Folia Mathematica Acta Universitatis Lodziensis Vol. 15, No. 1, pp. 41–62 c 2008 for University of L´ od´ z Press SOME TRIGONOMETRIC IDENTITIES RELATED TO POWERS OF COSINE AND SINE FUNCTIONS ROMAN WITU LA † , DAMIAN S LOTA ††‡ 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 con- nected with the following trigonometric identities previously presented in [8]: sin 2n (x) + cos 2n ( x - π 6 ) + cos 2n ( x + π 6 ) + cos 2n (x)+ + cos 2n ( x - π 3 ) + cos 2n ( x + π 3 ) ≡ 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: C + n (x, ϕ) := cos n (x + ϕ) + cos n (x - ϕ), (2) C - n (x, ϕ) := cos n (x - ϕ) - cos n (x + ϕ), (3) S + n (x, ϕ) := sin n (x + ϕ) + sin n (x - ϕ), (4) S - n (x, ϕ) := sin n (x + ϕ) - sin n (x - ϕ). (5) for n ∈ N. The term ”decomposition” means here the trigonometric poly- nomial 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 iden- tities, 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.
SOME TRIGONOMETRIC IDENTITIES RELATED TOPOWERS OF COSINE AND SINE FUNCTIONS
ROMAN WITU LA†, DAMIAN S LOTA††‡
Abstract. Some decomposition of certain basic symmetric functions of n-thpower of cosine and sine functions is presented here. Next the applications
of these decompositions to generating many new binomial and trigonometricidentities are discussed.
1. Introduction
In this paper we wish to investigate different trigonometric identities con-nected with the following trigonometric identities previously presented in [8]:
sin2n(x) + cos2n(x− π
6
)+ cos2n
(x+ π
6
)+ cos2n(x)+
+ cos2n(x− π
3
)+ cos2n
(x+ π
3
)≡ const ⇔ n = 1, 2, . . . , 5.(1)
It is natural to relate these purpose to decompositions of the following foursymmetric functions of the n-th powers of cosine and sine functions:
C+n (x, ϕ) := cosn(x+ ϕ) + cosn(x− ϕ),(2)
C−n (x, ϕ) := cosn(x− ϕ)− cosn(x+ ϕ),(3)
S+n (x, ϕ) := sinn(x+ ϕ) + sinn(x− ϕ),(4)
S−n (x, ϕ) := sinn(x+ ϕ)− sinn(x− ϕ).(5)
for n ∈ N. The term ”decomposition” means here the trigonometric poly-nomial form of these functions. We note that some other decomposition ofthe functions (2)–(5) using the Chebyshev polynomial of the second kind inthe separate paper [7] is discussed.
The basic decompositions of functions (2)–(5) are presented in Section 2and in two next sections are applied to generating of trigonometric iden-tities, starting from the simplest forms (see Section 3 and Lemma 8) to
‡Institute of Mathematics, Silesian University of Technology, Kaszubska 23, Gliwice44-100. Poland. E-mail: [email protected].
‡‡Institute of Mathematics, Silesian University of Technology, Kaszubska 23, Gliwice44-100. Poland. E-mail: [email protected].
AMS subject classifications: 26C05, 11B37, 11B83.
42 Roman Witu la, Damian S lota
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 thesedecompositions to obtained some nontrivial binomial identities is also given.
2. The basic decomposition
Lemma 1 (see [3]). The following two classical identities hold
2n−2C+n (x, ϕ) =
bn/2c∑k=0
(n
k
)cos
((n− 2k)ϕ
)cos
((n− 2k)x
)−
− 12
(⌊n2
⌋−
⌊n− 1
2
⌋) (n
bn2 c
)(6)
and
(7) 2n−2C−n (x, ϕ) =b(n−1)/2c∑
k=0
(n
k
)sin
((n− 2k)ϕ
)sin
((n− 2k)x
).
The proof of the above identities by induction follows and will be omittedhere.
