Prove that the function given by $f(x) = \cos x$ is neither increasing nor decreasing in $(0, 2\pi)$.

(N/A) Given the function $f(x) = \cos x$.
Step $1$: Find the derivative of the function.
$f^{\prime}(x) = \frac{d}{dx}(\cos x) = -\sin x$.
Step $2$: Analyze the sign of $f^{\prime}(x)$ in the interval $(0, 2\pi)$.
$(a)$ For $x \in (0, \pi)$,$\sin x > 0$,which implies $f^{\prime}(x) = -\sin x < 0$. Thus,the function $f$ is strictly decreasing on $(0, \pi)$.
$(b)$ For $x \in (\pi, 2\pi)$,$\sin x < 0$,which implies $f^{\prime}(x) = -\sin x > 0$. Thus,the function $f$ is strictly increasing on $(\pi, 2\pi)$.
Step $3$: Conclusion.
Since the function is decreasing in the interval $(0, \pi)$ and increasing in the interval $(\pi, 2\pi)$,it is neither increasing nor decreasing on the entire interval $(0, 2\pi)$.