How to prove that $|\cos x-\cos y|\leq |x-y| \ \forall x,y \in \mathbb{R}$?

Question

As above, how would you go about proving this inequality?

Answer 1

Assume initially that $x \le y$. From the FTC we have $$ \cos(y) - \cos(x) = \int_x^y \cos'(z) dz = -\int_x^y \sin(z) dz $$ and so $$ |\cos(y) - \cos(x)| \le \left\vert -\int_x^y \sin(z) dz \right\vert \le \int_x^y |\sin(z)| dz \le \int_x^y dz = y-x = |y-x|. $$ If $y \le x$ a similar argument works, which I'll leave as an exercise.

Answer 2

Mean value theorem: there exists $c$ between $x$ and $y$ such that $\;\cos x-\cos y=-\sin c(x-y)$, hence $$\lvert\cos x-\cos y\rvert=\lvert\sin c\rvert\lvert x-y\rvert\le\lvert x-y\rvert. $$

Answer 3

$$\left|\cos x -\cos y\right|=\left|−2\sin\frac{​​​x+y}{2}​​\cdot\sin\frac{​​​x-y}{2}​​\right|$$ $$=2\left|\sin\frac{​​​x+y}{2}​​​\right|\left|\sin\frac{​​​x-y}{2}​​​\right|\le 2\cdot 1\cdot\left|\frac{​​​x-y}{2}​​​\right|=\left|x - y\right|.$$

Answer 4

$\left|f(x)-f(y)\right|\leq \left(\sup\limits_{z\in [x,y]}\left|f'(z)\right|\right)\cdot \left|x-y\right|$