The sequence of the following identities can easily be deduced from iden-tities (6) and (7) (the formulas (12)–(14) are known see [1], the formula (15)is probably new):
22n−2 S+2n(x, ϕ) =
=n∑k=0
(−1)n−k(
2nk
)cos
(2(n− k)ϕ
)cos
(2(n− k)x
)− 1
2
(2nn
)
=n−1∑k=0
(−1)n−k(
2nk
)cos
(2(n− k)ϕ
)cos
(2(n− k)x
)+
12
(2nn
);(8)
22n−3 S+2n−1(x, ϕ) =
=n−1∑k=0
(−1)n−k−1
(2n− 1k
)cos
((2n− 2k − 1)ϕ
)sin
((2n− 2k − 1)x
);(9)
22n−3 S−2n−1(x, ϕ) =
=n−1∑k=0
(−1)n−k−1
(2n− 1k
)sin
((2n− 2k − 1)ϕ
)cos
((2n− 2k − 1)x
);(10)
Some Trigonometric Identities Related to... 43
22n−2 S−2n(x, ϕ) =n−1∑k=0
(−1)n−k−1
(2nk
)sin
(2(n− k)ϕ
)sin
(2(n− k)x
);(11)
22n−2 sin2n−1(x) =n−1∑k=0
(−1)n−k−1
(2n− 1k
)sin
((2n− 2k − 1)x
);(12)
22n−1 sin2n(x) = 22n−2 S+2n(x, 0) =
=n∑k=0
(−1)n−k(
2nk
)cos
(2(n− k)x
)− 1
2
(2nn
);(13)
2n−1 cosn(x) = 2n−2 C+n (x, 0) =
=bn/2c∑k=0
(n
k
)cos
((n− 2k)x
)− 1
2
(⌊n2
⌋−
⌊n− 1
2
⌋) (n
bn/2c
);(14)
22n−2 C+2n
(x+ π
8 ,π4
)= cos2n
(x− π
8
)+ cos2n
(x+ 3
8π)
=
=n∑k=0
(2nk
)cos
((n− k)π2
)cos
(2(n− k)x+ (n− k)π4
)− 1
2
(2nn
)=
=12
(2nn
)−
(2nn− 2
)cos
(4x+ π
2
)+
(2nn− 4
)cos
(8x+ π
)−
−(
2nn− 6
)cos
(12x+ 3
2π)
+(
2nn− 8
)cos
(16x+ 2π
)− . . .
=12
(2nn
)+
(2nn− 2
)sin(4x)−
(2nn− 4
)cos(8x) +
+(
2nn− 6
)sin(12x) +
(2nn− 8
)cos(16x) +
+(
2nn− 10
)sin(20x)−
(2n
n− 12
)cos(24x) +
+(
2nn− 14
)sin(28x) +
(2n
n− 16
)cos(32x) + . . .(15)
In the Tables 1 and 2 below the first seven decompositions of the type (6)–(11) of every function C+
n , C−n , S+n and S−n are presented.
44 Roman Witu la, Damian S lota
Table. 1. The first seven decompositions of every function C+n and C−
n
n C+n (x, ϕ) C−n (x, ϕ)
1 2 cos(ϕ) cos(x) 2 sin(ϕ) sin(x)
2 1 + cos(2ϕ) cos(2x) sin(2ϕ) sin(2x)
3 12
`3 cos(ϕ) cos(x) + cos(3ϕ) cos(3x)) 1
2
`3 sin(ϕ) sin(x) + sin(3ϕ) sin(3x))
4 14
`3 + 4 cos(2ϕ) cos(2x) + cos(4ϕ) cos(4x)
´14
`4 sin(2ϕ) sin(2x) + sin(4ϕ) sin(4x)
´5 1
8
`10 cos(ϕ) cos(x) + 5 cos(3ϕ) cos(3x)+ 1
8
`10 sin(ϕ) sin(x) + 5 sin(3ϕ) sin(3x)+
+ cos(5ϕ) cos(5x)´
+ sin(5ϕ) sin(5x)´
6 116
`10 + 15 cos(2ϕ) cos(2x)+ 1
16
`15 sin(2ϕ) sin(2x)+
+6 cos(4ϕ) cos(4x) + cos(6ϕ) cos(6x)´
+6 sin(4ϕ) sin(4x) + sin(6ϕ) sin(6x)´
7 132
`35 cos(ϕ) cos(x) + 21 cos(3ϕ) cos(3x)+ 1
32
`35 sin(ϕ) sin(x) + 21 sin(3ϕ) sin(3x)+
+7 cos(5ϕ) cos(5x) + cos(7ϕ) cos(7x)´
+7 sin(5ϕ) sin(5x) + sin(7ϕ) sin(7x)´
Table. 2. The first seven decompositions of every function S+n and S−
