Uniform Continuity of Cos Function without use Lipschitz Condition

Question

To show that the cos function is uniform continuous on R, instead of showing that the function satisfies that Lipschitz condition ($|cos(x_1)-cos(x_2)| \le |x_1-x_2|$ for all $x_1, x_2$ in $R$), is there an alternative way to show that cos function is uniformly continuous on R. Is it possible to show the uniform continuity making use of the fact that the function is uniformly continuous on $[0, 2 \pi]$?

Answer 1

Hint: Notice if $x,y \in \mathbb{R}$, then we can apply the Lagrange's theorem to get a bound for $|\cos y - \cos x|$. For instance, by Lagrange theorem there exists $\xi \in \mathbb{R}$ such that $$ |\cos y - \cos x| = \cos' (\xi) |y - x| \leq |y-x|$$.

Now, say $\epsilon > 0$, therefore pick your $\delta = \epsilon$ , and the desired result follows!