A new trigonometric identity with applications

TL;DR

This paper introduces a novel trigonometric identity involving sums of cotangent and sine functions for odd integers, with applications to Chebyshev polynomials and Bernoulli polynomials, revealing new mathematical relationships.

Contribution

It presents a new trigonometric identity for odd integers and explores its implications for Chebyshev and Bernoulli polynomials, expanding mathematical understanding.

Findings

Proves a new identity involving trigonometric sums for odd integers.

Derives equivalent identities involving Chebyshev polynomials.

Establishes zero-sum identities for Bernoulli polynomials with specific parameters.

Abstract

In this paper we obtain a new curious identity involving trigonometric functions. Namely, for any positive odd integer $n$ we prove that $$\sum_{k=1}^n(-1)^k(\cot kx)\sin k(n-k)x=\frac{1-n}2,$$ which is equivalent to the identity $$\sum_{k=1}^n(-1)^kU_{n-k}(\cos kx)=-\frac{n+1}2,$$ where $U_m(z)$ stands for the $m$th Chebyshev polynomial of the second kind. As a consequence, for any positive odd integer $n$ and positive integer $m$ we obtain $$\sum_{k=1}^n(-1)^kk^{2m}B_{2m+1}\left(\frac{n-k}2\right)=0,$$ where $B_j(x)$ denotes the Bernoulli polynomial of degree $j$.

Equations

$$\sum _{k=1}^n(-1{)}^k(\cot kx)\sin k(n-k)x=\frac{1-n}{2}.$$

$$\sum _{k=1}^n(-1{)}^k{k}^{2m}{B}_{2m+1}\left(\frac{n-k}{2}\right)=0,$$

$$\sum _{k=1}^{2n+1}\frac{(-1{)}^k{\omega }^{k(3k+1)/2}}{1-{\omega }^{3k}}=-\frac{n+1}{2},$$

$$\sum _{k=1}^{2n+1}\left(\frac{y^k}{1+y^{3k}}+\frac{(-y{)}^k}{1-y^{3k}}\right)=-n-1,$$

$$\text{Re}\bigg{(}\sum_{k=1}^{n-1}\left(\frac{\zeta^{k}}{1+\zeta^{km}}-(-1)^{n+\delta}\frac{(-\zeta)^{k}}{1-\zeta^{km}}\right)\bigg{)}=(-1)^{n-1}\left\lfloor\frac{n}{2}\right\rfloor.$$

$${\rm Re}\bigg{(}\sum_{k=1}^{n}\frac{(-1)^{k}q^{-k(n-k)/2}}{1-q^{k}}\bigg{)}=-\frac{n+1}{4}.$$

$$\sum _{k=1}^n(-1{)}^k(\cot kx)\sin k(n-k)x=\frac{1-n}{2},$$

$$\sum _{k=1}^n(-1{)}^k{k}^{2m}{B}_{2m+1}\left(\frac{n-k}{2}\right)=0,$$

$${\rm Re}\bigg{(}\sum_{k=1}^{n}\frac{\zeta^{k(km+l)/2}}{1-\zeta^{km}}\bigg{)}=-\frac{n+1}{4}.$$

$$\sum _{k=1}^{2n+1}\frac{{\zeta }^{k(3k+1)/2}}{1-{\zeta }^{3k}}=-\frac{n+1}{2}.$$

$$\sum _{\genfrac{}{}{0ex}{}{1⩽k⩽n}{0⩽j<(n-k)/2}}(-1{)}^k{z}^{k(2j+k-n)}=0.$$

$$\sigma =$$

$$=$$

$$L=$$

$$=$$

$$\text{Re}(L)=$$

$$=$$

$$\sum _{k=1}^n(-1{)}^k\cos \frac{k(n-k)}{2}\theta$$

$$=$$

$$2\text{Re}(L)+1=\sum _{k=1}^n(-1{)}^k\cot \frac{k\theta }{2}\sin \frac{k(n-k)}{2}\theta$$

$$\text{Re}(L)=$$

$$=$$

$$2\text{Re}(L)=$$

$$=$$

$$\sum _{(n-k)/2<j⩽n-k}{z}^{k(2j+k-n)}=\sum _{0⩽s<(n-k)/2}{z}^{k(2(n-k-s)+k-n)}=\sum _{0⩽s<(n-k)/2}{z}^{-k(2s+k-n)}.$$

$$2\text{Re}(L)=$$

$$+\sum _{k=1}^n(-1{)}^k\sum _{0⩽j<(n-k)/2}(z^{k(2j+k-n)}+z^{-k(2j+k-n)})$$

$$=$$

$$2\text{Re}(L)=(-1{)}^n\frac{n+1}{2}=-\frac{n+1}{2}$$

$$\cot x=\sum _{j=0}^{\infty }\frac{(-1{)}^j{2}^{2j}{B}_{2j}{x}^{2j-1}}{(2j)!}$$

$$\sin x=\sum _{j=0}^{\infty }\frac{(-1{)}^j{x}^{2j+1}}{(2j+1)!}.$$

$$\frac{1-n}{2}=\sum _{k=1}^n(-1{)}^k\cot kx\sin k(n-k)x$$

$$=$$

$$\sum _{k=1}^n(-1{)}^k{k}^{2m}\sum _{j=0}^m\frac{2^{2j}{B}_{2j}}{(2j)!}\cdot \frac{(n-k{)}^{2m-2j+1}}{(2m-2j+1)!}=0,$$

Full text

[TABLE]

Consequently, for any positive odd integer $n$ and positive integer $m$ we have

[TABLE]

where $B_{j}(x)$ denotes the Bernoulli polynomial of degree $j$.

