Fancy Summation

Here is an Euler-themed solution. First of all, notice that $\cos^2 x_k = \frac{1}{1+\tan^2 x_k} = \frac{1}{1+x_k^2}$ , and so, $\sum_{k=1}^{\infty} \cos^2(x_k) = \sum_{k=1}^{\infty} \frac{1}{1+x_k^2}.$ Next, introduce the function $f(z) = \frac{\sin\sqrt{z}}{\sqrt{z}} - \cos\sqrt{z}$ . This is an entire function of order 1/2 . Moreover, a careful inspection shows that the only zeros of $f$ are either $0$ or $x_k^2$ for some $k \in \mathbb{N}$ , all of which are simple. So by the Hadamard factorization theorem , $f(z-1) = A(z-1)\prod_{k=1}^{\infty} \left( 1 - \frac{z}{x_k^2 + 1} \right).$ (Notice that the zeros of $f$ are either $1$ or $x_k^2+1$ for some $k \in \mathbb{N}$ .) Plugging $z = 0$ determines the value of $A$ as $1/e$ , and so, $\prod_{k=1}^{\infty} \left( 1 - \frac{z}{x_k^2 + 1} \right) = e\frac{f(z-1)}{z-1} = 1 - \frac{e^2-7}{4} z + \mathcal{O}(z^2).$ Now comparing the coefficient of $z$ in both sides, we obtain $\sum_{k=1}^{\infty} \frac{1}{1+x_k^2} = \frac{e^2-7}{4}.$

The function $f(z) \; = \; z\cos z - \sin z$ has a triple zero at $z=0$ , and simple zeros at $z = \pm x_n$ for $n \in \mathbb{N}$ . Moreover $f'(x) \; = \; -z\sin z$ . Thus the function $g(z) \; = \; \frac{z\sin z}{f(z)}$ is an odd function with a simple pole at $z=0$ and simple poles at $z = \pm x_n$ for $n \in \mathbb{N}$ . Moreover $\mathrm{Res}_{z=0} g(z) \; = \; -3 \hspace{2cm} \mathrm{Res}_{z=\pm x_n}g(z) \; = \; -1$ If we let $C_N$ be the square with vertices $N\pi+iN\pi,N\pi-iN\pi,-N\pi-iN\pi,-N\pi+iN\pi$ , then we can show that there exists $K > 0$ such that $|g(z)| \le K$ for all $z \in C_N$ for all positive integers $N$ , and this implies that $\lim_{N \to \infty}\int_{C_N} \frac{g(z)}{z^2+1}\,dz \; = \; 0$ Now $\mathrm{Res}_{z=i}\frac{g(z)}{z^2+1} \; = \; \frac{g(i)}{2i} \; = \; \tfrac14(e^2-1) \hspace{2cm} \mathrm{Res}_{z=-i} \frac{g(z)}{z^2+1} \; = \; \frac{g(-i)}{-2i} \; = \; \tfrac14(e^2-1)$ and so $\begin{aligned} \frac{1}{2\pi i}\int_{C_N} \frac{g(z)}{z^2+1}\,dz & = \; \mathrm{Res}_{z=0}\frac{g(z)}{z^2+1} + \mathrm{Res}_{z=i}\frac{g(z)}{z^2+1} + \mathrm{Res}_{z=-i}\frac{g(z)}{z^2+1} + 2\sum_{n=1}^N \mathrm{Res}_{z=x_n}\frac{g(z)}{z^2+1} \\ & = \; -3 + \tfrac14(e^2-1) + \tfrac14(e^2-1) - 2\sum_{n=1}^N \frac{1}{x_n^2+1} \\ & = \; \tfrac12(e^2-7) - 2\sum_{n=1}^N \cos^2x_n \end{aligned}$ Letting $N \to \infty$ tells us that $\sum_{n=1}^\infty \cos^2x_n \; = \; \boxed{\tfrac14(e^2-7)}$