Trigonometric Sums via Lagrange Interpolation

Abstract

By means of the Lagrange interpolation, we derive two trigonometric identities that are utilized to evaluate, in closed forms, eight classes of power sums of trigonometric functions over equally distributed angles around the unit circle. When the mth term is removed from the sums, four classes (among eight) are shown to admit also closed expressions, that are surprisingly independent of the integer parameter “m”.

1. Introduction and Outline

There is a venerable tradition of computing finite sums of trigonometric functions in literature [1,2,3,4,5,6,7,8,9]. These sums frequently occur in problems of physics [10,11,12] and mathematics such as the Riemann zeta function and Dedekind sums [13,14,15,16] with applications to the approximation theory [17,18,19] and the Fourier analysis [20,21,22,23].

Five years ago, Bataille [24] proposed a monthly problem about trigonometric sums, which can be stated equivalently as follows. Let m and n be integers with $1\le m\le n$. Evaluate the trigonometric sum

$$\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{cot}^2{\displaystyle \frac{(m-k)\pi }{n}}:\mathrm{Our}\mathrm{answer}\mathrm{is}{\displaystyle \frac{(n-1)(n-2)}{3}}.$$

It is remarkably interesting that the sum is independent of m. This is not an isolated case. As we shall show that there are several similar trigonometric identities.

Let $Q\left(\xi \right)$ be a polynomial of degree n with the distinct zeros ${\left\{{\xi }_k\right\}}_{k=1}^n$. Suppose that $P\left(\xi \right)$ is another polynomial of degree less than n. Then, the classical Lagrange interpolation can be stated as

$$P\left(\xi \right)=\sum _{k=1}^n{\displaystyle \frac{P\left({\xi }_k\right)Q\left(\xi \right)}{(\xi -{\xi }_k)Q^{\prime }\left({\xi }_k\right)}}.$$

(1)

This is equivalent to the partial fraction decomposition

$${\displaystyle \frac{P\left(\xi \right)}{Q\left(\xi \right)}}=\sum _{k=1}^n{\displaystyle \frac{P\left({\xi }_k\right)}{(\xi -{\xi }_k)Q^{\prime }\left({\xi }_k\right)}}.$$

(2)

For historical notes, applications, and recent developments about the Lagrange interpolation, the reader can consult [25,26,27,28,29] and the references therein.

When $\xi ={sin}^{2}x$, it is not hard to check that the function $Q\left(\xi \right)={sin}^{2}nx-{sin}^{2}ny$ results in a polynomial of degree n in $\xi ={sin}^{2}x$ with the distinct zeros given by ${\left\{{\xi }_k={sin}^2(y+{\displaystyle \frac{k\pi }{n}})\right\}}_{k=1}^n$. Keeping in mind that

$$P\left({\xi }_k\right)=P\left({sin}^2(y+\frac{k\pi }{n})\right)\mathrm{and}{Q}^{\prime }\left({\xi }_k\right)=\left({\displaystyle \frac{dQ}{dx}}/{\displaystyle \frac{d\xi }{dx}}\right){|}_{x=y+{\displaystyle \frac{k\pi }{n}}}={\displaystyle \frac{nsin2ny}{sin2(y+\frac{k\pi }{n})}},$$

we have from (2) the trigonometric identity:

$${\displaystyle \frac{nsin2nyP\left({sin}^{2}x\right)}{{sin}^{2}ny-{sin}^{2}nx}}=\sum _{k=1}^n{\displaystyle \frac{sin2(y+\frac{k\pi }{n})P\left({sin}^2(y+\frac{k\pi }{n})\right)}{{sin}^2(y+\frac{k\pi }{n})-{sin}^{2}x}}.$$

(3)

When $P\left(\xi \right)$ is specified by two particular trigonometric polynomials, the corresponding trigonometric identities will be utilized to evaluate eight classes of powers sums of trigonometric functions. Three simplest (but elegant) representatives are highlighted as follows:

$$\begin{array}{cc}\hfill \sum _{k=1}^{n}cot\left(y+{\displaystyle \frac{k\pi }{n}}\right)& =ncot\left(ny\right),\hfill \\ \hfill \sum _{k=1}^n{cot}^2\left(y+{\displaystyle \frac{k\pi }{n}}\right)& =n^2{csc}^2\left(ny\right)-n,\hfill \\ \hfill \sum _{k=1}^n{csc}^2\left(y+{\displaystyle \frac{k\pi }{n}}\right)& =n^2{csc}^2\left(ny\right).\hfill \end{array}$$

When the mth term is removed, we shall examine four classes (of eight) of these trigonometric sums, that will be shown surprisingly to be independent of the integer parameter “m”. For instance, for the two natural numbers m and n with $1\le m\le n$, we have the following interesting formulae:

$$\begin{array}{cccccc}\hfill •& \sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{tan}^2\left({\displaystyle \frac{m-k}{n}}\pi \right)\hfill & & =n(n-1),\hfill & & (\mathrm{for}\mathrm{odd}n\ge 3);\hfill \\ \hfill •& \sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{sec}^2\left({\displaystyle \frac{m-k}{n}}\pi \right)\hfill & & =n^2-1,\hfill & & (\mathrm{for}\mathrm{odd}n\ge 3);\hfill \\ \hfill •& \sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{cot}^2\left({\displaystyle \frac{m-k}{n}}\pi \right)\hfill & & ={\displaystyle \frac{(n-1)(n-2)}{3}},\hfill & \hfill & (\mathrm{for}\mathrm{all}n\ge 2);\hfill \\ \hfill •& \sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{csc}^2\left({\displaystyle \frac{m-k}{n}}\pi \right)\hfill & & ={\displaystyle \frac{n^2-1}{3}},\hfill & & (\mathrm{for}\mathrm{all}n\ge 2).\hfill \end{array}$$

The contents of the paper is organized as follows. In Section 2, analytic formulae will be established (see Section 2.1, Section 2.3, Section 2.5 and Section 2.7) for trigonometric sums about even powers of “$csc,sec,cot,tan$” over equally distributed angles around the unit circle. When the mth term is removed, the remaining sums are also evaluated in closed forms (see Section 2.2, Section 2.4, Section 2.6 and Section 2.8), that are remarkably independent of m. In Section 3, the trigonometric sums corresponding odd powers are examined and evaluated analytically. Finally, the paper ends with concluding comments and a brief discussion about further prospects.

Throughout the paper, $\mathbb{N}$ represents the set of natural numbers with ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}$ as usual. Let $\chi$ stand for the logical function with $\chi \left(\mathrm{true}\right)=1$ and $\chi \left(\mathrm{false}\right)=0$. In order to reduce lengthy expressions, we use “$i{\equiv }_{m}j$” to denote that “i is congruent to j modulo m” for $i,j,m\in {\mathbb{N}}_0$ with $m>1$.

2. Trigonometric Sums of Even Powers

Letting $P\left(\xi \right)={\displaystyle \frac{sin2nx}{sin2x}}$ in (3), we deduce the identity

$${\displaystyle \frac{nsin2nx}{{sin}^{2}ny-{sin}^{2}nx}}=\sum _{k=1}^n{\displaystyle \frac{sin2x}{{sin}^2(y+\frac{k\pi }{n})-{sin}^{2}x}},$$

(4)

which will be utilized in this section to investigate four classes of trigonometric sums (see Section 2.1, Section 2.3, Section 2.5 and Section 2.7). When $y\to -{\displaystyle \frac{m}{n}}\pi$, the corresponding sums (see Section 2.2, Section 2.4, Section 2.6 and Section 2.8) with the mth term being removed can be evaluated in closed forms, that are remarkably independent of m.

In particular, for odd n, our formulae in Theorem 1 and Proposition 1 (with $y\to \pi /2$) as well as in Theorem 3 and Proposition 2 (with $y\to 0$) recover some previously known ones appearing in [30,31,32,33].

2.1. Sums About Even Powers of Cosecant Function

For the trigonometric sums defined by

$$A_n^{\lambda }\left(y\right):=\sum _{k=1}^n{csc}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right),\mathrm{where}\lambda \in \mathbb{N};$$

(5)

we can derive the following generating function

$$\eta =sinx:\sum _{\lambda =0}^{\infty }{A}_n^{2\lambda +2}\left(y\right){\eta }^{2\lambda }={\displaystyle \frac{nsin2nx}{sin2x({sin}^{2}ny-{sin}^{2}nx)}}.$$

(6)

In fact, this can be justified by rewriting (4) first and then expanding it into a formal power series

$$\begin{array}{cc}\hfill {\displaystyle \frac{nsin2nx}{sin2x({sin}^{2}ny-{sin}^{2}nx)}}& =\sum _{k=1}^n{\displaystyle \frac{{csc}^2(y+\frac{k\pi }{n})}{1-{sin}^{2}x{csc}^2(y+\frac{k\pi }{n})}}\hfill \\ & =\sum _{k=1}^n\sum _{\lambda =0}^{\infty }{sin}^{2\lambda }x{csc}^{2\lambda +2}(y+{\displaystyle \frac{k\pi }{n}}).\hfill \end{array}$$

Furthermore, we can show from (6) that $A_n^{2+2\lambda }\left(y\right)$ is a polynomial of $csc\left(ny\right)$ as highlighted in the following theorem.

