This came up as I was thinking about how to solve an assignment problem I've been given. I'm not posting the assignment problem here as I want to think about it myself, but one approach I am currently working on is to try to show that $f:\mathbb{R}\to\mathbb{R},\quad f(x) = 1-\sin(x)$ is a contraction mapping.
At the moment I have a feeling it isn't. For example take $x_1 < 0 < x_2$, then $ \left|\frac{(1-\sin x_1)-(1-\sin x_2)}{x_1-x_2}\right| = \left|\frac{\sin(x_1) - \sin(x_2)}{x_1-x_2}\right| \to 1$ as $x_1 \to 0^-, x_2 \to 0^+$.
Hence there does not exist a non-negative $k<1 \in \mathbb{R}$ such that $|\sin(x_1)-\sin(x_2)| \leq k|x_1-x_2|$ for all $x_1,x_2 \in \mathbb{R}$. (Just take $x_1,x_2$ arbitrarily close to $0$.) So $f$ is not a contraction mapping.
That was the idea. Thinking just a little bit further, we see that this occurs because |f'(x)| = 1 at $x=0$. So perhaps we have a necessary and sufficient condition for a function on the reals to be a contraction mapping, namely that \forall x\in\mathbb{R},|f'(x)|\leq k<1 for some non-negative real $k$. Is this the case? Or have I missed some subtle (or not-so-subtle) point?