n
n S+n (x, ϕ) S−n (x, ϕ)
1 2 cos(ϕ) sin(x) 2 sin(ϕ) cos(x)
2 1− cos(2ϕ) cos(2x) sin(2ϕ) sin(2x)
3 12
`3 cos(ϕ) sin(x)− cos(3ϕ) sin(3x)) 1
2
`3 sin(ϕ) cos(x)− sin(3ϕ) cos(3x))
4 14
`3− 4 cos(2ϕ) cos(2x) + cos(4ϕ) cos(4x)
´14
`4 sin(2ϕ) sin(2x)− sin(4ϕ) sin(4x)
´5 1
8
`10 cos(ϕ) sin(x)− 5 cos(3ϕ) sin(3x)+ 1
8
`10 sin(ϕ) cos(x)− 5 sin(3ϕ) cos(3x)+
+ cos(5ϕ) sin(5x)´
+ sin(5ϕ) cos(5x)´
6 116
`10− 15 cos(2ϕ) cos(2x)+ 1
16
`15 sin(2ϕ) sin(2x)−
+6 cos(4ϕ) cos(4x)− cos(6ϕ) cos(6x)´
−6 sin(4ϕ) sin(4x) + sin(6ϕ) sin(6x)´
7 132
`35 cos(ϕ) sin(x)− 21 cos(3ϕ) sin(3x)+ 1
32
`35 sin(ϕ) cos(x)− 21 sin(3ϕ) cos(3x)+
+7 cos(5ϕ) sin(5x)− cos(7ϕ) sin(7x)´
+7 sin(5ϕ) cos(5x)− sin(7ϕ) cos(7x)´
At the end of this section it will be given the application some of abovedecompositions to generating an interesting sets of binomial identities (seealso [2, 4]).
Corollary 1. Since for x, ϕ→ 0 we have
C−n (x, ϕ) =(
1− (x− ϕ)2
2!+ . . .
)n
−(
1− (x+ ϕ)2
2!+ . . .
)n
≈
≈ n
2
((x+ ϕ)2 − (x− ϕ)2
)= 2nxϕ,
Some Trigonometric Identities Related to... 45
so by (6) the following formula hold:
limx,ϕ→0
2n−2C−n (x, ϕ)xϕ
= limx,ϕ→0
b(n−1)/2c∑k=0
(n
k
)sin
((n− 2k)ϕ
)ϕ
sin((n− 2k)x
)x
,
i.e.,
n 2n−1 =b(n−1)/2c∑
k=0
(n
k
) (n− 2 k
)2.
Corollary 2. From (12) it can be easily derived the formula
22n
(sin(x)x
)2n+1
=
=∞∑r=1
n∑k=0
(−1)n−k+r
(2n+ 1k
)(2n− 2 k + 1)2r−1
(2 r − 1)!x2(r−n−1),
hence we get the sequence of the following binomial identities:n∑k=0
(−1)k(
2n+ 1k
) (2n− 2 k + 1
)2r−1 = 0
for r = 1, 2, . . . , n,n∑k=0
(−1)k(
2n+ 1k
) (2n− 2 k + 1
)2n+1 = 22n (2n+ 1)!,
n∑k=0
(−1)k(
2n+ 1k
) (2n− 2 k + 1
)2n+3 =13
(2n+ 1) 22n−1 (2n+ 3)!,
n∑k=0
(−1)k(
2n+ 1k
) (2n− 2 k + 1
)2n+5 =10n+ 3
360(2n+ 1) 22n (2n+ 5)!,
etc., since we have:
(16)(
sin(x)x
)n
= 1− n
6x2 +
n (5n− 2)360
x4 + . . .
Similarly, from (13) we deduce the formula
22n−1
(sin(x)x
)2n
+12
(2nn
)x−2n =
=∞∑r=0
n∑k=0
(−1)n−k+r
(2nk
) (2 (n− k)
)2r
(2 r)!x2(r−n),
46 Roman Witu la, Damian S lota
which by (16) implies the second sequence of the following binomial identi-ties:
n∑k=0
(−1)n−k(
2nk
)=
12
(2nn
),
n−1∑k=0
(−1)k(
2nk
)(n− k)2r = 0,
for r = 1, 2, . . . , n− 1,
n−1∑k=0
(−1)k(
2nk
) (n− k
)2n =12
(2n)!,
n−1∑k=0
(−1)k(
2nk
) (n− k
)2n+2 =n
24(2n+ 2)!,
n−1∑k=0
(−1)k(
2nk
) (n− k
)2n+4 =n (5n− 1)
2880(2n+ 4)!,
etc.
Corollary 3. From (14) we obtain
2n−1
(1− n
2x2 +
n (3n− 2)24
x4 + . . .