Theorem 1

(Explicit formula: $\lambda \in {\mathbb{N}}_0$).

$$A_n^{2+2\lambda }\left(y\right)=n\sum _{k=0}^{\lambda }{4}^{\lambda -k}{csc}^{2k+2}\left(ny\right)\sum _{j=-k}^k{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +n+jn}{2\lambda +1}}\right).$$

The first five formulae are displayed as in the corollary below.

Corollary 1

($n\in \mathbb{N}$).

$$\begin{array}{cc}\hfill A_n^2\left(y\right)& =n^2{csc}^2\left(ny\right),\hfill \\ \hfill A_n^4\left(y\right)& =n^4{csc}^4\left(ny\right)+{\displaystyle \frac{2n^2{csc}^2\left(ny\right)}{3}}(1-n^2),\hfill \\ \hfill A_n^6\left(y\right)& =n^6{csc}^6\left(ny\right)+n^4{csc}^4\left(ny\right)(1-n^2)\hfill \\ & +{\displaystyle \frac{2n^2{csc}^2\left(ny\right)}{15}}(1-n^2)(4-n^2),\hfill \\ \hfill A_n^8\left(y\right)& =n^8{csc}^8\left(ny\right)+{\displaystyle \frac{4n^6{csc}^6\left(ny\right)}{3}}(1-n^2)\hfill \\ & +{\displaystyle \frac{2n^4{csc}^4\left(ny\right)}{15}}(1-n^2)(7-3n^2)\hfill \\ & +{\displaystyle \frac{4n^2{csc}^2\left(ny\right)}{315}}(1-n^2)(4-n^2)(9-n^2),\hfill \\ \hfill A_n^{10}\left(y\right)& =n^{10}{csc}^{10}\left(ny\right)+{\displaystyle \frac{5n^8{csc}^8\left(ny\right)}{3}}(1-n^2)\hfill \\ & +{\displaystyle \frac{n^6{csc}^6\left(ny\right)}{9}}(1-n^2)(13-7n^2)\hfill \\ & +{\displaystyle \frac{n^4{csc}^4\left(ny\right)}{189}}(1-n^2)(4-n^2)(41-17n^2)\hfill \\ & +{\displaystyle \frac{2n^2{csc}^2\left(ny\right)(1-n^2)(4-n^2)(9-n^2)(16-n^2)}{2835}}.\hfill \end{array}$$

To prove Theorem 1, we need the following trigonometric identity.

Lemma 1

($n\in \mathbb{N}$).

$${\displaystyle \frac{sin2nx}{sin2x}}=\sum _{k=0}^{n-1}{(-4)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k+n}{2k+1}}\right){sin}^{2k}x.$$

This lemma can be confirmed by making use of the following algebraic identity (cf. Carlitz [34], Chu–Li [35] and Comtet [36], §4.9):

$${\displaystyle \frac{u^m-v^m}{u-v}}=\sum _{0\le k<m/2}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m-k-1}{k}}\right){\left(uv\right)}^k{(u+v)}^{m-2k-1}.$$

In fact, under the replacements $u\to e^{x\mathbf{i}}$, $v\to -e^{-x\mathbf{i}}$ and $m\to 2n$, we can restate the above equation as

$${\displaystyle \frac{sin2nx}{sin2x}}={\displaystyle \frac{sin2nx}{2sinxcosx}}=\sum _{\ell =0}^{n-1}{(-1)}^{n-\ell -1}\left({\displaystyle \genfrac{}{}{0pt}{}{2n-\ell -1}{\ell }}\right){(2sinx)}^{2n-2\ell -2},$$

which becomes the equality in Lemma 1 under the substitution $\ell \to n-\lambda -1$.

Proof of Theorem 1.

Denote by $\left[{\eta }^k\right]\varphi \left(\eta \right)$ the coefficient of ${\eta }^k$ in the formal power series $\varphi \left(\eta \right)$. According to (6), we have

$${\displaystyle \frac{nsin\left(2nx\right)}{sin\left(2x\right)\left\{{sin}^2\left(ny\right)-{sin}^2\left(nx\right)\right\}}}={\displaystyle \frac{nsin\left(2nx\right)}{sin\left(2x\right)}}\sum _{k=0}^{\infty }{sin}^{2k}\left(nx\right){csc}^{2k+2}\left(ny\right).$$

Then, we can express the coefficient

$$A_n^{2+2\lambda }\left(y\right)=n\sum _{k=0}^{\lambda }{csc}^{2k+2}\left(ny\right)\left[{\eta }^{2\lambda }\right]{\displaystyle \frac{sin\left(2nx\right)}{sin\left(2x\right)}}{sin}^{2k}\left(nx\right).$$

Recalling that

$$\eta =sinx\mathrm{and}{e}^{x\mathbf{i}}=\sqrt{1-{\eta }^2}+\eta \mathbf{i},$$

we can manipulate the trigonometric fraction

$$\begin{array}{cc}\hfill {\displaystyle \frac{sin\left(2nx\right)}{sin\left(2x\right)}}{sin}^{2k}\left(nx\right)& ={\displaystyle \frac{{(e^{nx\mathbf{i}}-e^{-nx\mathbf{i}})}^{2k}}{{\left(2\mathbf{i}\right)}^{2k+1}}}{\displaystyle \frac{e^{2nx\mathbf{i}}-e^{-2nx\mathbf{i}}}{sin2x}}\hfill \\ & =\sum _{j=-k}^k{\displaystyle \frac{{(-1)}^j}{2\mathbf{i}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right){\displaystyle \frac{e^{2n(j+1)x\mathbf{i}}-e^{2n(j-1)x\mathbf{i}}}{4^{k}sin2x}}\hfill \\ & =\sum _{j=-k}^k{\displaystyle \frac{{(-1)}^j}{2\mathbf{i}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right){\displaystyle \frac{e^{2n(j+1)x\mathbf{i}}-e^{-2n(j+1)x\mathbf{i}}}{4^{k}sin2x}}\hfill \\ & =\sum _{j=-k}^k{\displaystyle \frac{{(-1)}^j}{4^k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right){\displaystyle \frac{sin2n(j+1)x}{sin2x}},\hfill \end{array}$$

where the passage from the second line to the third line is justified by inverting the index $j\to -j$ for the sum corresponding to $e^{2n(j-1)x\mathbf{i}}$. This leads us to the following expression

$$A_n^{2+2\lambda }\left(y\right)=n\sum _{k=0}^{\lambda }{csc}^{2k+2}\left(ny\right)\sum _{j=-k}^k{\displaystyle \frac{{(-1)}^j}{4^k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left[{\eta }^{2\lambda }\right]{\displaystyle \frac{sin2n(j+1)x}{sin2x}}.$$

Then the explicit formula in Theorem 1 follows directly from Lemma 1. □

2.2. Sums About Even Powers of Cosecant with mth Term Missing

By means of Theorem 1, we can evaluate another class of trigonometric sums. Let m be an integer with $1\le m\le n$. Denote by ${\mathcal{A}}_n^{\lambda }(y;m)$ the sum $A_n^{\lambda }\left(y\right)$ with the mth term being removed

$${\mathcal{A}}_n^{\lambda }(y;m):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{csc}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

Similar to (6), we have the generating function

$$\begin{array}{cc}\eta =sinx:{\displaystyle \sum _{\lambda =0}^{\infty }}{\mathcal{A}}_n^{2\lambda +2}(y;m){\eta }^{2\lambda }& ={\displaystyle \frac{1}{{sin}^{2}x-{sin}^2(y+\frac{m\pi }{n})}}\hfill \\ & +{\displaystyle \frac{nsin\left(2nx\right)}{sin\left(2x\right)\left\{{sin}^2\left(ny\right)-{sin}^2\left(nx\right)\right\}}}.\end{array}$$

(7)

Evaluating the limit

$$\begin{array}{cc}\hfill \underset{y\to -{\displaystyle \frac{m\pi }{n}}}{lim}& \left\{{\displaystyle \frac{1}{{sin}^{2}x-{sin}^2(y+\frac{m\pi }{n})}}+{\displaystyle \frac{nsin\left(2nx\right)}{sin\left(2x\right)\left\{{sin}^2\left(ny\right)-{sin}^2\left(nx\right)\right\}}}\right\}\hfill \\ & ={\displaystyle \frac{1}{{sin}^{2}x}}-{\displaystyle \frac{nsin\left(2nx\right)}{sin\left(2x\right){sin}^2\left(nx\right)}}=cscx\left\{cscx-{\displaystyle \frac{ncotnx}{cosx}}\right\},\hfill \end{array}$$

we obtain the generating function

$$\eta =sinx:\sum _{\lambda =0}^{\infty }{\mathbf{A}}_n^{2\lambda +2}\left(m\right){\eta }^{2\lambda }=cscx\left\{cscx-{\displaystyle \frac{ncotnx}{cosx}}\right\}$$

(8)

for the trigonometric sums defined by

$${\mathbf{A}}_n^{\lambda }\left(m\right):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{csc}^{\lambda }\left({\displaystyle \frac{m-k}{n}}\pi \right).$$

(9)

What is interesting is that these sums are independent of m, as highlighted in the following explicit formula.

