I honestly feel like this function is not a contraction, but the exercise I am doing explicitly relies on it being a contraction and using the Banach fixed point theorem.
Show that $f(x)=\sin x+1$ is a contraction in a neighbourhood of $\pi$.
It being around $\pi$ actually makes it worse because I cannot put an upper bound on $\cos x$ less than $1$. Using the mean value theorem all I can show is that $f$ is a Lipschitz function of constant $1$. Any help would be greatly appreciated.
The result is wrong.
If $f$ was a contraction in a neighborhood of $\pi$, it would exist $0\le k \lt 1$ such that
$$\vert f(x) -f(\pi) \vert = \vert \sin x \vert \le k \vert x -\pi\vert $$ for $x$ in that neighborhood. A contradiction as
$$\lim\limits_{x \to \pi} \frac{\sin x}{x-\pi}=-1.$$