MVT to prove an inequality

Question

Use the Mean Value Theorem to prove that $|\cos a − \cos b| \leq |a − b|\,\forall a, b \in\mathbb R$.

Please advise if I am heading the right direction because I got stuck after doing this: $$\begin{align} f(x) &= \cos x \\ f'(x) &= -\sin x \\ c &= |a-b| \\ -\sin|a-b| &= \frac{|\cos a − \cos b|}{|a-b|} \end{align}$$

Answer 1

Note that by MVT we have

$$\frac{|\cos a − \cos b|}{|a-b|}= |\sin c|\le 1$$

for some $c\in(a,b)$.