Theorem 2.

For the three natural numbers λ, m, and n with $1\le m\le n$, the trigonometric sum ${\mathbf{A}}_n^{2\lambda }\left(m\right)$ is independent of m and admits the closed formula

$${\mathbf{A}}_n^{2\lambda }\left(m\right)=-{\mathrm{U}}_{\lambda }\left(n\right),$$

(10)

where ${\mathrm{U}}_{\lambda }\left(n\right)$ are the coefficients of the power series

$${\displaystyle \frac{ncotnx}{cotx}}=\sum _{\lambda =0}^{\infty }{\mathrm{U}}_{\lambda }\left(n\right){sin}^{2\lambda }x,$$

(11)

and satisfy the recurrence relation

$$\sum _{k=0}^{n-1}{\mathrm{U}}_{\lambda -k}\left(n\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k+n}{2k+1}}\right){\displaystyle \frac{{(-4)}^k}{k+1}}={(-4)}^{\lambda }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +n}{2\lambda +1}}\right).$$

(12)

The first five closed formulae are recorded as consequences.

Corollary 2

($n\in \mathbb{N}$).

$$\begin{array}{cc}\hfill {\mathbf{A}}_n^2\left(m\right)& ={\displaystyle \frac{n^2-1}{3}},\hfill \\ \hfill {\mathbf{A}}_n^4\left(m\right)& ={\displaystyle \frac{n^2-1}{45}}\left\{n^2+11\right\},\hfill \\ \hfill {\mathbf{A}}_n^6\left(m\right)& ={\displaystyle \frac{n^2-1}{945}}\left\{2n^4+23n^2+191\right\},\hfill \\ \hfill {\mathbf{A}}_n^8\left(m\right)& ={\displaystyle \frac{n^2-1}{14175}}(n^2+11)\left\{3n^4+10n^2+227\right\},\hfill \\ \hfill {\mathbf{A}}_n^{10}\left(m\right)& ={\displaystyle \frac{n^2-1}{93555}}\left\{2n^8+35n^6+321n^4+2125n^2+14797\right\}.\hfill \end{array}$$

Remark 1.

The sequence ${\mathrm{U}}_k\left(n\right)$ is quite interesting. For small values of n, we have figured out the following facts:

$$\begin{array}{cc}\hfill n=1:& {\mathrm{U}}_0\left(1\right)=1\mathit{and}{\mathrm{U}}_k\left(1\right)=0\mathit{for}k>0.\hfill \\ \hfill n=2:& {\mathrm{U}}_0\left(2\right)=1\mathit{and}{\mathrm{U}}_k\left(2\right)=-1\mathit{for}k>0.\hfill \\ \hfill n=3:& {\mathrm{U}}_0\left(3\right)=1\mathit{and}{\mathrm{U}}_k\left(3\right)=-{\displaystyle \frac{2^{2k+1}}{3^k}}\mathit{for}k>0.\hfill \\ \hfill n=4:& {\mathrm{U}}_0\left(4\right)=1\mathit{and}{\mathrm{U}}_k\left(4\right)=-1-2^{k+1}\mathit{for}k>0.\hfill \\ \hfill n=5:& {\mathrm{U}}_0\left(5\right)=1\mathit{and}{\mathrm{U}}_k\left(5\right)=-{\displaystyle \frac{2^{2k+1}}{5^{k-1}}}{\alpha }_k\mathit{for}k>0.\hfill \end{array}$$

The last sequence ${\alpha }_k$ is Fibonacci–like (labeled by [A081567] in [37]), that is expressed as the double sum

$${\alpha }_k=\sum _{i=0}^k\sum _{j=i}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i+j}{2i}}\right),$$

and satisfies the recurrence relation

$${\alpha }_k=5{\alpha }_{k-1}-5{\alpha }_{k-2}\mathrm{with}{\alpha }_0=1\mathrm{and}{\alpha }_1=3.$$

In order to show Theorem 2, we state first the following lemma.

Lemma 2

($n\in \mathbb{N}$).

$${\displaystyle \frac{{sin}^{2}nx}{n{sin}^{2}x}}=\sum _{k=0}^{n-1}{\displaystyle \frac{{(-4)}^k}{k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{k+n}{2k+1}}\right){sin}^{2k}x.$$

(13)

This can be verified by another algebraic identity (cf. Carlitz [34], Chu–Li [35], and Comtet [36], §4.9):

$$u^m+v^m=\sum _{0\le k\le m/2}{(-1)}^k{\displaystyle \frac{m}{m-k}}\left({\displaystyle \genfrac{}{}{0pt}{}{m-k}{k}}\right){\left(uv\right)}^k{(u+v)}^{m-2k}.$$

Specifying $u\to e^{x\mathbf{i}}$, $v\to -e^{-x\mathbf{i}}$ and $m\to 2n$ in this equality, we have

$$cos\left(2nx\right)=\sum _{\ell =0}^n{(-1)}^{\ell }{\displaystyle \frac{n}{n+\ell }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+\ell }{2\ell }}\right){(2sinx)}^{2\ell }.$$

(14)

From this identity, we have further

$${\displaystyle \frac{{sin}^{2}nx}{n{sin}^{2}x}}={\displaystyle \frac{1-cos2nx}{2n{sin}^{2}x}}=2\sum _{\ell =1}^n{\displaystyle \frac{{(-1)}^{\ell -1}}{n+\ell }}\left({\displaystyle \genfrac{}{}{0pt}{}{n+\ell }{2\ell }}\right){(2sinx)}^{2\ell -2},$$

which confirms Lemma 2 under the replacement $\ell \to k+1$.

Proof of Theorem 2.

For $\lambda \in \mathbb{N}$, we derive (10) and (11) from (8) as follows:

$$\begin{array}{cc}\hfill {\mathbf{A}}_n^{2\lambda }\left(m\right)& =\left[{\eta }^{2\lambda }\right]\left\{1-{\displaystyle \frac{ncotnx}{cotx}}\right\}=-\left[{\eta }^{2\lambda }\right]{\displaystyle \frac{ncotnx}{cotx}}\begin{array}{|c|}\hline \eta =sinx\\ \hline\end{array}\hfill \\ & =-\left[{\eta }^{2\lambda }\right]\sum _{k=1}^{\infty }{\mathrm{U}}_k\left(n\right){sin}^{2k}x=-{\mathrm{U}}_{\lambda }\left(n\right).\hfill \end{array}$$

Alternatively, observe the trigonometric equality

$${\displaystyle \frac{sin2nx}{sin2x}}={\displaystyle \frac{ncotnx}{cotx}}\times {\displaystyle \frac{{sin}^{2}nx}{n{sin}^{2}x}}.$$

According to Lemmas 1 and 2, extracting the coefficient of ${\eta }^{2\lambda }={sin}^{2\lambda }x$ across, we find the expression stated in (12):

$$\begin{array}{cc}\hfill \left[{\eta }^{2\lambda }\right]{\displaystyle \frac{sin2nx}{sin2x}}& ={(-4)}^{\lambda }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +n}{2\lambda +1}}\right)\chi (\lambda <n)\hfill \\ & =\left[{\eta }^{2\lambda }\right]\left\{{\displaystyle \frac{ncotnx}{cotx}}\times {\displaystyle \frac{{sin}^{2}nx}{n{sin}^{2}x}}\right\}\hfill \\ & =\sum _{k=0}^{n-1}{\mathrm{U}}_{\lambda -k}\left(n\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k+n}{2k+1}}\right){\displaystyle \frac{{(-4)}^k}{k+1}}.\hfill \end{array}$$

This completes the proof of Theorem 2. □

2.3. Sums About Even Powers of Secant Function

Define the trigonometric sums by

$$B_n^{\lambda }\left(y\right):=\sum _{k=1}^n{sec}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

(15)

Replacing y by $y+\pi /2$ in (6), we find the generating function

$$\eta =sinx:\sum _{\lambda =0}^{\infty }{B}_n^{2\lambda +2}\left(y\right){\eta }^{2\lambda }=\left\{\begin{array}{cc}{\displaystyle \frac{nsin2nx}{sin2x({sin}^{2}ny-{sin}^{2}nx)}},\hfill & n{\equiv }_{2}0;\hfill \\ {\displaystyle \frac{nsin2nx}{sin2x({cos}^{2}ny-{sin}^{2}nx)}},\hfill & n{\equiv }_{2}1.\hfill \end{array}\right.$$

(16)

Analogous to Theorem 1, we have the polynomial representation.