)=
=12
(⌊n− 12
⌋−
⌊n2
⌋) (n
bn/2c
)+
∞∑r=0
bn/2c∑k=0
(−1)2r(n
k
)(n− 2 k)2r
(2 r)!x2r,
which implies:
bn/2c∑k=0
(n
k
) (n− 2 k
)2 = n 2n−1,
bn/2c∑k=0
(n
k
) (n− 2 k
)4 = n (3n− 2) 2n−1,
etc.
Corollary 4. We have the formula (see [4, 6]):
Pn(x) = 2−nbn/2c∑k=0
(−1)k(n− k
k
) (2n− 2 kn− k
)xn−2k,
Some Trigonometric Identities Related to... 47
where Pn(x) denotes the n-th Legendre polynomial. Hence, by (14) for evenn ∈ N, we get∫ π
0P2n
(cosϕ
)dϕ = π 2−4n
n∑k=0
(−4)k(
2n− k
k
) (4n− 2 k2n− k
) (2n− 2 kn− k
)
= π 2−4nn∑k=0
(−4)k(4n− 2 k
)!
k!((n− k)!
)2 (2n− k)!.(17)
On the other hand, we have ([6]):
P2n
(cosϕ
)= 2−4n
(2nn
)2
+n−1∑k=0
ak,n cos(2n− 2 k
)for some ak,n ∈ Q, which by (17) implies the identity:(
2nn
)2
=n∑k=0
(−4)k(4n− 2 k
)!
k!((n− k)!
)2 (2n− k)!
=n∑k=0
(−4)k(
2n− k
k
) (4n− 2 k2n− k
) (2n− 2 kn− k
).
3. Simple trigonometric identities
We mention only the simple form of trigonometric identities to discusshere. Simultaneously this approach leads in Section 4 to direct our con-siderations to only symmetric trigonometric identities with respect to thephase translations.
Lemma 2. Fix a, ϕ, ψ ∈ R. Then
fa,ϕ,ψ(x) := a
{sin2 xcos2 x
}+ sin2(x+ ϕ) + sin2(x+ ψ) ≡ const
(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 = k π and a ={−2 cos(2ϕ)
2 cos(2ϕ)
}.
Proof. First, we note that:
fa,ϕ,ψ(x) ≡ const =⇒ f ′a,ϕ,ψ(x) ≡ 0 =⇒ f ′a,ϕ,ψ(0) = 0.
48 Roman Witu la, Damian S lota
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:
fa,ϕ,ψ(x) := a
{sin2 xcos2 x
}+ 1
which implies a = 0. If ϕ+ ψ = k π then, we have:
fa,ϕ,ψ(x) ={a sin2 x+ S+
2 (x, ϕ)a cos2 x+ S+
2 (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:
a
{sin2 xcos2 x
}+ cos2(x+ ϕ) + cos2(x+ ψ) ≡ const
iff either:ϕ− ψ = (2k + 1)π2 and a = 0
or
ϕ+ ψ = k π and a ={−2 cos(2ϕ)
2 cos(2ϕ)
}.
Proof. Set ϕ := ϕ+ π2 and ψ := ψ + π
2 in the Lemma 2. �
Lemma 3. We have:1) a cos4(x) + S+
4 (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) + S+4 (x, ϕ) ≡ const
iff either a = 1 and ϕ = ±π3 + k π, k ∈ Z
or a = −2 and ϕ = k π, k ∈ Z.
Proof. The assertion from the following decomposition follows:
1)
a cos4(x) + S+4 (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) + S+4 (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 ∈ Zor 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
(− π
2
)⇔ sin3(ϕ) + sin3(ψ) = 0 ⇔
⇔ sin(ϕ) + sin(ψ) = 0 ⇔ sin(ϕ+ ψ
2
)cos
(ϕ− ψ
2
)= 0 ⇔
⇔ 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 ⇔
⇔ a = 0 ∨ sin(ϕ+ ψ
2
)sin
(ϕ− ψ
2
)= 0 ⇔
⇔ 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) = C+3 (x, ϕ) + a
2 C+3 (x, 0) =
= 14
(2 cos(3ϕ) + a
)cos(3x) + 3
4
(2 cos(ϕ) + a
)cos(x),
50 Roman Witu la, Damian S lota
which is independent on x iff{2 cos(3ϕ) + a = 02 cos(ϕ) + a = 0 ⇔
{a = −2 cos(ϕ)cos(3ϕ)− cos(ϕ) = 0 ⇔
⇔{a = −2 cos(ϕ)sin(ϕ) sin(2ϕ) = 0 ⇔
{a = −2 cos(ϕ)sin(2ϕ) = 0 ⇔
⇔{a = −2 cos(ϕ)ϕ = k π
2 for some k ∈ Z ⇔ either a = 2 (−1)l+1 and ϕ = l π, l ∈ 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)
= aC+3 (x, 0) + C+
3 (x, ϕ).
