Show that the function given by $f(x) = \sin x$ is decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

(N/A) The given function is $f(x) = \sin x$.
To determine the intervals of increase or decrease,we find the derivative of the function:
$f'(x) = \frac{d}{dx}(\sin x) = \cos x$.
We are considering the interval $x \in \left(\frac{\pi}{2}, \pi\right)$,which lies in the second quadrant.
In the second quadrant,the cosine function is negative,i.e.,$\cos x < 0$ for all $x \in \left(\frac{\pi}{2}, \pi\right)$.
Since $f'(x) = \cos x < 0$ for all $x \in \left(\frac{\pi}{2}, \pi\right)$,the function $f(x) = \sin x$ is strictly decreasing in the interval $\left(\frac{\pi}{2}, \pi\right)$.