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\begin{center}
\vskip 1cm{\LARGE\bf 
Some New Identities for Arctangents\\
\vskip .1in
and Chebyshev Polynomials
}
\vskip 1cm
\large
Jiejie Gao\\
School of Mathematics\\  
Northwest University\\
Xi'an, Shaanxi\\ 
China\\
\href{mailto:gaojiejie@stumail.nwu.edu.cn}{\tt gaojiejie@stumail.nwu.edu.cn}  \\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper, we use some properties of Chebyshev polynomials and trigonometric functions to study four classes of sums of products of two arctangents involving these polynomials.  We also give some infinite series identities concerning Fibonacci and Lucas numbers.	
\end{abstract}

\section{Introduction}\label{sec1}
For all integers $n\geq1$ and all real $x$, the Chebyshev polynomials of
the first and second kind, $T_n(x)$ and $U_n(x)$, are defined by the second-order linear recurrences
$$
T_{n+1}(x)=2xT_n(x)-T_{n-1}(x) \ \ \textrm{and} \ \
U_{n+1}(x)=2xU_n(x)-U_{n-1}(x),
$$
where the first two terms are 
$T_0(x)=1$, $T_1(x)=x$, $U_0(x)=1$ and $U_1(x)=2x$.

Explicit formulae for $T_n(x)$ and $U_n(x)$ can be expressed as follows:
$$
T_n(x) =\bigl(\alpha^n+\beta^n\bigr)/2 \ \ \textrm{and} \ \
U_n(x)=\bigl(\alpha^{n+1}-\beta^{n+1}\bigr)/2\sqrt{x^2-1},
$$
where $\alpha=\alpha(x)=x+\sqrt{x^2-1}$, $\beta=\beta(x)=x-\sqrt{x^2-1}$.

It is common knowledge that the Chebyshev polynomials $T_n(x)$ and $U_n(x)$ play an important role in approximation theory, and so many scholars have studied their properties and obtained many interesting conclusions. In particular, Kim and his team\cite{KimDolgyKim2018,KimDolgyKimSeo2016,KimKimJangDolgy2019,KimKimDolgyKwon2018,KimKimJangDolgy2018,KimKimJangJang2018} have done a lot of important research work. You can also find many papers on Chebyshev polynomials in the literature\cite{MaLv2017,ZhangHan2020,Zhang2002,ZhangChen2018,Adegoke2016,MahonHoradam1985,FrontczakGoy2021,HarneBadshahVerma2014}. For example, Ma and Lv \cite{MaLv2017} computed the reciprocal sums of Chebyshev polynomials. For $k=1$, $2$ and $3$, they considered the summations
$$
\sum_{a=1}^{q-1}{T_a^{-2k}\bigl(\cos\pi h/q\bigr)} \ \ \textrm{and} \ \
\sum_{a=1}^{q-1}{U_{a-1}^{-2k}\bigl(\cos\pi h/q\bigr)},
$$
where $q$ is an odd number and $h$ is an integer co-prime to $q$.

Zhang \cite{Zhang2002} studied convolution sums 
involving $T_n(x)$, and proved the following identities:
\begin{align*}
\sum_{a_1+a_2+\cdots+a_{k+1}=n+k+1}\prod_{i=1}^{k+1}T_{a_{i}}(x) =(1/2^{k} \cdot k!) \cdot \sum_{h=0}^{k+1}(-x)^h\cdot \binom{k+1}{h} \cdot U_{n+2k+1-h}^{(k)}(x),
\end{align*}
where $U_n^{(k)}(x)$ denotes the $k$-th derivative of $U_n(x)$ with respect to $x$,  and the summation is over all $\bigl(k+1\bigr)$-dimension non-negative integer coordinates $\bigl(a_1, a_2,\ldots, a_{k+1}\bigr)$ such that $a_1+a_2+\cdots+a_{k+1}=n+k+1$.

 Zhang and Chen \cite{ZhangChen2018} proved the following result:
\begin{align*}
&\sum_{a_1+a_2+\cdots+a_{h+1}=n}U_{a_{1}}(x)U_{a_{2}}(x)\cdots U_{a_{h+1}}(x)\\
&=(1/2^{h}\cdot h!) \cdot \sum_{j=1}^{h}C\bigl(h,j\bigr)/x^{2h-j}\sum_{i=0}^{n}\bigl(n-i+j\bigr)!/\bigl(n-i\bigr)! \cdot \binom{2h+i-j-1}{i} \cdot U_{n-i+j}(x)/x^{i},
\end{align*}
where $C(h,i)$ is the second order non-linear recurrence sequence defined by 
$C(h,0)=0$, $C(h,h)=1$, $C(h+1,1)=1\cdot3\cdot5\cdots(2h-1)=(2h-1)!$ and
\begin{align*}
C(h+1,i+1)=(2h-1-i)\cdot C(h,i+1)+C(h,i)
\end{align*}for all $1\leq i\leq h-1$.

Adegoke \cite{Adegoke2016} studied sums of arctangents involving Fibonacci numbers and Lucas numbers and obtained the following result:
\begin{align*}
\sum_{r=p}^{\infty}\arctan\lambda F_{2j}L_{4jr+2k}/\bigl(F_{4jr+2k}^2-F_{2j}^2+\lambda^2\bigr)&=\arctan\lambda/F_{4jp+2k-2j};\\
\sum_{r=p}^{\infty}\arctan\lambda F_{2j}F_{4jr+2k-1}/\bigl(L_{8jr+4k-2}-L_{4j}+\lambda^2\bigr)&=\arctan\lambda/L_{4jp+2k-2j-1},
\end{align*}
where $\lambda \in \mathbb{R}$, $j$, $k$, $p$ $\in \mathbb{Z}$ and $j\neq0$.