Lemma 5. We have:
1)ha,ϕ(x)cos(x)
≡ const ⇔ ha,ϕ(x)cos(x)
≡ 6 sin2(ϕ) cos(ϕ) ⇔
⇔ a+ T3
(cos(ϕ)
)= 0 ⇔ a = − cos(3ϕ);
2) 2 t(a) :=3
√a+
√a2 − 1 +
3
√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)
= 12
(cos(3ϕ) + a
) cos(3x)cos(x)
+ 32
(cos(ϕ) + a
)≡ const ⇔
⇔ a = − cos(3ϕ) andha,ϕ(x)cos(x)
≡ 32
(cos(ϕ)− cos(3ϕ)
),
which implies 1).2) We have
2T3
(t(a)
)=
(a+ + a−
)3 − 3(a+ + a−
)=
(a+ + a−
) [(a+ + a−
)2 − 3]
=
=(a+ + a−
) [(a+
)2 − 1 +(a−
)2]
=(a+
)3 +(a−
)3 = 2 a.
where a+ := 3√a+
√a2 − 1 and a− := 3
√a−
√a2 − 1.
Some Trigonometric Identities Related to... 51
3) We note that t(a) is an odd function. Moreover we have:
t′(a) =1
6√a2 − 1
(3
√a+
√a2 − 1− 3
√a−
√a2 − 1
).
Hence, we deduce that t(a) is a decreasing function on (−∞,−1] and conse-quently an increasing one on [1,∞). Should also be noticed, that t(1) = 1.
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− π
6
)+ cos2n
(x+ π
6
),
gn(x) := cos2n(x) + cos2n(x− π
3
)+ cos2n
(x+ π
3
).
We have the following identities:
n fn(x) gn(x)
1 32
32
2 98
98
3 332
`10− cos(6x)
´332
`10 + cos(6x)
´4 3
128
`35− 8 cos(6x)
´3
128
`35 + 8 cos(6x)
´5 27
512
`14− 5 cos(6x)
´27512
`14 + 5 cos(6x)
´6 3
2048
`462− 220 cos(6x) + cos(12x)
´3
2048
`462 + 220 cos(6x) + cos(12x)
´
54 Roman Witu la, Damian S lota
n fn(x) + gn(x) n fn(x) + gn(x)
1 3 7 32048
`858 + 7 cos(12x)
´2 9
48 45
16384
`429 + 8 cos(12x)
´3 15
89 51
32768
`715 + 24 cos(12x)
´4 105
6410 969
262144
`286 + 15 cos(12x)
´5 189
12811 1197
524288
`442 + 33 cos(12x)
´6 3
1024
`462 + cos(12x)
´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:
f12(x) + g12(x) = 34194304
(6118
(221 + 22 cos(12x)
)+ cos(24x)
).
Remark 1. We have also(C+
2
(x, π6
))n +(C+
2
(x, π3
))n −−
((−1)n + 1
) (12 cos(2x)
)n + n((−1)n − 1
) (12 cos(2x)
)n−1 =
=
0, for n = 1,2, for n = 2, 3,2 + 3 cos2(2x), for n = 4,2 + 5 cos2(2x), for n = 5.
The next result indicates the direction in which attempts at generalizingcertain results from Section 3 should follow.
Lemma 8. Let ϕ,ψ ∈ R and
Θn(x) = C+n (x, 0) + 2C+
n (x, ϕ) + 2C+n (x, ψ), n ∈ N, x ∈ R.
If Θn(x) ≡ const for some two different values of n ∈ N and at least for oneodd value of n ∈ N, then
Θn(x) = C+n (x, 0) + 2C+
n (x, 25 π) + 2C+
n (x, 45 π), n ∈ N, x ∈ R.
