Just $\sin(x+\sin(x+...))$

$y=\sin(x+y)$

$\frac{dy}{dx}=\cos(x+y)\left(1+\frac{dy}{dx}\right)$

$=\frac{\cos(x+y)}{1-\cos(x+y)}$

$\frac{q}{1-q}\text{ for }-1\leq q\leq1\text{ strictly increases}$

$\text{Therefore, its lowest value is }\frac{-1}{1--1}=-0.5$

Let $u=\sin(x + \sin (x + \sin (x + \cdots )))$ . Then

$\begin{aligned} u & = \sin(x + u) \\ \frac {du}{dx} & = \cos (x+u) + \cos (x+u)\frac {du}{dx} & \small \blue{\text{By chain rule}} \\ \implies \frac {du}{dx} & = \frac {\cos (x+u)}{1-\cos (x+u)} & \small \blue{\text{By half-angle tangent substitution}} \\ & = \frac {\frac {1-t^2}{1+t^2}}{1-\frac {1-t^2}{1+t^2}} = \frac {1-t^2}{2t^2} & \small \blue{\text{and let }t = \tan \frac {x+u}2} \\ & = \frac 1{2t^2} - \frac 12 \ge \boxed{-0.5} & \small \blue{\text{Minimum occurs when }t \to \infty} \end{aligned}$

Reference: Half-angle tangent substitution