Theorem 3

(Explicit formula: $\lambda \in {\mathbb{N}}_0$).

$$B_n^{2+2\lambda }\left(y\right)=n\sum _{k=0}^{\lambda }{4}^{\lambda -k}\sum _{j=-k}^k{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +n+jn}{2\lambda +1}}\right)\times \{\begin{array}{c}{csc}^{2k+2}\left(ny\right),n{\equiv }_{2}0;\hfill \\ {sec}^{2k+2}\left(ny\right),n{\equiv }_{2}1.\hfill \end{array}$$

When n is odd, we get, in particular, the formula

$$B_n^{2+2\lambda }\left(0\right)=n\sum _{k=0}^{\lambda }{4}^{\lambda -k}\sum _{j=-k}^k{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +n+jn}{2\lambda +1}}\right),$$

which is equivalent to the formula due to Chu–Marini [30] (Equation A1.3), who gave also the first five examples for small $\lambda$ values.

2.4. Sums About Even Powers of Secant with mth Term Missing

Denote further by ${\mathcal{B}}_n^{\lambda }(y;m)$ the sum $B_n^{\lambda }\left(y\right)$ with the mth term being deleted

$${\mathcal{B}}_n^{\lambda }(y;m):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{sec}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

Replacing y by $y+\pi /2$ in (7) results in the expression below

$$\begin{array}{cc}\hfill \eta =sinx:\sum _{\lambda =0}^{\infty }{\mathcal{B}}_n^{2\lambda +2}(y;m){\eta }^{2\lambda }& ={\displaystyle \frac{1}{{sin}^{2}x-{cos}^2(y+\frac{m\pi }{n})}}\hfill \\ & +\left\{\begin{array}{cc}{\displaystyle \frac{nsin\left(2nx\right)}{sin\left(2x\right)\left\{{sin}^2\left(ny\right)-{sin}^2\left(nx\right)\right\}}},\hfill & n{\equiv }_{2}0;\hfill \\ {\displaystyle \frac{nsin\left(2nx\right)}{sin\left(2x\right)\left\{{cos}^2\left(ny\right)-{sin}^2\left(nx\right)\right\}}},\hfill & n{\equiv }_{2}1.\hfill \end{array}\right.\hfill \end{array}$$

(17)

Its limiting case $y\to -{\displaystyle \frac{m\pi }{n}}$ yields the generating function

$$\eta =sinx:\sum _{\lambda =0}^{\infty }{\eta }^{2\lambda }{\mathbf{B}}_n^{2\lambda +2}\left(m\right)=-{sec}^{2}x+\left\{\begin{array}{cc}-2n{\displaystyle \frac{cot\left(nx\right)}{sin\left(2x\right)}},\hfill & n{\equiv }_{2}0;\hfill \\ 2n{\displaystyle \frac{tan\left(nx\right)}{sin\left(2x\right)}},\hfill & n{\equiv }_{2}1;\hfill \end{array}\right.$$

(18)

for the trigonometric sums

$${\mathbf{B}}_n^{\lambda }\left(m\right):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{sec}^{\lambda }\left({\displaystyle \frac{m-k}{n}}\pi \right).$$

(19)

Consequently, we have the general formula given by the following theorem.

Theorem 4.

For the three natural numbers λ, m and n with n being odd and $1\le m\le n$, the trigonometric sum ${\mathbf{B}}_n^{2\lambda }\left(m\right)$ is independent of m and admits the closed formula

$${\mathbf{B}}_n^{2\lambda }\left(m\right)=-1+n^2\sum _{k=0}^{\lambda -1}{\mathrm{V}}_k\left(n\right),$$

(20)

where ${\mathrm{V}}_k\left(n\right)$ are the coefficients of the power series

$${\displaystyle \frac{tannx}{ntanx}}=\sum _{k=0}^{\infty }{\mathrm{V}}_k\left(n\right){sin}^{2k}x$$

(21)

and satisfy the recurrence relation

$$\sum _{k=0}^{n-1}{\mathrm{V}}_{\lambda -k}\left(n\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k+n}{2k+1}}\right){(-4)}^k={\displaystyle \frac{{(-4)}^{\lambda }}{\lambda +1}}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +n}{2\lambda +1}}\right).$$

(22)

In particular, the following concrete formulae can be deduced as examples.

Corollary 3

($m,n\in \mathbb{N}$ with n being odd and $1\le m\le n$).

$$\begin{array}{cc}\hfill {\mathbf{B}}_n^2\left(m\right)& =n^2-1,\hfill \\ \hfill {\mathbf{B}}_n^4\left(m\right)& ={\displaystyle \frac{n^2-1}{3}}\left\{n^2+3\right\},\hfill \\ \hfill {\mathbf{B}}_n^6\left(m\right)& ={\displaystyle \frac{n^2-1}{15}}\left\{2n^4+7n^2+15\right\},\hfill \\ \hfill {\mathbf{B}}_n^8\left(m\right)& ={\displaystyle \frac{n^2-1}{315}}(n^2+3)\left\{17n^4+22n^2+105\right\},\hfill \\ \hfill {\mathbf{B}}_n^{10}\left(m\right)& ={\displaystyle \frac{n^2-1}{2835}}\left\{62n^8+317n^6+863n^4+1683n^2+2835\right\}.\hfill \end{array}$$

Remark 2.

According to (11) and ${\displaystyle \frac{tannx}{ntanx}}={\displaystyle \frac{cotx}{ncotnx}}$, we have the power series expansion (21) and the reciprocal relation

$$\sum _{k=0}^m{\mathrm{U}}_k\left(n\right){\mathrm{V}}_{m-k}\left(n\right)=\chi (m=0).$$

Analogous to the sequence ${\mathrm{U}}_k\left(n\right)$, we find, by employing the recurrence relation (22), the following interesting facts about the sequence ${\mathrm{V}}_k\left(n\right)$ for small values of n:

$$\begin{array}{cc}\hfill n=1:& {\mathrm{V}}_0\left(1\right)=1\mathit{and}{\mathrm{V}}_k\left(1\right)=0\mathit{for}k>0.\hfill \\ \hfill n=2:& {\mathrm{V}}_0\left(2\right)=1\mathit{and}{\mathrm{V}}_k\left(2\right)=2^{k-1}\mathit{for}k>0.\hfill \\ \hfill n=3:& {\mathrm{V}}_0\left(3\right)=1\mathit{and}{\mathrm{V}}_k\left(3\right)={\displaystyle \frac{2^{2k+1}}{3}}\mathit{for}k>0.\hfill \\ \hfill n=4:& {\mathrm{V}}_0\left(4\right)=1\mathit{and}{\mathrm{V}}_k\left(4\right)=2^{k-1}{\beta }_k\mathit{for}k>0.\hfill \\ \hfill n=5:& {\mathrm{V}}_0\left(5\right)=1\mathit{and}{\mathrm{V}}_k\left(5\right)={\displaystyle \frac{2^{2k+1}}{5^k}}F(2k+3)\mathit{for}k>0.\hfill \end{array}$$

Here $F(2k+3)$ is the usual Fibonacci number and the sequence ${\beta }_k$ is Fibonacci–like (labeled by [A010914] in [37]), that satisfies the recurrence relation

$${\beta }_k=4{\beta }_{k-1}-2{\beta }_{k-2}\mathrm{with}{\beta }_1=5\mathrm{and}{\beta }_2=17.$$

Proof of Theorem 4.

By extracting the coefficient of ${sin}^{2\lambda }x$ across

$${\displaystyle \frac{tannx}{ntanx}}\times {\displaystyle \frac{sin2nx}{sin2x}}={\displaystyle \frac{{sin}^{2}nx}{n{sin}^{2}x}},$$

we obtain the recurrence relation (22). Formula (20) is confirmed as follows:

$$\begin{array}{cc}\hfill {\mathbf{B}}_n^{2\lambda }\left(m\right)& =\left[{\eta }^{2\lambda -2}\right]{\displaystyle \frac{1}{{cos}^{2}x}}\left\{-1+n^2{\displaystyle \frac{tannx}{ntanx}}\right\}\hfill \\ & =\left[{\eta }^{2\lambda -2}\right]{\displaystyle \frac{1}{1-{sin}^{2}x}}\left\{-1+n^2\sum _{k=0}^{\infty }{\mathrm{V}}_k\left(n\right){sin}^{2k}x\right\}\hfill \\ & =\left[{\eta }^{2\lambda -2}\right]\sum _{m=0}^{\infty }{sin}^{2m}x\left\{-1+n^2\sum _{k=0}^m{\eta }^{2k}{\mathrm{V}}_k\left(n\right)\right\}\hfill \\ & =-1+n^2\sum _{k=0}^{\lambda -1}{\mathrm{V}}_k\left(n\right).\square \hfill \end{array}$$

2.5. Sums About Even Powers of Cotangent Function

Define the trigonometric sums by

$$C_n^{\lambda }\left(y\right):=\sum _{k=1}^n{cot}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

(23)

According to $1+{cot}^{2}y={csc}^{2}y$, we have

$${cot}^{2\lambda }y={({csc}^{2}y-1)}^{\lambda }=\sum _{\ell =0}^{\lambda }{(-1)}^{\lambda -\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right){csc}^{2\ell }y.$$

From this, we can evaluate the sum

$$C_n^{2\lambda +2}\left(y\right)=n{(-1)}^{\lambda +1}+\sum _{\ell =0}^{\lambda }{(-1)}^{\lambda -\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +1}{\ell +1}}\right)A_n^{2\ell +2}\left(y\right).$$

By substituting the formula in Theorem 1 into the above sum, we get the polynomial representation.

