Prove that the function given by $f(x) = \cos x$ is increasing in $(\pi, 2\pi)$.

(N/A) To determine the intervals where the function $f(x) = \cos x$ is increasing or decreasing,we first find its derivative.
$f'(x) = \frac{d}{dx}(\cos x) = -\sin x$.
For the function to be increasing,we require $f'(x) > 0$.
This implies $-\sin x > 0$,which simplifies to $\sin x < 0$.
In the interval $(\pi, 2\pi)$,the sine function is negative (it lies in the third and fourth quadrants).
Since $\sin x < 0$ for all $x \in (\pi, 2\pi)$,it follows that $f'(x) = -\sin x > 0$ for all $x \in (\pi, 2\pi)$.
Therefore,the function $f(x) = \cos x$ is strictly increasing in the interval $(\pi, 2\pi)$.