Mahon and Horadam \cite{MahonHoradam1985} studied sums of arctangents
involving Pell and Pell-Lucas polynomials and  obtained the following result:
\begin{align*}
\sum_{r=1}^{n}\arctan2x/P_{2r-1}(x)&=\pi/2-\arctan1/P_{2n}(x);\\
\sum_{r=1}^{n}\arctan(-1)^{r-1}/P_{2r}(x)&=\arctan P_n(x)/P_{n+1}(x),
\end{align*}
where $x \in \mathbb{R}$, $n$ $\in \mathbb{Z}$ and $n\geq1$.

We observe that Chebyshev polynomials have similar properties to Pell and Pell-Lucas polynomials, so it is natural to wonder whether Chebyshev polynomials have similar identities. Inspired by \cite{Adegoke2016} and \cite{MahonHoradam1985}, in this paper, we use the properties of Chebyshev polynomials and trigonometric functions to study the sums of products of two arctangents involving Chebyshev polynomials, and give some infinite series identities for Fibonacci and Lucas numbers. We prove the following results:
\begin{theorem}\label{1} Let $n$, $k$ be positive integers. Then for all real $x>1$, we have the identities
	\begin{align*}
	&\sum_{k=1}^{n}\arctan2xT_{2k}(x)/\bigl(T^{2}_{2k}(x)+x^2-2\bigr)\arctan2\bigl(x^2-1\bigr)U_{2k-1}(x)/\bigl(T^{2}_{2k}(x)+x^2\bigr)\\
	& \quad\quad =\arctan^{2}1/x-\arctan^{2}1/T_{2n+1}(x);\\
	&\sum_{k=1}^{n}\arctan2xT_{2k-1}(x)/\bigl(T^{2}_{2k-1}(x)+x^2-2\bigr)\arctan2\bigl(x^2-1\bigr)U_{2k-2}(x)/\bigl(T^{2}_{2k-1}(x)+x^2\bigr)\\
	&\quad\quad =\pi^{2}/16-\arctan^{2}1/T_{2n}(x).
	\end{align*}
\end{theorem}

\begin{theorem}\label{2} Let $n$, $k$ be positive integers. Then for all real $x>1$, we have the identities
	\begin{align*}
	 &\sum_{k=1}^{n}\arctan2xU_{2k}(x)/\bigl(U^{2}_{2k}(x)-2\bigr)\arctan2T_{2k+1}(x)/U^{2}_{2k}(x)\qquad\qquad\qquad \qquad\qquad\\
	& \quad\quad =\arctan^{2}1/2x-\arctan^{2}1/U_{2n+1}(x);\qquad \qquad \qquad \qquad\qquad\\
	&\sum_{k=1}^{n}\arctan2xU_{2k-1}(x)/\bigl(U^{2}_{2k-1}(x)-2\bigr)\arctan2T_{2k}(x)/U^{2}_{2k-1}(x)\qquad \qquad\qquad \qquad \qquad\\
	& \quad\quad =\pi^{2}/16-\arctan^{2}1/U_{2n}(x).\qquad \qquad\qquad \qquad\qquad
	\end{align*}
\end{theorem}

\begin{theorem}\label{3} Let $n$, $k$ be positive integers. Then for all real $x>1$, we have the identities
	\begin{align*}
	&\sum_{k=1}^{n}\arctan\bigl(1-x^2\bigr)/2xT^{2}_k(x)\arctan\bigl(2T^{2}_k(x)+x^2-1\bigr)/\bigl(x^2-1\bigr)U_{2k-1}(x)\\
	& \quad\quad =\arctan^{2}T_n(x)/T_{n+1}(x)-\arctan^{2}1/x;\\
	&\sum_{k=1}^{n}\arctan2x^{2}\bigl(1-x^{2}\bigr)/\bigl(2x^{2}-1\bigr)T^{2}_k(x)\arctan\bigl(T^{2}_k(x)+2x^2\bigl(x^{2}-1\bigr)\bigr)/x\bigl(x^{2}-1\bigr)U_{2k-1}(x)\\
	& \quad\quad =\arctan^{2}T_{n-1}(x)/T_{n+1}(x)+\arctan^{2}T_n(x)/T_{n+2}(x)-\pi^{2}/16-\arctan^{2}1/\bigl(2x^{2}-1\bigr).
	\end{align*}
\end{theorem}

\begin{theorem}\label{4} Let $n$, $k$ be positive integers. Then for all real $x>1$, we have the identities
	
	\begin{align*}
	&\sum_{k=1}^{n}\arctan1/2xU^{2}_k(x)\arctan\bigl(2U^{2}_k(x)-1\bigr)/U_{2k+1}(x)\\
	& \quad\quad =\arctan^{2}U_n(x)/U_{n+1}(x)-\arctan^{2}1/2x;\\
	&\sum_{k=1}^{n}\arctan2x^{2}/\bigl(2x^{2}-1\bigr)U^{2}_k(x)\arctan\bigl(U^{2}_k(x)-2x^{2}\bigr)/xU_{2k+1}(x)\\
	& \quad\quad =\arctan^{2}U_n(x)/U_{n+2}(x)+\arctan^{2}U_{n-1}(x)/U_{n+1}(x)-\arctan^{2}1/\bigl(4x^{2}-1\bigr).
	\end{align*}
\end{theorem}