If Θn(x) ≡ const, for two different even values of n ∈ N, then
Θ2n(x) = C+2n(x, 0) + 2C+
2n(x,25 π) + 2C+
2n(x,45 π), 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 trigonometricsystem one of the following three conditions holds:
(25)
{cos(ϕ) + cos(ψ) = −1
2 ,
cos(2ϕ) + cos(2ψ) = −12 ,
Some Trigonometric Identities Related to... 55
whenever (−1)k + (−1)l = 0; or
(26)
{cos(ϕ) + cos(ψ) = −1
2 ,
cos(3ϕ) + cos(3ψ) = −12 ,
whenever (−1)k + (−1)l = −2; or
(27)
{cos(2ϕ) + cos(2ψ) = −1
2 ,
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) = 14 ,
cos2(ϕ) + cos2(ψ) = 34 ,
i.e.cos(ϕ) cos(ψ) = −1
4 .
So, system (25) is equivalent to the following system:
(28)
{cos(ϕ) + cos(ψ) = −1
2 ,
cos(ϕ) cos(ψ) = −14 .
The solutions cosϕ and cosψ of (28) form the set of the roots of polynomial
x2 + 12 x−
14 ,
i.e. the set
(29){
cosϕ, cosψ}
={
14 (−1±
√5)
}=
{cos(2
5π), cos(45π)
}.
Ad (26) We have:{cos(ϕ) + cos(ψ) = −1
2 ,
T3
(cos(ϕ)
)+ T3
(cos(ψ)
)= −1
2 ,
i.e. {cos(ϕ) + cos(ψ) = −1
2 ,
4(cos3(ϕ)) + cos3(ψ)
)= −2.
By transforming the second equation we obtain, respectively:
4(cos(ϕ) + cos(ψ)
) ((cos(ϕ) + cos(ψ)
)2 − 3 cos(ϕ) cos(ψ))
= −2,
4(− 1
2
) (14 − 3 cos(ϕ) cos(ψ)
)= −2,
cos(ϕ) cos(ψ) = −14 ,
so the system (28) holds and the equalities (29) are satisfied.
56 Roman Witu la, Damian S lota
Ad (27) Condition (27) from condition (25) for ϕ := 2ϕ and ψ := 2ψfollows. Hence, we get:
(30){
cos(2ϕ), cos(2ψ)}
={
cos(25π), cos(4
5π)}.
Now, the assertions of Lemma 8 from (29) and (30) follows. �
4. Some generalizations
Let us now step down to present the announced generalized trigonometricidentities of the same nature as identity (1). Each one should be precededby essential technical lemmas describing the values of some trigonometricsums.
Lemma 9. Let n, r ∈ N, (2n− 1) 6 | r. Then the following equality holds:
(31) σn(r) :=n−1∑k=0
exp(i
2 k r2n− 1
π
)=
=
12 −
i2 tan
(r π
2(2n− 1)
), whenever r ∈ 2N,
12 + i
2 cot(
r π
2(2n− 1)
), whenever r ∈ 2N− 1,
where 2N (2N− 1) denotes the set of even (odd) positive integers.
In the next Lemma, identity (1) shall be generalized. The Lemma isderived on the grounds of (6), (7) and Corollary 6.
Lemma 10. Let us set (for n, r ∈ N):
(34) Φ+r,n(x) := C+
r (x, 0) + 2n−1∑k=1
C+r
(x,
2kπ2n− 1
)
and
(35) Φ−r,n(x) :=n−1∑k=1
C−r
(x,
2kπ2n− 1
).
Then we have for r ∈ 2N− 1:
Φ+r,n(x) =
0, r < 2n− 1,
(2n− 1)(
rr−2n+1
2
)22−r cos
((2n− 1)x
), 2n− 1 ≤ r < 3(2n− 1),
(2n− 1)((
rr−2n+1
2
)22−r×
× cos((2n− 1)x
)+
+(
rr−6n+3
2
)22−r cos
(3(2n− 1)x
)), 3(2n− 1) ≤ r < 5(2n− 1),
58 Roman Witu la, Damian S lota
and for r ∈ 2N:
Φ+r,n(x) =
8>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>:
(2n− 1)`
rr/2
´21−r, r < 2(2n− 1),
(2n− 1)“`
rr/2
´21−r+
+
r
r−2(2n−1)2
!22−r cos
`2(2n− 1)x
´”, 2(2n− 1) ≤ r < 4(2n− 1),
(2n− 1)“`
rr/2
´21−r+
+
r
r−4n+22
!22−r cos
`2(2n− 1)x
´+
+
r
r−8n+42
!22−r cos
`4(2n− 1)x
´!, 4(2n− 1) ≤ r < 6(2n− 1).