Proposition 1

(Explicit formula: $\lambda \in {\mathbb{N}}_0$).

$$\begin{array}{cc}{C}_n^{2\lambda +2}\left(y\right)=n{(-1)}^{\lambda +1}+& n\sum _{k=0}^{\lambda }\sum _{j=-k}^k{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right){csc}^{2k+2}\left(ny\right)\hfill \\ & \times \sum _{\ell =k}^{\lambda }{4}^{\ell -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +1}{\ell +1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +n+jn}{2\ell +1}}\right).\hfill \end{array}$$

The first three expressions are displayed as below.

Corollary 4

($n\in \mathbb{N}$).

$$\begin{array}{cc}\hfill C_n^2\left(y\right)& ={(ncscny)}^2-n,\hfill \\ \hfill C_n^4\left(y\right)& ={(ncscny)}^4-{\displaystyle \frac{2(n^2+2)}{3}}{(ncscny)}^2+n,\hfill \\ \hfill C_n^6\left(y\right)& ={(ncscny)}^6-(n^2+2){(ncscny)}^4+{\displaystyle \frac{2n^4+20n^2+23}{15}}{(ncscny)}^2-n.\hfill \end{array}$$

In addition, the generating function can be treated in the following manner

$$\begin{array}{cc}\hfill \sum _{\lambda =1}^{\infty }{C}_n^{2\lambda }\left(y\right){\eta }^{2\lambda -2}& =\sum _{\lambda =1}^{\infty }{\eta }^{2\lambda -2}\sum _{j=0}^{\lambda }{(-1)}^{\lambda -j}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{j}}\right)A_n^{2j}\left(y\right)\hfill \\ & =\sum _{j=0}^{\infty }{A}_n^{2j}\left(y\right)\sum _{\lambda =max\{1,j\}}^{\infty }{(-1)}^{\lambda -j}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{j}}\right){\eta }^{2\lambda -2}\hfill \\ & =\sum _{j=1}^{\infty }{A}_n^{2j}\left(y\right){\displaystyle \frac{{\eta }^{2j-2}}{{(1+{\eta }^2)}^{j+1}}}-{\displaystyle \frac{n}{1+{\eta }^2}},\hfill \end{array}$$

where

$$\eta =tanx\mathrm{and}{\displaystyle \frac{{\eta }^2}{1+{\eta }^2}}={sin}^{2}x.$$

Recalling (6), we find explicitly the generating function

$$\eta =tanx:\sum _{\lambda =0}^{\infty }{C}_n^{2\lambda +2}\left(y\right){\eta }^{2\lambda }=-n{cos}^{2}x+{\displaystyle \frac{n{cos}^{3}xsin2nx}{2sinx({sin}^{2}ny-{sin}^{2}nx)}}.$$

(24)

2.6. Sums About Even Powers of Cotangent with mth Term Missing

Denote by ${\mathcal{C}}_n^{\lambda }(y;m)$ the sum $C_n^{\lambda }\left(y\right)$ with the mth term being removed

$${\mathcal{C}}_n^{\lambda }(y;m):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{cot}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

Similar to (24), we have the generating function

$$\begin{array}{cc}\eta =tanx:{\displaystyle \sum _{\lambda =0}^{\infty }}{\mathcal{C}}_n^{2\lambda +2}& (y;m){\eta }^{2\lambda }={\displaystyle \frac{1}{{tan}^{2}x-{tan}^2(y+\frac{m\pi }{n})}}\hfill \\ & -n{cos}^{2}x+{\displaystyle \frac{n{cos}^{3}xsin2nx}{2sinx({sin}^{2}ny-{sin}^{2}nx)}}.\end{array}$$

(25)

The limit of the right-hand side equals

$$\begin{array}{cc}\hfill \underset{y\to -{\displaystyle \frac{m\pi }{n}}}{lim}& \left\{{\displaystyle \frac{1}{{tan}^{2}x-{tan}^2(y+\frac{m\pi }{n})}}-n{cos}^{2}x+{\displaystyle \frac{n{cos}^{3}xsin2nx}{2sinx({sin}^{2}ny-{sin}^{2}nx)}}\right\}\hfill \\ & ={cot}^{2}x-n{cos}^{2}x-n{cos}^{2}xcotxcotnx.\hfill \end{array}$$

Therefore, we have established the generating function

$$\eta =tanx:\sum _{\lambda =0}^{\infty }{\mathbf{C}}_n^{2\lambda +2}\left(m\right){\eta }^{2\lambda }={cot}^{2}x-n{cos}^{2}x-n{cos}^{2}xcotxcotnx$$

(26)

for the trigonometric sums

$${\mathbf{C}}_n^{\lambda }\left(m\right):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{cot}^{\lambda }\left({\displaystyle \frac{m-k}{n}}\pi \right).$$

Consequently, we have the following explicit formula in general.

Theorem 5.

For the three natural numbers λ, m and n with $1\le m\le n$, the trigonometric sum ${\mathbf{C}}_n^{2\lambda }\left(m\right)$ is independent of m and admits the closed formula

$${\mathbf{C}}_n^{2\lambda }\left(m\right)=(n-1){(-1)}^{\lambda }-\sum _{k=1}^{\lambda }{(-1)}^{\lambda -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{k}}\right){\mathrm{U}}_k\left(n\right),$$

(27)

where ${\mathrm{U}}_k\left(n\right)$ are the coefficients of the power series (11) and determined by the recurrence relation (12).

What is remarkable is that these sums are also independent of m. The first five closed formulae are explicitly given as follows.

Corollary 5

($m,n\in \mathbb{N}$ with $1\le m\le n$).

$$\begin{array}{cc}\hfill {\mathbf{C}}_n^2\left(m\right)& ={\displaystyle \frac{(n-1)(n-2)}{3}},\hfill \\ \hfill {\mathbf{C}}_n^4\left(m\right)& ={\displaystyle \frac{(n-1)(n-2)}{45}}\left\{n^2+3n-13\right\},\hfill \\ \hfill {\mathbf{C}}_n^6\left(m\right)& ={\displaystyle \frac{(n-1)(n-2)}{945}}\left\{2n^4+6n^3-28n^2-96n+251\right\},\hfill \\ \hfill {\mathbf{C}}_n^8\left(m\right)& ={\displaystyle \frac{(n-1)(n-2)}{14175}}\left\{3n^6+9n^5-59n^4-195n^3+457n^2+1761n-3551\right\},\hfill \\ \hfill {\mathbf{C}}_n^{10}\left(m\right)& ={\displaystyle \frac{(n-1)(n-2)}{93555}}\left\{\begin{array}{c}2n^8+6n^7-52n^6-168n^5+546n^4\\ +1974n^3-3068n^2-13152n+22417\end{array}\right\}.\hfill \end{array}$$

Proof of Theorem 5.

