Prove that $\left | \cos x - \cos y \right | \leq \left | x - y \right |, \forall x,y \in \mathbb{R}.$ This is one of the problem in my text book where the section which the problem is stated talks about the mean value theorem and Rolle's theorem. By looking at this, I have no idea where to start. Can I have some hints??
 A: $\cos(x)$ is differentiable in $\mathbb{R}$.
By Mean value theorem $|\cos(x) - \cos(y)| = |sin(\xi)||x - y|$ where $\xi \in (x,y)$ or $(y,x)$.
$|\sin(x)| < 1$ $\forall$ $x \in \mathbb{R}$.
Now get your answer. 
A: This follows also from the standard $|\sin(x)| \leq |x|$:
$$\left| cos(x) - \cos(y)  \right|=2  \left| \sin(\frac{x+y}{2}) \sin(\frac{y-x}{2})  \right| \leq 2 \cdot  1 \cdot \left|\frac{y-x}{2}  \right|$$
A: By integration:
$$|\cos y - \cos x| = \left|-\int_x^y \sin t dt \right| \leq \int_x^y |\sin t dt| \leq \int_x^y 1 dt = |y-x|$$
A: $\newcommand{\+}{^{\dagger}}%
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By ${\it Mean Value Theorem}$,
$\exists\ \xi \in {\mathbb R}\quad \ni\quad \min\braces{x,y} < \xi < \max\braces{x,y}$ which satisfies
$$
{\cos\pars{x} - \cos\pars{y} \over x - y}  = -\sin\pars{\xi}
\quad\imp\quad 
\verts{\cos\pars{x} - \cos\pars{y} \over x - y}  = \verts{\sin\pars{\xi}} \leq 1
$$