In the above theorems, we only consider the case $x>1$. If $x<-1$, we have $-x>1$,
$
\alpha(x)=-(-x)+\sqrt{(-x)^2-1}=-\beta(-x),$ $
\beta(x)=-(-x)-\sqrt{(-x)^2-1}=-\alpha(-x).$
From the definition of $T_n(x)$ and $U_n(x)$, we obtain
$T_n(x)=(-1)^{n}\cdot T_n(-x),$ $
U_n(x)=(-1)^{n}\cdot U_n(-x).$
By observing the above theorems, we find that the results are
the same in both cases, so we do not discuss them here.
Taking $n\rightarrow\infty$, from our theorems we can deduce the following:
\begin{corollary}\label{5}
	Let $k$ be an integer. Then for all real $x>1$, we have
	\begin{align*}
	&\sum_{k=1}^{\infty}\arctan2xT_{2k}(x)/\bigl(T^{2}_{2k}(x)+x^2-2\bigr)\arctan2\bigl(x^2-1\bigr)U_{2k-1}(x)/\bigl(T^{2}_{2k}(x)+x^2\bigr)\\
	& \quad\quad =\arctan^{2}1/x;\\
	&\sum_{k=1}^{\infty}\arctan2xT_{2k-1}(x)/\bigl(T^{2}_{2k-1}(x)+x^2-2\bigr)\arctan2\bigl(x^2-1\bigr)U_{2k-2}(x)/\bigl(T^{2}_{2k-1}(x)+x^2\bigr)\\
	& \quad\quad =\pi^{2}/16.
	\end{align*}
\end{corollary}
\begin{corollary}\label{6}
	Let $k$ be an integer. Then for all real $x>1$, we have
	\begin{align*}
	\sum_{k=1}^{\infty}\arctan2xU_{2k}(x)/\bigl(U^{2}_{2k}(x)-2\bigr)\arctan2T_{2k+1}(x)/U^{2}_{2k}(x)&=\arctan^{2}1/2x;\\
	\sum_{k=1}^{\infty}\arctan2xU_{2k-1}(x)/\bigl(U^{2}_{2k-1}(x)-2\bigr)\arctan2T_{2k}(x)/U^{2}_{2k-1}(x)&=\pi^{2}/16.
	\end{align*}
\end{corollary}
\begin{corollary}\label{7}
	Let $k$ be an integer. Then for all real $x>1$, we have
	\begin{align*}
	&\sum_{k=1}^{\infty}\arctan\bigl(1-x^2\bigr)/2xT^{2}_k(x)\arctan\bigl(2T^{2}_k(x)+x^2-1\bigr)/\bigl(x^2-1\bigr)U_{2k-1}(x)\\
	& \quad\quad =\arctan^{2}\beta-\arctan^{2}1/x;\\
	&\sum_{k=1}^{\infty}\arctan2x^2\bigl(1-x^2\bigr)/\bigl(2x^2-1\bigr)T^{2}_k(x)\arctan\bigl(T^{2}_k(x)+2x^{2}\bigl(x^2-1\bigr)\bigr)/x\bigl(x^2-1\bigr)U_{2k-1}(x)\\
	& \quad\quad =2\arctan^{2}\beta^{2}-\pi^{2}/16-\arctan^{2}1/\bigl(2x^2-1\bigr),
	\end{align*}
where $\beta=\beta(x)=x-\sqrt{x^2-1}$.
\end{corollary}

\begin{corollary}\label{8}
	Let $k$ be an integer. Then for all real $x>1$, we have
	\begin{align*}
	&\sum_{k=1}^{\infty}\arctan1/2xU^{2}_k(x)\arctan\bigl(2U^{2}_k(x)-1\bigr)/U_{2k+1}(x)=\arctan^{2}\beta-\arctan^{2}1/2x;\\
	&\sum_{k=1}^{\infty}\arctan2x^{2}/\bigl(2x^2-1\bigr)U^{2}_k(x)\arctan\bigl(U^{2}_k(x)-2x^{2}\bigr)/xU_{2k+1}(x)\\
	& \quad\quad =2\arctan^{2}\beta^{2}-\arctan^{2}1/\bigl(4x^2-1\bigr),
	\end{align*}	
where $\beta=\beta(x)=x-\sqrt{x^2-1}$.
\end{corollary}

\begin{remark}
	Here we give several values of $T_n(x)$ and $U_n(x)$ as follows:
	\begin{displaymath}
	T_{n}\bigl(3/2\bigr)= L_{2n}/2;T_{n}\bigl(7/2\bigr)= L_{4n}/2;\qquad
	U_{n}\bigl(3/2\bigr)= F_{2n+2}; U_{n}\bigl(7/2\bigr)= F_{4n+4}/3.\\
	\end{displaymath}
	\begin{displaymath}
	T_{n}\bigl(\sqrt{5}/2\bigr)= \begin{cases}
	L_{n}/2, & \text{if $n$ is even;}\\
	\sqrt{5}F_{n}/2, & \text{if $n$ is odd.}
	\end{cases}\qquad
	U_{n}\bigl(\sqrt{5}/2\bigr)= \begin{cases}
	L_{n+1}, & \text{if $n$ is even;}\\
	\sqrt{5}F_{n+1}, & \text{if $n$ is odd.}
	\end{cases}\quad
	\end{displaymath}
	\begin{displaymath}
	T_{n}\bigl(\sqrt{5}\bigr)= \begin{cases}
	L_{3n}/2, & \text{if $n$ is even;}\\
	\sqrt{5}F_{3n}/2, & \text{if $n$ is odd.}
     \end{cases}	\qquad
	U_{n}\bigl(\sqrt{5}\bigr)= \begin{cases}
	L_{3n+3}/4, & \text{if $n$ is even;}\\
	\sqrt{5}F_{3n+3}/4, & \text{if $n$ is odd.}
	\end{cases}\quad
	\end{displaymath}
where $F_n$ and $L_n$ denote the famous Fibonacci and Lucas numbers.
\end{remark}

