why contraction mapping theorem fails in this case?

Question

Suppose $f(x) = \cos x$, then for $x \in [0, \frac{\pi}{3}]$, we know that there exists a unique fixed point as the mapping $f$ is contraction mapping on $[0, \frac{\pi}{3}]$. One can see this in plot as well. Now consider a smaller set, e.g., $x \in [0, \frac{\pi}{6}]$, the mapping $f$ still is a contraction mapping as $$ |f(x)-f(y)| = |\sin \zeta||x-y| \le \frac{1}{2}|x-y|. $$ However, if you plot it in the graph, you can observe that the graph $\cos x$ does not intersect with $h(x) = x$ in $[0, \frac{\pi}{6}]$, i.e., there is no fixed point in $[0, \frac{\pi}{6}]$. What is possibly going wrong here? What assumptions did I fail to satisfy for the contraction mapping principle?