−1)
e
+1


e−i 2 (2 n−1) =
− i 2rπ
=−


r
π
2
n−1

e
+1
cos 2 n−1




i

π
π
= − 12 1 + sec 2 rn−1
+ 2 tan 2 rn−1
,
for (n + r) ∈ 2N − 1.
Some Trigonometric Identities Related to...
59
Corollary 7. We have:
(39)
n+r
(−1)
+2
n−1
X
k
(−1) cos
k=1
(2 k − 1) r π
2n − 1
= − sec
rπ
2n − 1
and
(40)
2
n−1
X
(−1)k sin
k=0
(2 k − 1) r π
2n − 1
= (−1)n+r tan
rπ
.
2n − 1
Corollary 8. Using identities (6), (7) and Corollary 7 we find:
n−1
X
(2 k − 1) π
k +
(41) Ξ+
(x)
:=
=
(−1)
C
x,
r,n
r
2n − 1
k=1
= −21−r
br/2c X
k=0
r
k
(−1)n+r + sec
(r − 2 k)π
2n − 1
cos (r − 2 k) x
and
(42)
Ξ−
r,n (x)
:=
n−1
X
k
(−1)
Cr−
k=1
n+r
= (−1)
2
1−r
(2 k − 1) π
x,
2n − 1
br/2c X
k=0
r
k
tan
=
(r − 2 k) π
2n − 1
sin (r − 2 k) x .
Lemma 12. We have:
(43)
=
n−1
X
2nrπ
rπ
2krπ
1 − (−1)n ei 2 n−1
1 − (−1)n+r ei 2 n−1
(−1)k exp i
=
=
=
2rπ
2rπ
2n − 1
1 + ei 2 n−1
1 + ei 2 n−1
k=0

π
π

cos 2 (2rn−1)
 1 + ei 2 rn−1
rπ


e−i 2 (2 n−1) =

=
rπ

i 22n−1

π

1+e
cos 2 rn−1





1
rπ
i
rπ

for (n + r) ∈ 2N − 1,

 = 2 1 + sec 2 n−1 − 2 tan 2 n−1 ,



π
π

i 2 rn−1
−i sin 2 (2rn−1)

rπ
1
−
e


e−i 2 (2 n−1) =

=
rπ

i 22n−1

π

1+e
cos 2 rn−1





π
π
= 12 1 − sec 2 rn−1
− 2i tan 2 rn−1
,
for (n + r) ∈ 2N.
Corollary 9. We have:
n−1
X
rπ
2krπ
(44)
(−1)n+r−1 sec
=1+2
(−1)k cos
2n − 1
2n − 1
k=1
60
Roman Witula, Damian Slota
and
(45)
2
n−1
X
k
(−1) sin
k=0
2krπ
2n − 1
= − tan
rπ
.
2n − 1
Corollary 10. Using identities (6), (7) and Corollary 9 we obtain:
(46)
Ψ+
r,n (x)
:=
Cr+
x, 0 + 2
n−1
X
k
(−1)
Cr+
k=1
2−r
=2
br/2c X
k=0
r
k
(−1)n 12
1+2
n−1
X
l
(−1) cos
l=1
b 2r c
−
b r−1
2 c
r
2kπ
x,
2n − 1
=
2l(r − 2 k)π
2n − 1
cos (r − 2 k) x +
=
b 2r c
br/2c X r
(r − 2 k) π
= (−1)n+r−1 22−r
sec
cos (r − 2 k) x +
k
2n − 1
k=0
r + (−1)n 21 b 2r c − b r−1
2 c
b 2r c
+
and
(47)
Φ−
r,n (x)
:=
n−1
X
k
(−1)
Cr−
k=1
= −21−r
br/2c X
k=0
r
k
2kπ
x,
2n − 1
tan
=
(r − 2 k) π
2n − 1
sin (r − 2 k) x .
Lemma 13. We have:
n
irπ − 1
X
rπ e
(2 k − 1) r π
exp i
= ei 2 n i r π
=
2n
e n −1
k=1