From the first formula of Corollary \ref{5} and Corollary \ref{6}, we can also deduce the following four Corollaries about Fibonacci and Lucas numbers.

\begin{corollary}
	Let $k$ be an integer. Then for $x=\sqrt{5}/2$, we have the identities
	\begin{align*}
	\sum\limits_{k=1}^{\infty}\arctan2\sqrt{5}L_{2k}/\bigl(L^{2}_{2k}-3\bigr)\arctan2\sqrt{5}F_{2k}/\bigl(L^{2}_{2k}+5\bigr)&=\arctan^{2}2/\sqrt{5};\\
	\sum\limits_{k=1}^{\infty}\arctan\sqrt{5}L_{2k+1}/\bigl(L^{2}_{2k+1}-2\bigr)\arctan\sqrt{5}F_{2k+1}/L^{2}_{2k+1}&=\arctan^{2}1/\sqrt{5}.
	\end{align*}
\end{corollary}

\begin{corollary}
	Let $k$ be an integer. Then for $x=3/2$, we have the identities
	\begin{align*}
	\sum\limits_{k=1}^{\infty}\arctan6L_{4k}/\bigl(L^{2}_{4k}+1\bigr)\arctan10F_{4k}/\bigl(L^{2}_{4k}+9\bigr)&=\arctan^{2}2/3;\\
	\sum\limits_{k=1}^{\infty}\arctan3F_{4k+2}/\bigl(F^{2}_{4k+2}-2\bigr)\arctan L_{4k+2}/F^{2}_{4k+2}&=\arctan^{2}1/3.\qquad
	\end{align*}
\end{corollary}

\begin{corollary}
	Let $k$ be an integer. Then for $x=\sqrt{5}$, we have the identities
	\begin{align*}
	\sum\limits_{k=1}^{\infty}\arctan4\sqrt{5}L_{6k}/\bigl(L^{2}_{6k}+12\bigr)\arctan8\sqrt{5}F_{6k}/\bigl(L^{2}_{6k}+20\bigr)&=\arctan^{2}\sqrt{5}/5;\\
	\sum\limits_{k=1}^{\infty}\arctan8\sqrt{5}L_{6k+3}/\bigl(L^{2}_{6k+3}-32\bigr)\arctan16\sqrt{5}F_{6k+3}/L^{2}_{6k+3}&=\arctan^{2}\sqrt{5}/10.
	\end{align*}
\end{corollary}

\begin{corollary}
	Let $k$ be an integer. Then for $x=7/2$, we have the identities
	\begin{align*}
	\sum\limits_{k=1}^{\infty}\arctan14L_{8k}/\bigl(L^{2}_{8k}+41\bigr)\arctan30F_{8k}/\bigl(L^{2}_{8k}+49\bigr)&=\arctan^{2}2/7;\\
	\sum\limits_{k=1}^{\infty}\arctan21F_{8k+4}/\bigl(F^{2}_{8k+4}-18\bigr)\arctan9L_{8k+4}/F^{2}_{8k+4}&=\arctan^{2}1/7.
	\end{align*}
\end{corollary}

\section{Lemmas}\label{sec2}
To complete the proofs of our theorems, we need several simple lemmas. The proof of these lemmas requires the properties of the Chebyshev polynomials of
the first and second kind, $T_n(x)$ and $U_n(x)$. All these can be found in \cite{Koshy2019,Koshy2014}, and we do not repeat them. First, we have the following:
\begin{lemma}\label{9}
	Let $k$, $\lambda$ be positive integers with $k\geq\lambda$. Then for all real $x$, we have
	\begin{align*}
	T_{k+\lambda}(x)-T_{k-\lambda}(x)&=2\bigl(x^2-1\bigr)U_{k-1}(x)U_{\lambda-1}(x),\\
	T_{k+\lambda}(x)+T_{k-\lambda}(x)&=2T_k(x)T_{\lambda}(x),\\
	T_{k+\lambda}(x)T_{k-\lambda}(x)&=T^{2}_k(x)+T^{2}_{\lambda}(x)-1.
	\end{align*}	
\end{lemma}
\begin{proof}
	See \cite[p.\ 393]{Koshy2014}.
\end{proof}
\begin{lemma}\label{10}
	Let $k$, $\lambda$ be positive integers with $k\geq\lambda$. Then for all real $x$, we have
	\begin{align*}
	U_{k+\lambda}(x)-U_{k-\lambda}(x)&=2T_{k+1}(x)U_{\lambda-1}(x),\\
	U_{k+\lambda}(x)+U_{k-\lambda}(x)&=2U_k(x)T_{\lambda}(x),\\
	U_{k+\lambda}(x)U_{k-\lambda}(x)&=U^{2}_k(x)-U^{2}_{\lambda-1}(x).
	\end{align*}	
\end{lemma}
\begin{proof}
		See \cite[p.\ 388]{Koshy2014}.
\end{proof}