We have for r ∈ 2N− 1 and r ≤ 2n− 1:
(36) Φ−r,n(x) = 21−r(r−1)/2∑k=0
(r
k
)cot
((r − 2 k)π2 (2n− 1)
)sin
((r − 2 k)x
)and for r ∈ 2N and r ≤ 2(2n− 1):
(37) Φ−r,n(x) = −21−rr/2∑k=0
(r
k
)tan
((r − 2 k)π2 (2n− 1)
)sin
((r − 2 k)x
).
Lemma 11. The following identity hold:
n−1∑k=1
(−1)k exp(i(2 k − 1) r π
2n− 1
)=
1− (−1)n−1 ei2 (n−1) r π
2 n−1
1 + ei2 r π2 n−1
(−1) eir π
2 n−1 =
=1− (−1)n+r−1 e−i
r π2 n−1
1 + ei2 r π2 n−1
(−1) eir π
2 n−1 = −ei r π
2 n−1 − (−1)n+r−1
1 + ei2 r π2 n−1
=
=
−ei r π2 n−1 − 1
ei2rπ
2 n−1 + 1=−i sin
(rπ
2 (2n−1)
)cos
(r π
2n−1
) e−ir π
2 (2 n−1) =
= 12
(1− sec
(r π
2n−1
))− i
2 tan(
r π2n−1
), for (n+ r) ∈ 2N− 1,
−ei r π
2 n−1 + 1
ei2 r π2 n−1 + 1
= −cos
(r π
2 (2n−1)
)cos
(r π
2n−1
) e−ir π
2 (2 n−1) =
= − 12
(1 + sec
(r π
2n−1
))+ i
2 tan(
r π2n−1
), for (n+ r) ∈ 2N− 1.
(38)
Some Trigonometric Identities Related to... 59
Corollary 7. We have:
(39) (−1)n+r + 2n−1∑k=1
(−1)k cos(
(2 k − 1) r π2n− 1
)= − sec
(r π
2n− 1
)and
(40) 2n−1∑k=0
(−1)k sin(
(2 k − 1) r π2n− 1
)= (−1)n+r tan
(r π
2n− 1
).
Corollary 8. Using identities (6), (7) and Corollary 7 we find:
Ξ+r,n(x) :=
n−1∑k=1
(−1)kC+r
(x,
(2 k − 1)π2n− 1
)=(41)
= −21−rbr/2c∑k=0
(r
k
) ((−1)n+r + sec
((r − 2 k)π2n− 1
))cos
((r − 2 k)x
)and
Ξ−r,n(x) :=n−1∑k=1
(−1)k C−r
(x,
(2 k − 1)π2n− 1
)=(42)
= (−1)n+r 21−rbr/2c∑k=0
(r
k
)tan
((r − 2 k)π
2n− 1
)sin
((r − 2 k)x
).
Lemma 12. We have:
(43)n−1∑k=0
(−1)k exp(i
2 k r π2n− 1
)=
1− (−1)n ei2 n r π2 n−1
1 + ei2 r π2 n−1
=1− (−1)n+r ei
r π2 n−1
1 + ei2 r π2 n−1
=
=
1 + eir π
2 n−1
1 + ei2 r π2 n−1
=cos
(r π
2 (2n−1)
)cos
(r π
2n−1
) e−ir π
2 (2 n−1) =
= 12
(1 + sec
(r π
2n−1
))− i
2 tan(
r π2n−1
), for (n+ r) ∈ 2N− 1,
1− eir π
2 n−1
1 + ei2 r π2 n−1
=−i sin
(r π
2 (2n−1)
)cos
(r π
2n−1
) e−ir π
2 (2 n−1) =
= 12
(1− sec
(r π
2n−1
))− i
2 tan(
r π2n−1
), for (n+ r) ∈ 2N.
Corollary 9. We have:
(44) (−1)n+r−1 sec(
r π
2n− 1
)= 1 + 2
n−1∑k=1
(−1)k cos(
2 k r π2n− 1
)
60 Roman Witu la, Damian S lota
and
(45) 2n−1∑k=0
(−1)k sin(
2 k r π2n− 1
)= − tan
(r π
2n− 1
).
