How do we prove $|\cos x-\cos y|\le|x-y|$ for all $x,y\in\mathbb{R}$? My thought:
Fix any point $y_0\in\mathbb{R}$, since we know $\cos' x=-\sin x$ and $|-\sin|\le1$, thus $|\cos' x|=\lim_{x\to y_0}\frac{|\cos x-\cos y_0|}{|x-y_0|}=...\le 1$. I think my issue is how to translate $\lim_{x\to y_0}\frac{|\cos x-\cos y_0|}{|x-y_0|}$ into $\frac{\cos x-\cos y}{x-y}$. Could someone give some insight?
 A: Hint You're on the right track. If you instead write the derivative relationship in terms of integrals, you get $$|\cos x - \cos y| = \left\vert\int_x^y \sin x \,dx \right\vert \leq \cdots .$$

 $$\cdots \leq \left\vert\int_x^y |\sin x| \,dx\right\vert .$$

A: Let $x, y \in \mathbb{R}$ be arbitrary, and without loss of generality, assume $y < x$.
The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(t) = \cos(t)$ is continuous and differentiable in $\mathbb{R}$, so it is obviously continuous in $[y, x]$ and differentiable in $(y, x)$.
By the Mean Value Theorem, there exists a point $c \in (y, x)$ such that
$$f^{\prime}(c) = -\sin(c) = \dfrac{\cos(x)-\cos(y)}{x-y}\text{.}$$
However, recalling that $|-\sin(w)| = |\sin(w)| \leq 1$ for all $w \in \mathbb{R}$, it follows that 
$$\left|\dfrac{\cos(x)-\cos(y)}{x-y}\right| \leq 1$$
hence
$$|\cos(x)-\cos(y)| \leq |x-y|$$
and the claim follows since $x$ and $y$ were arbitrarily chosen.
If $y > x$, then a nearly-identical argument to the above leads to the same conclusion.
If $y = x$, equality holds (since $|\cos(x)-\cos(y)| = |x-y| = 0$).
A: hint
$$|\cos (x)-\cos (y)|=$$
$$2|\sin (\frac {x+y}{2})\sin (\frac {x-y}{2})|$$
and $$|\sin (a)|\le |a|$$
A: Hints:
$$|\cos x-\cos y|=|\sin \xi| \cdot |x-y||\le|x-y|$$
A: We will prove this for all $0<x,\ y <\frac{\pi}{2}$
If $\pi :\mathbb{R}^2\rightarrow \mathbb{R},\ \pi (a,b)=a$, then $$
|\pi (P)-\pi (Q)| \leq |P-Q|$$
If $$P=(\cos\ x,\sin\ x),\ Q=(\cos\ y,\sin\ y),$$ then $$ |x-y| \geq
|P-Q| \geq |\pi(P)-\pi (Q) |$$