Apart from that, we also need to use the following
properties of the arctangent (see \cite[p.\ 80]{AbramowitzStegun1964}):
\begin{align*}
\arctan x + \arctan y &=\arctan\bigl(x+y\bigr)/\bigl(1-xy\bigr) \quad\bigl(xy>1\bigr),\\
\arctan x - \arctan y &=\arctan\bigl(x-y\bigr)/\bigl(1+xy\bigr) \quad\bigl(xy<-1\bigr).
\end{align*}

\section{Proofs of the theorems}\label{sec3}
In this section, we use the two basic lemmas and the properties of 
arctangents to prove our main results. First we prove Theorem \ref{1}. From Lemma \ref{9} and the properties of arctangents, we have
\begin{align}\label{11}
\arctan1/T_{k-\lambda}(x)+\arctan1/T_{k+\lambda}(x)&=\arctan2T_k(x)T_{\lambda}(x)/\bigl(T^{2}_k(x)+T^{2}_{\lambda}(x)-2\bigr),\qquad
\end{align}
\begin{align}\label{12}
\arctan1/T_{k-\lambda}(x)-\arctan1/T_{k+\lambda}(x)&=\arctan2\bigl(x^2-1\bigr)U_{k-1}(x)U_{\lambda-1}(x)/\bigl(T^{2}_k(x)+T^{2}_{\lambda}(x)\bigr).
\end{align}
Taking $\lambda=1$ in \eqref{11}, \eqref{12}  and replacing $k$ by $2k$, we have
\begin{align*}
\arctan1/T_{2k-1}(x)+\arctan1/T_{2k+1}(x)&=\arctan2xT_{2k}(x)/\bigl(T^{2}_{2k}(x)+x^2-2\bigr),\\
\arctan1/T_{2k-1}(x)-\arctan1/T_{2k+1}(x)&=\arctan2\bigl(x^2-1\bigr)U_{2k-1}(x)/\bigl(T^{2}_{2k}(x)+x^2\bigr).
\end{align*}
Multiplying the two equations and summing for $k$ from $1$ to $n$, we may deduce the identity
\begin{align}\label{13}
&\sum_{k=1}^{n}\arctan2xT_{2k}(x)/\bigl(T^{2}_{2k}(x)+x^2-2\bigr)\arctan2\bigl(x^2-1\bigr)U_{2k-1}(x)/\bigl(T^{2}_{2k}(x)+x^2\bigr)\nonumber\\
&\quad\quad =\arctan^{2}1/x-\arctan^{2}1/T_{2n+1}(x).
\end{align}
This proves the first formula in Theorem \ref{1}. Similarly, replacing $k$ by $2k-1$, from the method of proving \eqref{13}, we can deduce the second formula. This proves Theorem \ref{1}.

\bigskip

Now we prove Theorem \ref{2}. From Lemma \ref{10} and the properties of 
arctangents, we have
\begin{align}\label{14}
\arctan1/U_{k-\lambda}(x)+\arctan1/U_{k+\lambda}(x)&=\arctan2U_k(x)T_{\lambda}(x)/\bigl(U^{2}_k(x)-U^{2}_{\lambda-1}(x)-1\bigr),
\end{align}
\begin{align}\label{15}
\arctan1/U_{k-\lambda}(x)-\arctan1/U_{k+\lambda}(x)&=\arctan2T_{k+1}(x)U_{\lambda-1}(x)/\bigl(U^{2}_k(x)-U^{2}_{\lambda-1}(x)+1\bigr).
\end{align}
Taking $\lambda=1$ in \eqref{14}, \eqref{15} and replacing $k$ by $2k$, we have
\begin{align*}
\arctan1/U_{2k-1}(x)+\arctan1/U_{2k+1}(x)&=\arctan2xU_{2k}(x)/\bigl(U^{2}_{2k}(x)-2\bigr),\\
\arctan1/U_{2k-1}(x)-\arctan1/U_{2k+1}(x)&=\arctan2T_{2k+1}(x)/U^{2}_{2k}(x).
\end{align*}
Multiplying the two identities and summing for $k$ from $1$ to $n$, we have
\begin{align}\label{16}
&\sum_{k=1}^{n}\arctan2xU_{2k}(x)/\bigl(U^{2}_{2k}(x)-2\bigr)\arctan2T_{2k+1}(x)/U^{2}_{2k}(x)\nonumber\\
& \quad\quad =\arctan^{2}1/2x-\arctan^{2}1/U_{2n+1}(x).
\end{align}

From Lemma \ref{10} and \eqref{16}, we can easily obtain the second formula. This proves Theorem \ref{2}.