Corollary 10. Using identities (6), (7) and Corollary 9 we obtain:
Ψ+r,n(x) := C+
r
(x, 0
)+ 2
n−1∑k=1
(−1)k C+r
(x,
2 k π2n− 1
)=(46)
= 22−rbr/2c∑k=0
(r
k
)(1 + 2
n−1∑l=1
(−1)l cos(
2l(r − 2 k)π2n− 1
) )cos
((r − 2 k)x
)+
+ (−1)n 12
(b r2c − b
r−12 c
) (r
b r2c
)=
= (−1)n+r−1 22−rbr/2c∑k=0
(r
k
)sec
((r − 2 k)π
2n− 1
)cos
((r − 2 k)x
)+
+ (−1)n 12
(b r2c − b
r−12 c
) (r
b r2c
)and
Φ−r,n(x) :=n−1∑k=1
(−1)k C−r
(x,
2 k π2n− 1
)=(47)
= −21−rbr/2c∑k=0
(r
k
)tan
((r − 2 k)π
2n− 1
)sin
((r − 2 k)x
).
Lemma 13. We have:n∑k=1
exp(i(2 k − 1) r π
2n
)= ei
r π2 n
ei r π − 1ei
r πn − 1
=
=
0, whenever r is even
and n6 | r ∨ (n|r ∧ rn ∈ 2N− 1),
−2 eir π2 n
eir πn − 1
= i csc(r π2n
), whenever r is odd.
(48)
Corollary 11. We have:
(49)n∑k=1
cos(
(2 k − 1) r π2n
)=
n n|r ∧ r2n ∈ 2N,
−n n|r ∧ r2n ∈ 2N− 1,
0 otherwise.
Some Trigonometric Identities Related to... 61
and
(50)n∑k=1
sin(
(2 k − 1) r π2n
)=
{0 whenever r is even,csc
(r π2n
)whenever r is odd.
Corollary 12. By (6), (7) and Corollary 11 we get:
(51) ∆+r,n(x) :=
n∑k=1
C+r
(x,
(2 k − 1)π2n
)= 0, whenever r is odd,
and
(52) ∆−r,n(x) :=
n∑k=1
C−r
(x,
(2 k − 1)π2n
)=
=
0 whenever r is even,
22−rbr/2c∑k=0
(rk
)csc
((r−2 k)π
2n
)sin
((r − 2 k)x
)whenever r is odd.
Lemma 14. Let n ∈ N, r ∈ Z and n 6 | r. Then we have:n−1∑k=1
(−1)k exp(i(2 k − 1) r π
2n
)=
1− (−1)n−1 ei(n−1) r π
n
1 + eir πn
(−ei r π2 n ) =(53)
=−1− (−1)n+r e−i
r πn
2 cos( r π2n )=
=
−(1 + e−ir πn )
2 cos( r π2n )= −e−i r π
2 n , whenever (n+ r) is even,
−1 + e−ir πn
2 cos( r π2n )=−i sin( r π2n )
cos( r π2n )e−i
r π2 n =
= cos( r π2n )− sec( r π2n )− i sin( r π2n ), whenever (n+ r) is odd.
Corollary 13. We have:n−1∑k=1
(−1)k cos(
(2 k − 1) r π2n
)=(54)
=
{− cos( r π2n), whenever (n+ r) is even,
cos( r π2n)− sec( r π2n), whenever (n+ r) is odd,
and
(55)n−1∑k=1
(−1)k sin(
(2 k − 1) r π2n
)= (−1)n+r sin
(r π2n
).
62 Roman Witu la, Damian S lota
Corollary 14. We have:
Θ+r,n(x) :=
n−1∑k=1
(−1)k C+r
(x,
(2 k − 1)π2n
)=
=
−21−rbr/2c∑k=0
(rk
)cos
((r−2 k)π
2n
)cos ((r − 2 k)x) , for r ∈ 2N,
22−rbr/2c∑k=0
(rk
) [csc
((r−2 k)π
2n
)−
− sec(
(r−2 k)π2n
) ]cos
((r − 2 k)x
), for r ∈ 2N− 1.
and
Θ−r,n(x) :=
n−1∑k=1
(−1)k C−r
(x,
(2 k − 1)π2n
)=(56)
= (−1)n+r
br/2c∑k=0
(r
k
)sin
(r π2n
)sin
((r − 2 k)x
).
Remark 2. All the trigonometric identities of (31)–(56), on the bases ofidentities (8)–(11), also may be given for S+
n and S−n functions.
References
[1] I. S. Gradshteyn, I. M. Ryzhik, Tables of Integrals, Series, and Products, AcademicPress, 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, SovetskayaNauka, 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 Rus-
sian).[7] R. Witu la, D. S lota, Decomposition of Certain Symmetric Functions of Powers of
Cosine and Sine Functions, Int. J. Pure Appl. Math. 50 (2009), pp. 1-12.[8] R. Witu la, D. S lota, On Modified Chebyshev Polynomials J. Math. Anal. Appl. 324
