Step-by-Step Solution

1. Step 1: Find the derivative of the function \( f(x) \).

   Given \( f(x) = \frac{16\sin x}{4 + \cos x} - x \), we need to find \( f'(x) \).

   Using the quotient rule for differentiation, \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), we differentiate \( \frac{16\sin x}{4 + \cos x} \):

   \( \frac{d}{dx} \left( \frac{16\sin x}{4 + \cos x} \right) = \frac{(4 + \cos x)(16\cos x) - (16\sin x)(-\sin x)}{(4 + \cos x)^2} \)

   \( = \frac{64\cos x + 16\cos^2 x + 16\sin^2 x}{(4 + \cos x)^2} \)

   \( = \frac{64\cos x + 16(\cos^2 x + \sin^2 x)}{(4 + \cos x)^2} \)

   \( = \frac{64\cos x + 16}{(4 + \cos x)^2} \)

   \( = \frac{16(4 + \cos x)}{(4 + \cos x)^2} \)

   \( = \frac{16}{4 + \cos x} \)

   Now, differentiate \( -x \):

   \( \frac{d}{dx} (-x) = -1 \)

   So, \( f'(x) = \frac{16}{4 + \cos x} - 1 \)

2. Step 2: Simplify \( f'(x) \).

   \( f'(x) = \frac{16 - (4 + \cos x)}{4 + \cos x} \)

   \( = \frac{12 - \cos x}{4 + \cos x} \)

3. Step 3: Analyze the sign of \( f'(x) \) in the interval \( (\frac{\pi}{2}, \pi) \).

   In the interval \( (\frac{\pi}{2}, \pi) \), \( \cos x \) is negative and \( -1 \leq \cos x < 0 \).

   Therefore, \( 12 - \cos x > 0 \) because \( 12 - \cos x \geq 12 - (-1) = 13 > 0 \).

   Also, \( 4 + \cos x > 0 \) because \( 4 + \cos x \geq 4 + (-1) = 3 > 0 \).

   However, we need to show that \( f'(x) < 0 \) for the function to be strictly decreasing.

   Let's re-examine \( f'(x) = \frac{12 - \cos x}{4 + \cos x} - 1 \). There was an error in the initial calculation of f'(x). The correct derivative is:

   \(f'(x) = \frac{16}{4 + \cos x} - 1 = \frac{16 - (4 + \cos x)}{4 + \cos x} = \frac{12 - \cos x}{4 + \cos x}\)

   To determine if the function is strictly decreasing, we need to check if \(f'(x) < 0\) in the interval \((\frac{\pi}{2}, \pi)\).

   Since \(-1 \le \cos x < 0\) in the interval \((\frac{\pi}{2}, \pi)\), we have:

   \(12 - \cos x > 0\) (because \(12 - \cos x \ge 12 - (-1) = 13 > 0\))

   \(4 + \cos x > 0\) (because \(4 + \cos x \ge 4 - 1 = 3 > 0\))

   Thus, \(f'(x) = \frac{12 - \cos x}{4 + \cos x} > 0\) in the interval \((\frac{\pi}{2}, \pi)\).

   There must be an error in the problem statement. Let's reconsider the original function:

   \(f(x) = \frac{16\sin x}{4 + \cos x} - x\)

   \(f'(x) = \frac{16(4 + \cos x)\cos x - 16\sin x(-\sin x)}{(4 + \cos x)^2} - 1\)

   \(f'(x) = \frac{64\cos x + 16\cos^2 x + 16\sin^2 x}{(4 + \cos x)^2} - 1\)

   \(f'(x) = \frac{64\cos x + 16}{(4 + \cos x)^2} - 1\)

   \(f'(x) = \frac{16(4\cos x + 1)}{(4 + \cos x)^2} - 1\)

   \(f'(x) = \frac{16(4\cos x + 1) - (4 + \cos x)^2}{(4 + \cos x)^2}\)

   \(f'(x) = \frac{64\cos x + 16 - (16 + 8\cos x + \cos^2 x)}{(4 + \cos x)^2}\)

   \(f'(x) = \frac{56\cos x - \cos^2 x}{(4 + \cos x)^2}\)

   \(f'(x) = \frac{\cos x(56 - \cos x)}{(4 + \cos x)^2}\)

   In the interval \((\frac{\pi}{2}, \pi)\), \(\cos x < 0\), and \(56 - \cos x > 0\), and \((4 + \cos x)^2 > 0\).

   Therefore, \(f'(x) < 0\) in the interval \((\frac{\pi}{2}, \pi)\).

4. Step 4: Conclusion

   Since \( f'(x) < 0 \) in the interval \( (\frac{\pi}{2}, \pi) \), the function \( f(x) \) is strictly decreasing in this interval.