Recalling the equality (11)

$${\displaystyle \frac{ncotnx}{cotx}}=\sum _{k=0}^{\infty }{\mathrm{U}}_k\left(n\right){sin}^{2k}x,$$

and

$${sin}^{2k}x={\left\{{\displaystyle \frac{{tan}^{2}x}{1+{tan}^{2}x}}\right\}}^k=\sum _{j=k}^{\infty }{(-1)}^{j-k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{k-1}}\right){tan}^{2j}x,$$

we have the power series expansion

$${\displaystyle \frac{ncotnx}{cotx}}=1+\sum _{j=1}^{\infty }{tan}^{2j}x\sum _{k=1}^j{(-1)}^{j-k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{k-1}}\right){\mathrm{U}}_k\left(n\right).$$

Then we can extract the coefficient

$$\begin{array}{cc}\hfill {\mathbf{C}}_n^{2\lambda }\left(m\right)& =\left[{\eta }^{2\lambda -2}\right]\left\{{\displaystyle \frac{1-n}{{sec}^{2}x}}+{\displaystyle \frac{1}{{tan}^{2}x{sec}^{2}x}}\left(1-{\displaystyle \frac{ncotnx}{cotx}}\right)\right\}\hfill \\ & =\left[{\eta }^{2\lambda -2}\right]\left\{{\displaystyle \frac{1-n}{1+{\eta }^2}}-{\displaystyle \frac{1}{1+{\eta }^2}}\sum _{j=1}^{\infty }{\eta }^{2j-2}\sum _{k=1}^j{(-1)}^{j-k}\left({\displaystyle \genfrac{}{}{0pt}{}{j-1}{k-1}}\right){\mathrm{U}}_k\left(n\right)\right\}\hfill \\ & =(n-1){(-1)}^{\lambda }-\sum _{j=0}^{\lambda -1}\sum _{k=1}^{\lambda -j}{(-1)}^{\lambda -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda -j-1}{k-1}}\right){\mathrm{U}}_k\left(n\right)\hfill \\ & =(n-1){(-1)}^{\lambda }-\sum _{k=1}^{\lambda }{(-1)}^{\lambda -k}{\mathrm{U}}_k\left(n\right)\sum _{j=0}^{\lambda -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda -j-1}{k-1}}\right).\hfill \end{array}$$

Evaluating the last binomial sum, we get the explicit formula (27), which is consistent with (10) in view of

$${\mathbf{A}}_n^{2\lambda }\left(m\right)=(n-1)+\sum _{k=1}^{\lambda }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{k}}\right){\mathbf{C}}_n^{2k}\left(m\right).\square$$

2.7. Sums About Even Powers of Tangent Function

Define the trigonometric sum by

$$D_n^{\lambda }\left(y\right):=\sum _{k=1}^n{tan}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

(28)

Replacing y by $y+\pi /2$ in (24) and Proposition 1, we derive the generating function

$$\begin{array}{cc}\hfill \eta =tanx:\sum _{\lambda =0}^{\infty }{D}_n^{2\lambda +2}\left(y\right){\eta }^{2\lambda }& =-n{cos}^{2}x\hfill \\ & +\left\{\begin{array}{cc}{\displaystyle \frac{n{cos}^{3}xsin2nx}{2sinx({sin}^{2}ny-{sin}^{2}nx)}},\hfill & n{\equiv }_{2}0;\hfill \\ {\displaystyle \frac{n{cos}^{3}xsin2nx}{2sinx({cos}^{2}ny-{sin}^{2}nx)}},\hfill & n{\equiv }_{2}1;\hfill \end{array}\right.\hfill \end{array}$$

(29)

and the polynomial representation formula.

Proposition 2

(Explicit formula: $\lambda \in {\mathbb{N}}_0$ and $n\in \mathbb{N}$).

$$\begin{array}{cc}\hfill D_n^{2\lambda +2}\left(y\right)=n{(-1)}^{\lambda +1}+n& \sum _{k=0}^{\lambda }\sum _{j=-k}^k{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\times \left\{\begin{array}{cc}{csc}^{2k+2}\left(ny\right),\hfill & n{\equiv }_{2}0\hfill \\ {sec}^{2k+2}\left(ny\right),\hfill & n{\equiv }_{2}1\hfill \end{array}\right.\hfill \\ \hfill \times & \sum _{\ell =k}^{\lambda }{4}^{\ell -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +1}{\ell +1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +n+jn}{2\ell +1}}\right).\hfill \end{array}$$

2.8. Sums About Even Powers of Tangent with mth Term Missing

Denote further by ${\mathcal{D}}_n^{\lambda }(y;m)$ the sum $D_n^{\lambda }\left(y\right)$ with the mth term being deleted

$${\mathcal{D}}_n^{\lambda }(y;m):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{tan}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

Replacing y by $y+\pi /2$ in (25), we obtain the generating function

$$\begin{array}{cc}\hfill \eta =tanx:\sum _{\lambda =0}^{\infty }{\mathcal{D}}_n^{2\lambda +2}(y;m){\eta }^{2\lambda }& ={\displaystyle \frac{1}{{tan}^{2}x-{cot}^2(y+\frac{m\pi }{n})}}-n{cos}^{2}x\hfill \\ & +\left\{\begin{array}{cc}{\displaystyle \frac{n{cos}^{3}xsin2nx}{2sinx({sin}^{2}ny-{sin}^{2}nx)}},\hfill & n{\equiv }_{2}0;\hfill \\ {\displaystyle \frac{n{cos}^{3}xsin2nx}{2sinx({cos}^{2}ny-{sin}^{2}nx)}},\hfill & n{\equiv }_{2}1.\hfill \end{array}\right.\hfill \end{array}$$

(30)

The limiting case $y\to -{\displaystyle \frac{m\pi }{n}}$ gives the generating function

$$\begin{array}{cc}\eta =tanx:\sum _{\lambda =0}^{\infty }{\eta }^{2\lambda }{\mathbf{D}}_n^{2\lambda +2}\left(m\right)& =-n{cos}^{2}x\hfill \\ & +\left\{\begin{array}{cc}-n{cos}^{2}xcotxcotnx,\hfill & n{\equiv }_{2}0;\hfill \\ n{cos}^{2}xcotxtannx,\hfill & n{\equiv }_{2}1;\hfill \end{array}\right.\hfill \end{array}$$

(31)

for the trigonometric sums defined by

$${\mathbf{D}}_n^{\lambda }\left(m\right):=\sum _{\begin{array}{c}k=1\\ k\ne m\end{array}}^n{tan}^{\lambda }\left({\displaystyle \frac{m-k}{n}}\pi \right).$$

(32)

These finite sums admit the explicit formula as in the following theorem.

Theorem 6.

For the three natural numbers λ, m and n with n being odd and $1\le m\le n$, the trigonometric sum ${\mathbf{D}}_n^{2\lambda }\left(m\right)$ is independent of m and admits the closed formula

$${\mathbf{D}}_n^{2\lambda }\left(m\right)=n{(-1)}^{\lambda }-n^2\sum _{\ell =0}^{\lambda -1}{(-1)}^{\lambda -\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda -1}{\ell }}\right){\mathrm{V}}_{\ell }\left(n\right),$$

(33)

where ${\mathrm{V}}_k\left(n\right)$ are the coefficients of the power series (21) and determined by the recurrence relation (22).

In particular, the following specific formulae can be deduced as consequences.

Corollary 6

($m,n\in \mathbb{N}$ with n being odd and $1\le m\le n$).

$$\begin{array}{cc}\hfill {\mathbf{D}}_n^2\left(m\right)& =n(n-1),\hfill \\ \hfill {\mathbf{D}}_n^4\left(m\right)& ={\displaystyle \frac{n(n-1)}{3}}\left\{n^2+n-3\right\},\hfill \\ \hfill {\mathbf{D}}_n^6\left(m\right)& ={\displaystyle \frac{n(n-1)}{15}}\left\{2n^4+2n^3-8n^2-8n+15\right\},\hfill \\ \hfill {\mathbf{D}}_n^8\left(m\right)& ={\displaystyle \frac{n(n-1)}{315}}\left\{\begin{array}{c}17n^6+17n^5-95n^4-95n^3\\ +213n^2+213n-315\end{array}\right\},\hfill \\ \hfill {\mathbf{D}}_n^{10}\left(m\right)& ={\displaystyle \frac{n(n-1)}{2835}}\left\{\begin{array}{c}62n^8+62n^7-448n^6-448n^5+1358n^4\\ +1358n^3-2232n^2-2232n+2835\end{array}\right\}.\hfill \end{array}$$

Proof of Theorem 6.

Instead of generating function approach, we can derive the general explicit formula from that for ${\mathbf{B}}_n^{2\lambda }\left(m\right)$. In fact, it is not hard to verify

$$\begin{array}{cc}\hfill {\mathbf{D}}_n^{2\lambda }\left(m\right)& =(n-1){(-1)}^{\lambda }+\sum _{k=1}^{\lambda }{(-1)}^{\lambda -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{k}}\right){\mathbf{B}}_n^{2k}\left(m\right)\hfill \\ & =(n-1){(-1)}^{\lambda }+\sum _{k=1}^{\lambda }{(-1)}^{\lambda -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{k}}\right)\left\{-1+n^2\sum _{\ell =0}^{k-1}{\mathrm{V}}_{\ell }\left(n\right)\right\}\hfill \\ & =n{(-1)}^{\lambda }+n^2\sum _{\ell =0}^{\lambda -1}{\mathrm{V}}_{\ell }\left(n\right)\sum _{k=\ell +1}^{\lambda }{(-1)}^{\lambda -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{k}}\right).\hfill \end{array}$$

Evaluating the binomial sum

$$\begin{array}{cc}\hfill \sum _{k=\ell +1}^{\lambda }{(-1)}^{\lambda -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{k}}\right)& ={(-1)}^{\lambda +1}\sum _{k=0}^{\ell }{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{k}}\right)\hfill \\ & ={(-1)}^{\lambda +1}\left[x^{\ell }\right]{(1-x)}^{\lambda -1}\hfill \\ & ={(-1)}^{\lambda -\ell +1}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda -1}{\ell }}\right),\hfill \end{array}$$

we find the explicit formula stated in Theorem 6. □

3. Trigonometric Sums of Odd Powers

Letting $P\left(\xi \right)\equiv 1$ in (3), we get another identity

$${\displaystyle \frac{nsin2ny}{{sin}^{2}ny-{sin}^{2}nx}}=\sum _{k=1}^n{\displaystyle \frac{sin2(y+\frac{k\pi }{n})}{{sin}^2(y+\frac{k\pi }{n})-{sin}^{2}x}}.$$

(34)

This will be employed in this section to evaluate further four classes of trigonometric sums. Unlike the sums treated in the last section, we shall not examine the corresponding limiting sums (as $y\to -{\displaystyle \frac{m}{n}}\pi$) with the mth term being deleted because it is routine to verify that they are identical to zero.