Now we prove Theorem \ref{3}. From Lemma \ref{9} and the properties of
arctangents, we have
\begin{align}\label{17}
\arctan T_k(x)/T_{k+\lambda}(x)-\arctan T_{k-\lambda}(x)/T_k(x)&=\arctan\bigl(1-T_{\lambda}^{2}(x)\bigr)/2T^{2}_k(x)T_{\lambda}(x),
\end{align}
\begin{align}\label{18}
&\arctan T_k(x)/T_{k+\lambda}(x)+\arctan T_{k-\lambda}(x)/T_k(x)\nonumber\\
&=\arctan\bigl(2T^{2}_k(x)+T^{2}_{\lambda}(x)-1\bigr)/\bigl(x^2-1\bigr)U_{2k-1}(x)U_{{\lambda}-1}(x).\qquad\qquad\qquad\qquad\qquad
\end{align}
Taking $\lambda=1$ in \eqref{17} and \eqref{18}, we have
\begin{align*}
\arctan T_k(x)/T_{k+1}(x)-\arctan T_{k-1}(x)/T_k(x)&=\arctan\bigl(1-x^2\bigr)/2xT^{2}_k(x),\\
\arctan T_k(x)/T_{k+1}(x)+\arctan T_{k-1}(x)/T_k(x)&=\arctan\bigl(2T^{2}_k(x)+x^2-1\bigr)/\bigl(x^2-1\bigr)U_{2k-1}(x).
\end{align*}
Multiplying the above two formulae and summing for $k$ from $1$ to $n$, we have the following identity
\begin{align}\label{19}
&\sum_{k=1}^{n}\arctan\bigl(1-x^{2}\bigr)/2xT^{2}_k(x)\arctan\bigl(2T^{2}_k(x)+x^{2}-1\bigr)/\bigl(x^2-1\bigr)U_{2k-1}(x)\nonumber\\
& \quad\quad =\arctan^{2}T_n(x)/T_{n+1}(x)-\arctan^{2}1/x.
\end{align}
This proves the first formula in Theorem \ref{3}. Taking $\lambda=2$ in \eqref{17} and \eqref{18}, from the method of proving \eqref{19}, we can easily deduce the second formula. This proves Theorem \ref{3}.

Analogously, from Lemma \ref{10} and the method of proving Theorem \ref{3}, we can easily obtain Theorem \ref{4}.
This completes the proofs of our all results.
\section{Applications}\label{sec4}
To further illustrate the application of our results, we take some specific values to obtain identities about Fibonacci and Lucas numbers.

$\bullet$ Taking $x=\sqrt{5}/2$, from the first formula of Corollary \ref{5} and Corollary \ref{6}, we have the following identities
\begin{align*}
&\sum\limits_{k=1}^{\infty}\arctan\sqrt{5}T_{2k}\bigl(\sqrt{5}/2\bigr)/\bigl(T^{2}_{2k}\bigl(\sqrt{5}/2\bigr)-3/4\bigr)\arctan U_{2k-1}\bigl(\sqrt{5}/2\bigr)/\bigl(2T^{2}_{2k}\bigl(\sqrt{5}/2\bigr)+5/2\bigr)\\
& \quad\quad =\arctan^{2}2/\sqrt{5};\\
&\sum\limits_{k=1}^{\infty}\arctan\sqrt{5}U_{2k}\bigl(\sqrt{5}/2\bigr)/\bigl(U^{2}_{2k}\bigl(\sqrt{5}/2\bigr)-2\bigr)\arctan2T_{2k+1}\bigl(\sqrt{5}/2\bigr)/U^{2}_{2k}\bigl(\sqrt{5}/2\bigr)\\
& \quad\quad =\arctan^{2}1/\sqrt{5}.
\end{align*}
Note that
\begin{align*}
 T_{2k-1}\bigl(\sqrt{5}/2\bigr)=\sqrt{5}F_{2k-1}/2,\quad  T_{2k+1}\bigl(\sqrt{5}/2\bigr)=\sqrt{5}F_{2k+1}/2, \quad  T_{2k}\bigl(\sqrt{5}/2\bigr)=L_{2k}/2,\\
 U_{2k-1}\bigl(\sqrt{5}/2\bigr)=\sqrt{5}F_{2k},\qquad\quad U_{2k-2}\bigl(\sqrt{5}/2\bigr)=L_{2k-1},\qquad\quad
 U_{2k}\bigl(\sqrt{5}/2\bigr)=L_{2k+1}.
\end{align*}
Substituting the values into the above two identities, respectively, we have
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan2\sqrt{5}L_{2k}/\bigl(L^{2}_{2k}-3\bigr)\arctan2\sqrt{5}F_{2k}/\bigl(L^{2}_{2k}+5\bigr)&=\arctan^{2}2/\sqrt{5};\\
\sum\limits_{k=1}^{\infty}\arctan\sqrt{5}L_{2k+1}/\bigl(L^{2}_{2k+1}-2\bigr)\arctan\sqrt{5}F_{2k+1}/L^{2}_{2k+1}&=\arctan^{2}1/\sqrt{5}.
\end{align*}
Similarly, using the above method, we also have
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan10F_{2k-1}/\bigl(5F^{2}_{2k-1}-3\bigr)\arctan2L_{2k-1}/\bigl(5F^{2}_{2k-1}+5\bigr)&=\pi^{2}/16;\\
\sum\limits_{k=1}^{\infty}\arctan5F_{2k}/\bigl(5F^{2}_{2k}-2\bigr)\arctan L_{2k}/5F^{2}_{2k}&=\pi^{2}/16.
\end{align*}

