Show that the function $\displaystyle f(x)=\frac{16 \sin x}{4+\cos x}-x$, is strictly decreasing in $\displaystyle \left(\frac{\pi}{2}, \pi\right)$.

$$\begin{aligned} & f^{\prime}(x)=\frac{16[4+\cos x] \cos x+16 \sin ^{2} x}{(4+\cos x)^{2}}-1 \\ & =\frac{\cos x(56-\cos x)}{(4+\cos x)^{2}} \\ & \text { in }\left(\frac{\pi}{2}, \pi\right), \cos x<0 \Rightarrow f^{\prime}(x)<0 \\ & \therefore f(x) \text { in strictly decreasing in }\left(\frac{\pi}{2}, \pi\right) \end{aligned} $$