0,
whenever r is even



and n6 | r ∨ (n|r ∧ nr ∈ 2N − 1),
π
=
(48)
i 2r n
−2 e


 i rπ
= i csc 2r πn ,
whenever r is odd.
e n −1
Corollary 11. We have:
(49)
n
X
k=1
cos
(2 k − 1) r π
2n

 n
−n
=

0
r
n|r ∧ 2n
∈ 2N,
r
n|r ∧ 2n ∈ 2N − 1,
otherwise.
Some Trigonometric Identities Related to...
61
and
n
X
(50)
(2 k − 1) r π
2n
sin
k=1
=
0
csc
rπ
2n
whenever r is even,
whenever r is odd.
Corollary 12. By (6), (7) and Corollary 11 we get:
n
X
(2 k − 1) π
+
= 0,
whenever r is odd,
(51)
∆+
(x)
:=
C
x,
r,n
r
2n
k=1
and
(52)
∆−
r,n (x)
:=
n
X
Cr−
k=1
=

 0
 22−r
br/2c
P
k=0
r
k
csc
(2 k − 1) π
x,
2n
(r−2 k) π
2n
=
whenever r is even,
sin (r − 2 k) x whenever r is odd.
Lemma 14. Let n ∈ N, r ∈ Z and n 6 | r. Then we have:
(53)
n−1
X
k=1
(2 k − 1) r π
(−1) exp i
2n
k
1 − (−1)n−1 ei
=
rπ
1 + ei n
(n−1) r π
n
rπ
(−ei 2 n ) =
rπ
−1 − (−1)n+r e−i n
=
2 cos( 2r πn )

rπ
−(1 + e−i n )
rπ


= −e−i 2 n ,

rπ


2
cos(
)

2n


rπ
=
−i sin( 2r πn ) −i r π
−1 + e−i n

=
e 2n =

rπ


cos( 2r πn )
 2 cos( 2 n )



= cos( 2r πn ) − sec( 2r πn ) − i sin( 2r πn ),
=
Corollary 13. We have:
n−1
X
(2 k − 1) r π
k
(54)
=
(−1) cos
2n
k=1
(
− cos( 2r πn ),
=
cos( 2r πn ) − sec( 2r πn ),
whenever (n + r) is even,
whenever (n + r) is odd.
whenever (n + r) is even,
whenever (n + r) is odd,
and
(55)
n−1
X
k=1
(−1)k sin
(2 k − 1) r π
2n
= (−1)n+r sin
r π 2n
.
62
Roman Witula, Damian Slota
Corollary 14. We have:
n−1
X
(2 k − 1) π
+
k +
Θr,n (x) :=
(−1) Cr x,
=
2n
k=1

br/2c
P r

(r−2 k) π

1−r

cos ((r − 2 k) x) ,
−2
cos

2n
k


k=0

br/2c
P r h
=
(r−2 k) π
2−r
csc
−
2


2n
k


k=0
i



− sec (r−2 k) π
cos (r − 2 k) x ,
2n
for r ∈ 2N,
for r ∈ 2N − 1.
and
(56)
Θ−
r,n (x)
:=
n−1
X
k
(−1)
Cr−
k=1
n+r
= (−1)
br/2c X
k=0
(2 k − 1) π
x,
2n
r
k
sin
r π
2n
=
sin (r − 2 k) x .
Remark 2. All the trigonometric identities of (31)–(56), on the bases of
identities (8)–(11), also may be given for Sn+ and Sn− functions.
References
[1] I. S. Gradshteyn, I. M. Ryzhik, Tables of Integrals, Series, and Products, Academic
Press, New York, 1980.
[2] A. Harmanci, Two elementary commutativity theorems for rings, Acta Math. Acad.
Sci. Hungar. 29 (1977), pp. 23–29.
[3] P. S. Modenov, Problems on a Special Course of Elementary Mathematics, Sovetskaya
Nauka, Moscow, 1957 (in Russian).
[4] J. Riordan, Combinatorial Identities, Wiley, New York, 1968.
[5] T. Rivlin, Chebyshev Polynomials from Approximation Theory to Algebra and Number
Theory, 2nd ed., Wiley, New York, 1990.
[6] P. K. Suetin, Classical orthogonal polynomials, Izdat. Nauka, Moscow, 1976 (in Russian).
[7] R. Witula, D. Slota, Decomposition of Certain Symmetric Functions of Powers of
Cosine and Sine Functions, Int. J. Pure Appl. Math. 50 (2009), pp. 1-12.
[8] R. Witula, D. Slota, On Modified Chebyshev Polynomials J. Math. Anal. Appl. 324
(2006), pp. 321–343.