More functions to solve

Since $f(n+1)^2 + g(n+1)^2 \; = \; f(n)^2 + g(n)^2$ for all $n \ge 1$ , we deduce by induction that $f(n)^2 + g(n)^2 \; = \; 1$ for all $n \ge 1$ , and hence we can find a function $X \,:\, \mathbb{N} \to \mathbb{R}$ such that $f(n) \; = \; \cos(X(n)) \hspace{1cm} g(n) \; = \; \sin(X(n)) \hspace{2cm} n \in \mathbb{N}$ The defining conditions for $f$ and $g$ now tell us that $\cos(X(n+1)) \; = \; \cos\big(X(n) + g(n)\big) \hspace{2cm} \sin(X(n+1)) \; = \; \sin\big(X(n) + g(n)\big) \hspace{2cm} n \in \mathbb{N}$ and hence we deduce that $X(n+1) \; = \; X(n) + g(n) + 2k(n)\pi \; = \; X(n) + \sin(X(n)) + 2k(n)\pi \hspace{2cm} n \in \mathbb{N}$ for some integers $k(n)$ . Since the values of $k(n)$ do not affect the values of $f(n)$ and $g(n)$ , we may as well assume that $k(n) = 0$ for all $n$ , so that $X(n+1) \; = \; X(n) + \sin(X(n)) \hspace{2cm} n \in \mathbb{N}$ Now the function $F\,:\, [0,\pi] \to \mathbb{R}$ defined by $F(x) = x + \sin x$ is continuous and strictly increasing, with $F(0)=0$ and $F(\pi)=\pi$ . Thus $0 < X(n) < \pi\;\;\;\Rightarrow\;\;\;0 < X(n+1) = F(X(n)) < \pi$ Also, since $\sin(X(n)) > 0$ , we see that $X(n) < X(n+1)$ . Thus we deduce that the sequence $\big(X(n)\big)_{n \ge 1}$ is strictly increasing, with $0 < X(n) < \pi$ for all $n \ge 1$ . Thus $Z = \lim_{n \to \infty}X(n)$ exists, with $0 < Z \le \pi$ . Since $F$ is continuous and $X(n+1) = F(X(n))$ for all $n\ge 1$ , we deduce that $Z = F(Z)$ , so that $\sin Z = 0$ , and hence $Z = \pi$ . Hence $\lim_{n \to \infty}f(n) \,=\, \cos\big(\lim_{n \to \infty}X(n)\big) \,=\, \cos Z = \boxed{-1}$