Relatively simple real analysis related to mean value theorem.

Prove that |cos(x)-cos(y)| =< |x-y| for all x,y in the reals. This should be equivalent to showing that |x-y|-|cos(x)-cos(y)| >= 0. I also notice that if I divide by |x-y|, the inequality looks vaguely like the definition of derivative for the function cos(x): |(cos(x)-cos(y))/(x-y)|. The issue with all of these, however, is that I don't see how the mean value theorem applies. The fact I'm working with multiple variables makes me want to use partial derivatives, but those are nowhere to be seen in an introductory analysis course; I don't immediately see them being of any use here, either.

Re: Relatively simple real analysis related to mean value theorem.

I would use the Mean Value Theorem here. $\displaystyle \displaystyle \begin{align*} \frac{f(b) - f(a)}{b - a} = f'(c) \end{align*}$ for some $\displaystyle \displaystyle \begin{align*} c \in [a, b] \end{align*}$.

First note that if $\displaystyle \displaystyle \begin{align*} f(X) = \cos{(X)} \end{align*}$ then $\displaystyle \displaystyle \begin{align*} f'(X) = -\sin{(X)} \end{align*}$, and so $\displaystyle \displaystyle \begin{align*} \left| f'(X) \right| \leq 1 \end{align*}$ for all $\displaystyle \displaystyle \begin{align*} X \end{align*}$.

Then we can have $\displaystyle \displaystyle \begin{align*} f(b) = f(x) = \cos{(x)} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} f(a) = f(y) = \cos{(y)} \end{align*}$, and so

$\displaystyle \displaystyle \begin{align*} \frac{\cos{(x)} - \cos{(y)}}{x - y} &= -\sin{(c)} \\ \left| \frac{\cos{(x)} - \cos{(y)}}{x - y} \right| &= \left| -\sin{(c)} \right| \\ \frac{\left| \cos{(x)} - \cos{(y)} \right|}{\left|x - y \right|} &\leq 1 \\ \left| \cos{(x)} - \cos{(y)} \right| &\leq \left| x - y \right| \end{align*}$