$\bullet$ Take $x=3/2$ and note that
\begin{align*}
T_{2k+1}\bigl(3/2\bigr)=L_{4k+2}/2,\quad T_{2k-1}\bigl(3/2\bigr)=L_{4k-2}/2,\quad T_{2k}\bigl(3/2\bigr)=L_{4k}/2,\\
 U_{2k+1}\bigl(3/2\bigr)=F_{4k+4},\qquad\quad U_{2k-1}\bigl(3/2\bigr)=F_{4k},\qquad  U_{2k}\bigl(3/2\bigr)=F_{4k+2}.
\end{align*}
Substituting the values into the first formula of Corollary \ref{5}, Corollary \ref{6}, Corollary \ref{7} and Corollary \ref{8}, respectively, we have
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan6L_{4k}/\bigl(L^{2}_{4k}+1\bigr)\arctan10F_{4k}/\bigl(L^{2}_{4k}+9\bigr)&=\arctan^{2}2/3;\\
\sum\limits_{k=1}^{\infty}\arctan3F_{4k+2}/\bigl(F^{2}_{4k+2}-2\bigr)\arctan L_{4k+2}/F^{2}_{4k+2}&=\arctan^{2}1/3;\\
\sum\limits_{k=1}^{\infty}\arctan-5/3L^{2}_{2k}\arctan\bigl(2L^{2}_{2k}+5\bigr)/5F_{4k} &=\arctan^{2}\bigl(2/\sqrt{5}\bigr)/4-\arctan^{2}2/3;\\
\sum\limits_{k=1}^{\infty}\arctan1/3F^{2}_{2k+2}\arctan\bigl(2F^{2}_{2k+2}-1\bigr)/F_{4k+4} &=\arctan^{2}\bigl(2/\sqrt{5}\bigr)/4-\arctan^{2}1/3.
\end{align*}
Similarly, from the method above, we also have
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan6L_{4k-2}/\bigl(L^{2}_{4k-2}+1\bigr)\arctan10F_{4k-2}/\bigl(L^{2}_{4k-2}+9\bigr)&=\pi^{2}/16;\\
\sum\limits_{k=1}^{\infty}\arctan3F_{4k}/\bigl(F^{2}_{4k}-2\bigr)\arctan L_{4k}/F^{2}_{4k}&=\pi^{2}/16;
\end{align*}
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan-45/7L^{2}_{2k}\arctan\bigl(2L^{2}_{2k}+45\bigr)/15F_{4k}&=2\arctan^{2}\bigl(7-3\sqrt{5}\bigr)/2-\pi^{2}/16-\arctan^{2}2/7;\\
\sum\limits_{k=1}^{\infty}\arctan9/7F^{2}_{2k+2}\arctan\bigl(2F^{2}_{2k+2}-9\bigr)/3F_{4k+4}&=2\arctan^{2}\bigl(7-3\sqrt{5}\bigr)/2-\arctan^{2}1/8.
\end{align*}

$\bullet$ Take $x=\sqrt{5}$ and note that
\begin{align*}
T_{2k-1}\bigl(\sqrt{5}\bigr)=\sqrt{5}F_{6k-3}/2,\quad T_{2k+1}\bigl(\sqrt{5}\bigr)=\sqrt{5}F_{6k+3}/2,\quad T_{2k}\bigl(\sqrt{5}\bigr)=L_{6k}/2,\\ U_{2k-1}\bigl(\sqrt{5}\bigr)=\sqrt{5}F_{6k}/4,\quad  U_{2k-2}\bigl(\sqrt{5}\bigr)=L_{6k-3}/4,\qquad
 U_{2k}\bigl(\sqrt{5}\bigr)=L_{6k+3}/4.
\end{align*}
Substituting the values into the first formula of Corollary \ref{5} and Corollary \ref{6}, respectively, we have the following identities
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan4\sqrt{5}L_{6k}/\bigl(L^{2}_{6k}+12\bigr)\arctan8\sqrt{5}F_{6k}/\bigl(L^{2}_{6k}+20\bigr)&=\arctan^{2}\sqrt{5}/5;\\
\sum\limits_{k=1}^{\infty}\arctan8\sqrt{5}L_{6k+3}/\bigl(L^{2}_{6k+3}-32\bigr)\arctan16\sqrt{5}F_{6k+3}/L^{2}_{6k+3}&=\arctan^{2}\sqrt{5}/10.
\end{align*}
Analogously, from the above calculation method, we also obtain
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan20F_{6k-3}/\bigl(5F^{2}_{6k-3}+36\bigr)\arctan8L_{6k-3}/\bigl(5F^{2}_{6k-3}+20\bigr)&=\pi^{2}/16; \\
\sum\limits_{k=1}^{\infty}\arctan40F_{6k}/\bigl(5F^{2}_{6k}-32\bigr)\arctan16L_{6k}/5F^{2}_{6k}&=\pi^{2}/16.
\end{align*}

$\bullet$ Take $x=7/2$ and note that
\begin{align*}
T_{2k+1}\bigl(7/2\bigr)=L_{8k+4}/2,\quad T_{2k-1}\bigl(7/2\bigr)=L_{8k-4}/2,\quad T_{2k}\bigl(7/2\bigr)=L_{8k}/2, \\
U_{2k-2}\bigl(7/2\bigr)=F_{8k-4}/3,\quad  U_{2k-1}\bigl(7/2\bigr)=F_{8k}/3,\quad  U_{2k}\bigl(7/2\bigr)=F_{8k+4}/3.
\end{align*}
Substituting the values into the first formula of Corollary \ref{5} and Corollary \ref{6}, respectively, we have the identities
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan14L_{8k}/\bigl(L^{2}_{8k}+41\bigr)\arctan30F_{8k}/\bigl(L^{2}_{8k}+49\bigr)&=\arctan^{2}2/7;\\
\sum\limits_{k=1}^{\infty}\arctan21F_{8k+4}/\bigl(F^{2}_{8k+4}-18\bigr)\arctan9L_{8k+4}/F^{2}_{8k+4}&=\arctan^{2}1/7.
\end{align*}
Analogously, from the above calculation method, we also have
\begin{align*}
\sum\limits_{k=1}^{\infty}\arctan14L_{8k-4}/\bigl(L^{2}_{8k-4}+41\bigr)\arctan30F_{8k-4}/\bigl(L^{2}_{8k-4}+49\bigr)&=\pi^{2}/16;\\
\sum\limits_{k=1}^{\infty}\arctan21F_{8k}/\bigl(F^{2}_{8k}-18\bigr)\arctan9L_{8k}/F^{2}_{8k}&=\pi^{2}/16.
\end{align*}

\section{Acknowledgment}

Funding: This work is supported by the N. S. F. (12126357) of P. R. China.


\begin{thebibliography}{99}
	
	\bibitem{AbramowitzStegun1964}
	M. Abramowitz and I. A. Stegun,
	\newblock \emph{Handbook of Mathematical
		Functions with Formulas, Graphs, and Mathematical Tables},
	\newblock Vol.\ 55, US Government printing office, 1964.
	
