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 $b$

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