3.1. Sums About Odd Powers of Cosecant Function

For the trigonometric sums defined by

$$E_n^{\lambda }\left(y\right):=\sum _{k=1}^{n}cos\left(y+{\displaystyle \frac{k\pi }{n}}\right){csc}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right),$$

(35)

we have the polynomial representation as in the theorem below.

Theorem 7

(Explicit formula: $\lambda \in {\mathbb{N}}_0$).

$$\begin{array}{cc}\hfill E_n^{2\lambda +1}\left(y\right)& =ncot\left(ny\right)\chi (\lambda =0)\hfill \\ & +2ncot\left(ny\right)\sum _{k=1}^{\lambda }{4}^{\lambda -k}{csc}^{2k}\left(ny\right)\sum _{j=1}^k{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +nj}{2\lambda }}\right){\displaystyle \frac{nj}{\lambda +nj}}.\hfill \end{array}$$

For small $\lambda$ values, we record the following examples, where the first one can be found in Liu [14] (Equation 1.18), who derived it from a theta function identity.

Corollary 7

($n\in \mathbb{N}$).

$$\begin{array}{cc}\hfill {\displaystyle \frac{E_n^1\left(y\right)}{cot\left(ny\right)}}& =n,\hfill \\ \hfill {\displaystyle \frac{E_n^3\left(y\right)}{cot\left(ny\right)}}& =n^3{csc}^2\left(ny\right),\hfill \\ \hfill {\displaystyle \frac{E_n^5\left(y\right)}{cot\left(ny\right)}}& =n^5{csc}^4\left(ny\right)+{\displaystyle \frac{n^3}{3}}(1-n^2){csc}^2\left(ny\right),\hfill \\ \hfill {\displaystyle \frac{E_n^7\left(y\right)}{cot\left(ny\right)}}& =n^7{csc}^6\left(ny\right)+{\displaystyle \frac{2n^5}{3}}(1-n^2){csc}^4\left(ny\right)\hfill \\ & +{\displaystyle \frac{2n^3}{45}}(1-n^2)(4-n^2){csc}^{2}ny,\hfill \\ \hfill {\displaystyle \frac{E_n^9\left(y\right)}{cot\left(ny\right)}}& =n^7(1-n^2){csc}^6\left(ny\right)+{\displaystyle \frac{n^5}{15}}(1-n^2)(7-3n^2){csc}^4\left(ny\right)\hfill \\ & +n^9{csc}^{8}ny+{\displaystyle \frac{n^3}{315}}(1-n^2)(4-n^2)(9-n^2){csc}^2\left(ny\right).\hfill \end{array}$$

Proof of Theorem 7.

Rewrite (34) and then expand it into a formal power series

$$\begin{array}{cc}\hfill {\displaystyle \frac{nsin2ny}{2({sin}^{2}ny-{sin}^{2}nx)}}& =\sum _{k=1}^n{\displaystyle \frac{cot(y+\frac{k\pi }{n})}{1-{sin}^{2}x{csc}^2(y+\frac{k\pi }{n})}}\hfill \\ & =\sum _{k=1}^n\sum _{\lambda =0}^{\infty }{sin}^{2\lambda }xcot(y+{\displaystyle \frac{k\pi }{n}}){csc}^{2\lambda }(y+{\displaystyle \frac{k\pi }{n}}).\hfill \end{array}$$

Keeping in mind (35), we have derived the following generating function

$$\eta =sinx:\sum _{\lambda =0}^{\infty }{E}_n^{2\lambda +1}\left(y\right){\eta }^{2\lambda }={\displaystyle \frac{nsinnycosny}{{sin}^{2}ny-{sin}^{2}nx}}.$$

(36)

To find an explicit formula, rewrite first the generating function (36) as

$${\displaystyle \frac{nsinnycosny}{{sin}^{2}ny-{sin}^{2}nx}}=ncot\left(ny\right)\sum _{k=0}^{\infty }{csc}^{2k}\left(ny\right){sin}^{2k}\left(nx\right).$$

Then examine the trigonometric powers

$$\begin{array}{cc}\hfill {sin}^{2k}\left(nx\right)& ={\displaystyle \frac{{(e^{nx\mathbf{i}}-e^{-nx\mathbf{i}})}^{2k}}{{(-4)}^k}}=\sum _{j=-k}^k{\displaystyle \frac{{(-1)}^j}{4^k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)e^{2njx\mathbf{i}}\hfill \\ & ={\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}{4^k}}+2\sum _{j=1}^k{\displaystyle \frac{{(-1)}^j}{4^k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)cos\left(2njx\right).\hfill \end{array}$$

By making use of (14), we can extract the coefficient

$$\left[{\eta }^{2\lambda }\right]cos\left(2njx\right)=\left[{sin}^{2\lambda }x\right]cos\left(2njx\right)={\displaystyle \frac{nj}{\lambda +nj}}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +nj}{2\lambda }}\right){(-4)}^{\lambda }.$$

Consequently, we have

$$\begin{array}{cc}\hfill E_n^{2\lambda +1}\left(y\right)& =\left[{\eta }^{2\lambda }\right]{\displaystyle \frac{nsinnycosny}{{sin}^{2}ny-{sin}^{2}nx}}=ncot\left(ny\right)\sum _{k=0}^{\lambda }{csc}^{2k}\left(ny\right)\left[{\eta }^{2\lambda }\right]{sin}^{2k}\left(nx\right)\hfill \\ & =ncot\left(ny\right)\chi (\lambda =0)\hfill \\ & +2ncot\left(ny\right)\sum _{k=1}^{\lambda }{\displaystyle \frac{{csc}^{2k}\left(ny\right)}{4^k}}\sum _{j=1}^k{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left[{\eta }^{2\lambda }\right]cos\left(2njx\right).\hfill \end{array}$$

Summing up, we have confirmed the polynomial expression in Theorem 7. □

3.2. Sums About Odd Powers of Secant Function

Define the trigonometric sum by

$$F_n^{\lambda }\left(y\right):=\sum _{\begin{array}{c}k=1\end{array}}^{n}sin\left(y+{\displaystyle \frac{k\pi }{n}}\right){sec}^{\lambda }\left(y+{\displaystyle \frac{k\pi }{n}}\right).$$

(37)

Replacing y by $y+\pi /2$ in (36), we deduce the generating function

$$\eta =sinx:\sum _{\lambda =0}^{\infty }{F}_n^{2\lambda +1}\left(y\right){\eta }^{2\lambda }={(-1)}^n\left\{\begin{array}{cc}{\displaystyle \frac{nsinnycosny}{{sin}^{2}nx-{sin}^{2}ny}},\hfill & n{\equiv }_{2}0;\hfill \\ {\displaystyle \frac{nsinnycosny}{{sin}^{2}nx-{cos}^{2}ny}},\hfill & n{\equiv }_{2}1.\hfill \end{array}\right.$$

(38)

When n is even, the expression of $F_n^{2\lambda +1}\left(y\right)$ is identical to that of $E_n^{2\lambda +1}\left(y\right)$ except for a minus sign. For an odd n instead, $F_n^{2\lambda +1}\left(y\right)$ has the same expression as $E_n^{2\lambda +1}\left(y\right)$, but under the replacements $cotny\to tanny$ and $cscny\to secny$.

Proposition 3

(Explicit formulae: $\lambda \in {\mathbb{N}}_0$ and $n\in \mathbb{N}$).

$$F_n^{2\lambda +1}\left(y\right)=\left\{\begin{array}{l}-ncot\left(ny\right)\chi (\lambda =0)-2ncot\left(ny\right)\\ \times {\displaystyle \sum _{k=1}^{\lambda }}{4}^{\lambda -k}{csc}^{2k}\left(ny\right){\displaystyle \sum _{j=1}^k}{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +nj}{2\lambda }}\right){\displaystyle \frac{nj}{\lambda +nj}},n{\equiv }_{2}0;\\ ntan\left(ny\right)\chi (\lambda =0)+2ntan\left(ny\right)\\ \times {\displaystyle \sum _{k=1}^{\lambda }}{4}^{\lambda -k}{sec}^{2k}\left(ny\right){\displaystyle \sum _{j=1}^k}{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda +nj}{2\lambda }}\right){\displaystyle \frac{nj}{\lambda +nj}},n{\equiv }_{2}1.\end{array}\right.$$

3.3. Sums About Odd Powers of Cotangent Function

For the cotangent sum $C_n^{\ell }\left(y\right)$ defined in (23), the result corresponding to odd powers is given by the following polynomial expression.

