Uniform convergence of $\cos(f_n)$.

If a sequence of functions $\{f_n\}$ converges uniformly to $f$ on $D \subset \mathbb{R}$, then the sequence $\{\cos(f_n) \}$ also converges uniformly on $D$.

Let $\epsilon > 0$. Since $\{f_n\}$ converges uniformly, $\exists N \in \mathbb{N}$ such that $\forall x \in D$, $n \geq N, \implies |f_n(x) - f(x)| < \epsilon$.

After this, I seriously have no idea how to proceed. Any help would be useful!

I think that nothing is special about cosine here; it could be any (uniformly) continuous function. Let $\varepsilon > 0$. Then for some $\delta > 0$, $|g(x)-g(y)|<\varepsilon$ for all $|x-y|<\delta$. Then just use $\delta$ with your sequence of functions, so that for large $n$, $|f_n(x)-f(x)|<\delta$, so $|g(f_n(x))-g(f(x))|<\varepsilon$.