2020 Mathematics Subject Classification. Primary 05A19, 33B10; Secondary 11B68.

The first author is the corresponding author. The two authors were supported by the Natural Science Foundation of China (grants 11971222 and 12071208 respectively).

1. Introduction

Let $\mathbb{Z}^{+}$ denote the set of all positive integers. J.-C. Liu and F. Petrov [LP, (2.11)] showed that if $\omega=e^{2\pi i/(3n+2)}$ with $n\in\mathbb{Z}^{+}$ then

[TABLE]

which has the equivalent form (cf. [LP, (2.17)])

[TABLE]

where $y=e^{2\pi i/(6n+4)}$. Motivated by this, Z.-W. Sun [S] conjectured that if $m,n\in\{2,3,\ldots\}$ and $\delta\in\{0,1\}$, then for any primitive $(m(n-\delta)-(-1)^{\delta})$-th root $\zeta$ of unity we have the identity

[TABLE]

This was confirmed by Nemo and Sun in the cases $\delta=0$ and $\delta=1$ respectively. see [S] for the detailed proofs.

Inspired by the above work, we establish the following new result.

Theorem 1.1**.**

Let $n$ be any positive odd integer. Then, for any complex number $q$ with $|q|=1$ and $q^{k}\not=1$ for all $k=1,\ldots,n$, we have

[TABLE]

Equivalently, we have the trigonometric identity

[TABLE]

where $x$ is any real number not in the set $\{\frac{m}{k}\pi:\ m\in\mathbb{Z},\ k=1,\ldots,n\}$.

Corollary 1.1**.**

Suppose that $n$ is positive odd integer and $m$ is a positive integer. Then we have

[TABLE]

where $B_{j}(x)$ denotes the Bernoulli polynomial of degree $j$.

With the help of Theorem 1.1, we obtain the following result.

Theorem 1.2**.**

Let $l,m,n\in\mathbb{Z}^{+}$ with $l\equiv m\ ({\rm{mod}}\ 2)$ and $n\equiv 1\ ({\rm{mod}}\ 2)$. Then, for any primitive $(mn+l)$-th root $\zeta$ of unity, we have

[TABLE]

Applying Theorem 1.2 with $l=1$ and $m=3$, we immediately get the following consequence.

Corollary 1.2**.**

Let $n$ be a nonnegative integer and let $\zeta$ be a primitive $(6n+4)$-th root of unity. Then

[TABLE]

It is interesting to compare our (1.8) with Liu and Petrov’s (1.1). Actually, we first found (1.8) motivated by (1.1) and then discovered the more general Theorem 1.2 and related Theorem 1.1.

2. Proof of Theorem 1.1

Lemma 2.1**.**

Let $n$ be a positive integer and let $z$ be any complex number. Then

[TABLE]

Proof. Let $\sigma$ denote the left-hand side of (2.1). Then

[TABLE]

and hence $\sigma=0$. ∎

Proof of Theorem 1.1. Write $q=e^{i\theta}=\cos\theta+i\sin\theta$ with $\theta$ real, and let $L$ denote the sum in (1.4). Then

[TABLE]

and hence

[TABLE]

Note that

[TABLE]

since $n$ is odd. Therefore

[TABLE]

and hence (1.4) is equivalent to (1.5) with $x=\theta/2$.

By (2.2), we also have

[TABLE]

Set $z=e^{ix}$. Then

[TABLE]

For each $k=1,\ldots,n$, clearly

[TABLE]

Thus

[TABLE]

Combining this with Lemma 2.1, we obtain that

[TABLE]

and hence (1.4) follows.

The proof of Theorem 1.1 is now complete. ∎

3. Proofs of Corollary 1.1 and Theorem 1.2

Proof of Corollary 1.1. It is well known that

[TABLE]

and

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So, by (1.5) we have

[TABLE]

Comparing the coefficients of $x^{2m}$ in the both sides of the above equality, we obtain

[TABLE]

which is equivalent to the desired identity (1.6). ∎

Proof of Theorem 1.2. Clearly $L=mn+l$ is even. For $k=1\ldots,n$, we have

[TABLE]

Thus

[TABLE]

Note that

[TABLE]

and $q=\zeta^{m}$ is a primitive $L_{0}$-th root of unity. Applying Theorem 1.1 we see that the real part of

[TABLE]

is $-(n+1)/4$. This concludes the proof of Theorem 1.2. ∎