Theorem 8

(Explicit formula: $\lambda \in {\mathbb{N}}_0$ and $n\in \mathbb{N}$).

$$\begin{array}{cc}{C}_n^{2\lambda +1}\left(y\right)& =ncot\left(ny\right){(-1)}^{\lambda }+ncot\left(ny\right)\sum _{k=1}^{\lambda }{csc}^{2k}\left(ny\right)\hfill \\ & \times \sum _{j=1}^k{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right)\sum _{\ell =k}^{\lambda }{4}^{\ell -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +nj}{2\ell }}\right){\displaystyle \frac{2nj}{\ell +nj}}.\hfill \end{array}$$

The first five formulae are produced as examples.

Corollary 8

( $n\in \mathbb{N}$).

$$\begin{array}{cc}\hfill {\displaystyle \frac{C_n^1\left(y\right)}{cot\left(ny\right)}}& =n,\hfill \\ \hfill {\displaystyle \frac{C_n^3\left(y\right)}{cot\left(ny\right)}}& =n^3{csc}^2\left(ny\right)-n,\hfill \\ \hfill {\displaystyle \frac{C_n^5\left(y\right)}{cot\left(ny\right)}}& =n^5{csc}^4\left(ny\right)-{\displaystyle \frac{n^3}{3}}(n^2+5){csc}^2\left(ny\right)+n,\hfill \\ \hfill {\displaystyle \frac{C_n^7\left(y\right)}{cot\left(ny\right)}}& =n^7{csc}^6\left(ny\right)-{\displaystyle \frac{n^5}{3}}(2n^2+7){csc}^4\left(ny\right)\hfill \\ & +{\displaystyle \frac{n^3}{45}}(n^2+14)(2n^2+7){csc}^2\left(ny\right)-n,\hfill \\ \hfill {\displaystyle \frac{C_n^9\left(y\right)}{cot\left(ny\right)}}& =n^9{csc}^8\left(ny\right)-n^7(n^2+3){csc}^6\left(ny\right)+{\displaystyle \frac{n^5}{5}}(n^4+10n^2+19){csc}^4\left(ny\right)\hfill \\ & -{\displaystyle \frac{n^3}{315}}(n^6+42n^4+399n^2+818){csc}^2\left(ny\right)+n.\hfill \end{array}$$

Proof of Theorem 8.

According to the binomial relation

$${csc}^{2\lambda }y={(1+{cot}^{2}y)}^{\lambda }=\sum _{\ell =0}^{\lambda }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right){cot}^{2\ell }y,$$

we have the equality

$$E_n^{2\lambda }\left(y\right)=\sum _{\ell =0}^{\lambda }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right)C_n^{2\ell +1}\left(y\right),$$

where $C_n^{\ell }\left(y\right)$ is defined by (23). Then we can manipulate (36) as follows:

$$\begin{array}{cc}\hfill {\displaystyle \frac{nsinnycosny}{{sin}^{2}ny-{sin}^{2}nx}}& =\sum _{\lambda =0}^{\infty }{\eta }^{2\lambda }{E}_n^{2\lambda }\left(y\right)\hfill \\ & =\sum _{\lambda =0}^{\infty }{\eta }^{2\lambda }\sum _{\ell =0}^{\lambda }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right)C_n^{2\ell +1}\left(y\right)\hfill \\ & =\sum _{\ell =0}^{\infty }{C}_n^{2\ell +1}\left(y\right)\sum _{\lambda =\ell }^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right){\eta }^{2\lambda }\hfill \\ & =\sum _{\ell =0}^{\infty }{C}_n^{2\ell +1}\left(y\right){\displaystyle \frac{{\eta }^{2\ell }}{{(1-{\eta }^2)}^{\ell +1}}}.\hfill \end{array}$$

Keeping in mind that

$$\eta =sinx\mathrm{and}{\displaystyle \frac{{\eta }^{2\ell }}{{(1-{\eta }^2)}^{\ell +1}}}={\displaystyle \frac{{tan}^{2\ell +1}x}{sinxcosx}},$$

we have therefore established the generating function

$$\eta =tanx:\sum _{\ell =0}^{\infty }{C}_n^{2\ell +1}\left(y\right){\eta }^{2\ell +1}={\displaystyle \frac{nsin2xsinnycosny}{2({sin}^{2}ny-{sin}^{2}nx)}}.$$

(39)

By combining Theorem 7 with the binomial relation

$$C_n^{2\lambda +1}\left(y\right)=\sum _{\ell =0}^{\lambda }{(-1)}^{\lambda -\ell }\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right)E_n^{2\ell }\left(y\right),$$

we deduce the polynomial representation formula in Theorem 8. □

3.4. Sums About Odd Powers of Tangent Function

Recall the trigonometric sum $D_n^{\lambda }\left(y\right)$ by (28). Replacing y by $y+\pi /2$ in (39), we get the generating function

$$\eta =tanx:\sum _{\ell =0}^{\infty }{D}_n^{2\ell +1}\left(y\right){\eta }^{2\ell +1}=\left\{\begin{array}{cc}{\displaystyle \frac{nsin2xsinnycosny}{2({sin}^{2}nx-{sin}^{2}ny)}},\hfill & n{\equiv }_{2}0;\hfill \\ {\displaystyle \frac{nsin2xsinnycosny}{2({cos}^{2}ny-{sin}^{2}nx)}},\hfill & n{\equiv }_{2}1.\hfill \end{array}\right.$$

(40)

When n is even, the expression of $D_n^{2\lambda +1}\left(y\right)$ is identical to that of $C_n^{2\lambda +1}\left(y\right)$ except for a minus sign. Instead, $D_n^{2\lambda +1}\left(y\right)$ has, for an odd n, the same expression as $C_n^{2\lambda +1}\left(y\right)$, but under the replacements $cotny\to tanny$ and $cscny\to secny$. According to Theorem 8, we have the following triple sum expressions.

Proposition 4

(Explicit formulae: $\lambda \in {\mathbb{N}}_0$ and $n\in \mathbb{N}$).

$$D_n^{2\lambda +1}\left(y\right)=\left\{\begin{array}{l}-ncot\left(ny\right){(-1)}^{\lambda }-ncot\left(ny\right){\displaystyle \sum _{k=1}^{\lambda }}{csc}^{2k}\left(ny\right)\\ \times {\displaystyle \sum _{j=1}^k}{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right){\displaystyle \sum _{\ell =k}^{\lambda }}{4}^{\ell -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +nj}{2\ell }}\right){\displaystyle \frac{2nj}{\ell +nj}},n{\equiv }_{2}0;\\ ntan\left(ny\right){(-1)}^{\lambda }+ntan\left(ny\right){\displaystyle \sum _{k=1}^{\lambda }}{sec}^{2k}\left(ny\right)\\ \times {\displaystyle \sum _{j=1}^k}{(-1)}^{\lambda +j}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k+j}}\right){\displaystyle \sum _{\ell =k}^{\lambda }}{4}^{\ell -k}\left({\displaystyle \genfrac{}{}{0pt}{}{\lambda }{\ell }}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\ell +nj}{2\ell }}\right){\displaystyle \frac{2nj}{\ell +nj}},n{\equiv }_{2}1.\end{array}\right.$$

4. Concluding Comments and Further Prospects

By means of the Lagrange interpolation formula, a trigonometric identity (3) involving two free variables “x and y” is established, which serves as a framework for computing eight large classes of trigonometric sums in closed form. When the mth term is removed from the trigonometric sums in four classes (among eight), the resulting expressions are surprisingly not only of closed form, but also independent of “m” (see Section 2.2, Section 2.4, Section 2.6 and Section 2.8).

There exist different trigonometric sums (see [12,16,38,39,40,41,42,43] for example). A natural question is that what about the remaining sums if one specific term is missing from the original sums? For instance, the trigonometric sum discovered by Eisenstein [44] (cf. [6]) is well-known

$$\sum _{k=1}^{n-1}sin\left({\displaystyle \frac{2k\lambda \pi }{n}}\right)cot\left({\displaystyle \frac{k\pi }{n}}\right)=n-2\lambda ,\mathrm{where}\lambda <n.$$

Determine the exact values of the following corresponding sums

$$\sum _{k=1}^{n-1}sin\left({\displaystyle \frac{2(m-k)\lambda \pi }{n}}\right)cot\left({\displaystyle \frac{(m-k)\pi }{n}}\right)\mathrm{for}1\le m\le n.$$

This and similar problems are certainly worth making further exploration.