Combinatorial identities concerning trigonometric functions and Fibonacci / Lucas numbers

: In this work, by means of the generating function method and the De Moivre’s formula, we derive some interesting combinatorial identities concerning trigonometric functions and Fibonacci / Lucas numbers. One of them confirms the formula proposed recently by Svinin (2022)


Introduction and motivation
The Fibonacci and Lucas numbers are defined by the following recurrence relations [5,6] with the explicit formulae of Binet forms It is interesting that they can be expressed by binomial coefficients, which can be found in Koshy [8,Eqs 12.1 and 13.5] and [1] In fact, there are two general binomial identities, which can be found in Carlitz [2, Page 23], Comtet [3, §4.9] and [5] 0≤k<n/2 When specifying u = α and v = β, they reduce to the formulae mentioned above.Recently, Svinin [10] proposed a problem, where including demands to show that which can be rewritten as When making attempts to resolve (1.1), we find that a series of similar identities, concerning trigonometric function or Fibonacci/Lucas numbers, can be established.In the process of proving some of these formulae, we will use the De Moivre's formula (cos θ + i sin θ) n = cos nθ + i sin nθ, by which we can get the following identities (cos θ + i sin θ) n + (cos θ − i sin θ) n = 2 cos(nθ), (1.2) (cos θ + i sin θ) n − (cos θ − i sin θ) n = 2i sin(nθ), (1.3) where i denotes the imaginary unit.
In the following sections, we will evaluate, by means of generating functions and binomial linear relations, the following binomial sums Specifically, in the next section, we shall consider the sums where δ ∈ {0, 1}.Then in Section 3, we will examine the binomial sums Finally in Section 4, the paper will end with two formulae, one of which is equivalent to (1.1).For convenience, throughout the paper we shall make use of the notations: for a real number x, the symbol ⌊x⌋ denotes the greatest integer ≤ x.When n is a natural number, the symbol i ≡ n j stands for "i is congruent to j modulo n".For a formal power series f (x), its coefficient of x n denote by [x n ] f (x).Considering two formal power series f (x) and g(x), the following two equalities are well known [4,9] which will be used frequently in the proofs in subsequent sections. (2.1) Proof.By exchanging the summation order, we have Making the replacement n → n + k − δ, the last equation becomes Keeping in mind the formula [11,Eq 5.57] we get the generating function

Four combinatorial identities concerning trigonometric functions
Firstly, we establish four combinatorial identities concerning trigonometric functions.
According to partial fraction decomposition, it can be decomposed into where A, B and C are three parameters to be determined.Multiplying both sides of the equation by 1 − 2z and taking the limit at z → 1 2 yields Similarly, multiplying both sides of the equation by 1 + iz and taking the limit at z → i yields Multiplying both sides of the equation by 1 − iz and taking the limit at z → −i yields Therefore, we can decompose the rational fraction Keeping in mind (1.4) and the fact that x n , we can determine the coefficient of z n By means of the relations (1.2) and (1.3), we can get the formulae which results in the following binomial summation formula Noting that when δ = 0 and x = 2 which yields another combinatorial identity below.
Then by means of the binomial relation below and after a little manipulation, we can derive, from (2.4) and (2.3), the identity in Theorem 2. □ Theorem 3 (n ∈ N). where .
By using partial fractional decomposition, we get the generating function .
Computing the coefficient of z n , we obtain By means of the relations (1.2) and (1. 3), we have we can establish the following combinatorial identity: Keeping in mind the relations and we can derive the identity below: where θ 1 = arctan .
By means of the partial fractional decomposition, we can get the generating function Then we can derive, by utilizing the relations the following summation formula: where According to the relation which leads to the following binomial sum: (2.12) where θ 1 = arctan 2 2 .Proof.Letting δ = 1 and x = 36 in (2.1), we have which results in, by the partial fractional decomposition, the generating function Combining it with the relations we can obtain the following identity: where According to the relation between the generating functions of δ = 1 and δ = 0 which yields the summation formula below: Now, employing the binomial relation (2.5), we can get, from (2.14) and (2.13), the identity stated in Theorem 5. □
Proof.Exchanging the summation order, we can evaluate By making the replacement n → n + 2k, the last equation becomes By means of the formula we get the generating function

Three formulae concerning trigonometric functions
Analogously, in this section we establish another three formulae concerning trigonometric functions.
Proof.Letting x = −2 in Lemma 7, we have the generating function .
By extracting the coefficient of z n , we get By means of the identities [7, Eqs.1.90 and 1.96] we obtain the following identity Then the desired identity follows by using the binomial relation (3.7).□ Theorem 9 (n ∈ N).
Proof.Letting x = 18 in Lemma 7, we have the generating function .
Evaluating the coefficient of z n , we get Keeping in mind that we have where tan t = √ 5, which yield the identity The desired identity follows by means of (3.7).□ Theorem 10 (n ∈ N).
Proof.Letting x = −36 in Lemma 7, we have the generating function By means of the following results we can get the identity Then the proof completes by the aid of (3.7).□

Four identities concerning Fibonacci and Lucas numbers
By means of the partial fractional decomposition, we can get the generating function Evaluating the coefficient of z n , we get , which results in the following identity: Theorem 11 (n ∈ N).