	\bibitem{Koshy2019}
	T. Koshy,
	\newblock \emph{Fibonacci and Lucas Numbers with Applications},
	\newblock John Wiley \& Sons, 2019.
	
	\bibitem{Koshy2014}
	T. Koshy,
	\newblock \emph{Pell and Pell-Lucas Numbers with Appliations},
	\newblock Springer, 2014.
	
	\bibitem{KimDolgyKim2018}
	T. Kim, D. Dolgy, and D. Kim,
	\newblock Representing sums of finite products of Chebyshev polynomials of the second kind and Fibonacci polynomials in terms of Chebyshev polynomials,
	\newblock \emph{Adv. Stud. Contemp. Math.} \textbf{28} (2018), 321--336.
	
	\bibitem{KimDolgyKimSeo2016}
	T. Kim, D. V. Dolgy, D. S. Kim, and J. J. Seo,
	\newblock Convolved Fibonacci numbers and their applications, preprint, 2016.
	\newblock Available at \url{https://arxiv.org/abs/1607.06380}.
	
	\bibitem{KimKimJangDolgy2019}
	T. Kim, D. S. Kim, L. C. Jang, and D. V. Dolgy,
	\newblock Representing by several orthogonal polynomials for sums of finite products of Chebyshev polynomials of the first kind and Lucas polynomials,
	\newblock  \emph{Adv. Difference. Equ.} \textbf{2019} (2019), 1--15.
	
	\bibitem{KimKimDolgyKwon2018}
	T. Kim, D. S. Kim, D. V. Dolgy, and J. Kwon,
	\newblock Representing sums of finite products of Chebyshev polynomials of the first kind and Lucas polynomials by Chebyshev polynomials,
	\newblock \emph{Mathematics} \textbf{7} (2018), 26.
	
	\bibitem{KimKimJangDolgy2018}
	T. Kim, D. S. Kim, L. C. Jang, and D. V. Dolgy,
	\newblock Representation by Chebyshev polynomials for sums of finite products of Chebyshev polynomials,
	\newblock \emph{Symmetry} \textbf{10} (2018), 742.
	
	\bibitem{KimKimJangJang2018}	
	T. Kim, D. S. Kim, L. C. Jang, and G. W. Jang,
	\newblock  Fourier series for functions related to Chebyshev polynomials of the first kind and Lucas polynomials,
	\newblock \emph{Mathematics} \textbf{6} (2018), 276.
	
	\bibitem{MaLv2017}
	Y. K. Ma and X. X. Lv,
	\newblock Some identities involving the reciprocal sums of one-kind Chebyshev polynomials,
	\newblock \emph{Math. Probl. Eng.} \textbf{2017} (2017).
	
	\bibitem{ZhangHan2020}
	W. P. Zhang and D. Han,
	\newblock On the Chebyshev polynomials and some of their reciprocal sums,
	\newblock \emph{Symmetry} \textbf{12} (2020), 704.
	
	\bibitem{Zhang2002}
	W. P. Zhang,
	\newblock On Chebyshev polynomials and Fibonacci numbers,
	\newblock \emph{Fibonacci Quart.} \textbf{40} (2002), 424--428.
	
	\bibitem{ZhangChen2018}
	Y. X. Zhang and Z. Y. Chen,
	\newblock A new identity involving the Chebyshev polynomials,
	\newblock \emph{Mathematics} \textbf{6} (2018), 244.
	
	\bibitem{Adegoke2016}
	K. Adegoke,
	\newblock  Infinite arctangent sums involving Fibonacci and Lucas numbers, preprint, 2016.
	\newblock Available at \url{https://arxiv.org/abs/1603.08097}.
	
	\bibitem{MahonHoradam1985}
	B. Mahon and A. Horadam,
	\newblock Inverse trigonometrical summation formulas involving Pell polynomials,
	\newblock \emph{Fibonacci Quart.} \textbf{23} (1985), 319--324.
	
	\bibitem{FrontczakGoy2021}
	R. Frontczak and T. Goy,
	\newblock Chebyshev-Fibonacci polynomial relations using generating functions, preprint, 2021.
	\newblock Available at \url{https://arxiv.org/abs/2103.08015}.
	
	\bibitem{HarneBadshahVerma2014}
	S. Harne, V. Badshah, and V. Verma,
	\newblock Some identities involving Chebyshev polynomials, Fibonacci numbers and Lucas numbers,
	\newblock \emph{Int. J. Contemp. Math. Sci.} \textbf{9} (2014), 553--561.
\end{thebibliography}


\bigskip
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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11B37; Secondary 33B10, 11B39.

\noindent \emph{Keywords: } 
Chebyshev polynomial, trigonometric function, arctangent,
Fibonacci number, Lucas number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032} and
\seqnum{A000045}.)

\bigskip
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\bigskip

\vspace*{+.1in}
\noindent
Received August 31 2022;
revised versions received  September 16 2022; December 20 2022.
Published in {\it Journal of Integer Sequences}, January 1 2023.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in

\end{document}
