Show that $|b-a|\geq|\cos a-\cos b|$ for all real numbers $\,a\,$ and $\,b$

Even faster you can use fundamental theorem of calculus and assuming for example $a≤b$ $$ \begin{align*} |\cos(a)-\cos(b)| &= \left|\int_a^b \sin(x)\,\mathrm{d}x\right| ≤ \int_a^b \left|\sin(x)\right|\mathrm{d}x \\ &≤ \int_a^b 1\,\mathrm{d}x = |b-a| \end{align*} $$

We know that $$cos(A)-cos(B)=2sin(\frac{A+B}{2})sin(\frac{B-A}{2})$$ $$ x \geq \sin(x) ; x\geq0 \Rightarrow |x|\geq |\sin(x)|$$ $$\sin(x)\leq1 $$ For $x \neq 0$ $$|\frac{x}{\sin(x)}|\geq 1$$

Let $\frac{b-a}{2}=x$

$$|\frac{\frac{b-a}{2}}{\sin(\frac{b-a}{2})}|\geq 1$$

$$|\frac{\frac{b-a}{2}}{\sin(\frac{b-a}{2})}|\geq \sin(\frac{a+b}{2}) $$

After some rearranging,

$$|b-a|\geq|2\sin(\frac{b-a}{2})\sin(\frac{a+b}{2})|$$

$$ |b-a| \geq |cos(a)-cos(b)|$$ For a=b case, x=0 and that becomes a limit in this case $\lim_{x\rightarrow 0}\frac{x}{\sin(x)}=1$ which is true in this expression since $\sin(\frac{a+b}{2})$ is bounded;

$$|\frac{\frac{b-a}{2}}{\sin(\frac{b-a}{2})}|\geq \sin(\frac{a+b}{2})$$

Let $A,B$ be two points on the unit circle with coordinates $$A=(x_1,y_1),B=(x_2,y_2).$$ Then $$|x_1-x_2|\leq d(A,B)\leq {\rm length~of~arc~joining~}A{\rm~and~}B,$$ where $d(A,B)$ is the length of cord $AB$. Note that angle is just a way of measuring arc length on the unit circle.