# Uniform Continuity of Cos Function without use Lipschitz Condition

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 $[0, 2 \pi]$?

-
Since the derivative of this function is bounded,use the Lagrange mean-value theorem,you can get what you need. – 89085731 Nov 5 '12 at 2:10
Hanna, I think you have the right idea. You know that $\cos{x}$ is periodic, and continuous functions are uniformly continuous on compact sets. Therefore, for every $\epsilon$, the $\delta$ you pick (using the usual symbols for the definition) will be the same for every interval of length $2\pi$. – Christopher A. Wong Nov 5 '12 at 2:13

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!
 Sin $|\cos y| \leq 1$ for all $y$ – Citizen Nov 5 '12 at 2:39 Note that a bounded derivative is all that's needed to prove a function is uniformly continuous so really all you need to notice is that the derivative is bounded on the domain to get uniform continuity. Lagrange's mean value theorem is the reason why this is all you need to notice... – sponsoredwalk Nov 5 '12 at 2:40 Thanks all for the reply. Hanna – hanna Nov 5 '12 at 23:45