$\forall x,y \in \mathbb{R}:\vert\cos x- \cos y\vert\leq\vert x-y\vert$ Prove that $$\forall x,y \in \mathbb{R}:\vert\cos x- \cos y\vert\leq\vert x-y\vert.$$
My try : $ f(x) =\cos x$ and use the mean value theorem.
 A: $$\begin{align} |\cos x- \cos y|&=\left| \int_x^y \sin(t) dt \right|\\
&= \left| \int_{\min(x,y)}^{\max(x,y)}\sin(t) dt \right|\\ 
&\leq \int_{\min(x,y)}^{\max(x,y)} |\sin(t)| dt \\ 
&\leq  \int_{\min(x,y)}^{\max(x,y)} 1 dt \\ &
= |x-y|\end{align}$$
A: Hint
Let $f(x)=\cos(x)$.
Then you know that 
$$f'(x)=-\sin(x)$$
so
$$\forall x\in\mathbb R,\quad \vert f'(x)\vert \leq 1.$$
But
$$f'(a)=\lim_{h\to 0} \frac{f(a+h)-f(a)}h.$$
So you can show that you always have
$$\vert f(x)-f(y)\vert\leq (\max \vert f'\vert)\cdot\vert x-y\vert.$$
Can you take it from here?
You can also look here.
A: since $\left| \sin { x }  \right| \le \left| x \right| $
$$|\cos  x-\cos  y|≤\left| 2\sin { \frac { y-x }{ 2 } \sin { \frac { x+y }{ 2 }  }  }  \right| \le 2\left| \sin { \frac { x-y }{ 2 }  }  \right| \left| \sin { \frac { x+y }{ 2 }  }  \right| =2\left| \frac { x-y }{ 2 }  \right| \cdot 1=\left| x-y \right| $$
A: Use the mean value theorem on $f(u)=\cos u$ with the interval $[y,x]$ or $[x,y]$.
$$-1 \leq \frac{\cos x-\cos y}{x-y}=-\sin (c) \leq 1$$
Because for any $c \in \mathbb{R}$ we have $-1 \leq -\sin (c) \leq 1$.
So we have,
$$|\frac{\cos x-\cos y}{x-y}| \leq 1$$
$$|\cos x-\cos y| \leq |x-y|$$
