Show that $f(x)=\cos(x)$ is Lipschitz continuous function.

By the Mean Value Theorem, there is a $c \in (x,y)$ such that $$\textrm{cos}(x)-\textrm{cos}(y)=(-\textrm{sin}(c))(x-y).$$

So we have that $$|\textrm{cos}(x)-\textrm{cos}(y)|=|(-\textrm{sin}(c))(x-y)| \leq 1 \cdot |x-y|.$$

Thus $f(x)=\textrm{cos}(x)$ is Lipschitz.

For all $x$, $|\sin x|\leq |x|$. Using Sum to Product formula:

$|\cos(x)-\cos(y)|=\left|-2\sin\left(\dfrac{x+y}{2}\right)\;\sin\left(\dfrac{x-y}{2}\right)\right|\leq2\left|\sin\left(\dfrac{x-y}{2}\right)\right|\leq2\left|\dfrac{x-y}{2}\right|=|x-y|$

A function with bounded derivative is always Lipschitz continuous, for we have

$\vert f(x) - f(y) \vert = \left \vert \displaystyle \int_x^y f'(s) \; ds \right \vert \le \left \vert \displaystyle \int_x^y \vert f'(s) \vert \; ds \right \vert; \tag 1$

if now

$\vert f'(s) \vert \le M, \tag 2$

then

$\vert f(x) - f(y) \vert \le \left \vert \displaystyle \int_x^y \vert f'(s) \vert \; ds \right \vert \le \left \vert \displaystyle \int_x^y M \; ds \right \vert \le M \vert y - x \vert, \tag 3$

which shows that $f(x)$ is Lipschitz continuous with Lipschitz constant at most $M$.

Since

$\forall x \in \Bbb R, \; \vert (\cos x)' \vert = \vert \sin x \vert \le 1, \tag 4$

it follows that $\cos x$ is Lipschitz continuous with Lipschitz constant at most $1$.