Show that $|b-a|\geq|\cos a-\cos b|$ for all real numbers $\,a\,$ and $\,b$ $\mathbf{Question:}$ Show that $|b-a|\geq|\cos a-\cos b|$ for all real numbers a and b.
$\mathbf{My\ attempt:}$
The Mean Value Theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists $c \in (a,b)$ such that $f'(c)=\frac {f(b)-f(a)}{b-a}.$
Using the MVT where $f(c)=\cos c:$
$$
f'(c)=\frac {f(b)-f(a)}{b-a}
$$
$$
f'(c)(b-a)=f(b)-f(a)
$$
$$
(-\sin c)(b-a)=\cos(b)-\cos(a)
$$
$$
(\sin c)(b-a)=\cos(a)-\cos(b)
$$
Taking the absolute value of both sides:
$$
|\sin c||b-a|= |\cos(a)-\cos(b)|
$$
Because $|\sin c |\leq 1,\,$ we can bound $|\sin c|$ by $1$
$$
1 \cdot|b-a| \geq |\cos(a)-\cos(b)|
$$
Thus $|b-a|\geq|\cos a-\cos b|$ for all real numbers $a$ and $b$
 A: Even faster you can use fundamental theorem of calculus and assuming for example $a≤b$
$$
\begin{align*}
|\cos(a)-\cos(b)| &= \left|\int_a^b \sin(x)\,\mathrm{d}x\right| ≤ \int_a^b \left|\sin(x)\right|\mathrm{d}x
\\
&≤ \int_a^b 1\,\mathrm{d}x = |b-a|
\end{align*}
$$ 
A: 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;
$$|\frac{\frac{b-a}{2}}{\sin(\frac{b-a}{2})}|\geq \sin(\frac{a+b}{2})$$
A: 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.
