To determine the value of \(\cos \theta\) given \(\sin \theta = \frac{16}{17}\) and that \(\frac{\pi}{2} < \theta < \pi\), we can use the Pythagorean Identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
We start by substituting the value for \(\sin \theta\):
\[ \left(\frac{16}{17}\right)^2 + \cos^2 \theta = 1 \]
Calculating \(\left(\frac{16}{17}\right)^2\):
\[ \frac{256}{289} \]
Now substitute this into the Pythagorean Identity:
\[ \frac{256}{289} + \cos^2 \theta = 1 \]
Next, we need to express \(1\) with the same denominator:
\[ 1 = \frac{289}{289} \]
So we have:
\[ \frac{256}{289} + \cos^2 \theta = \frac{289}{289} \]
Now, solving for \(\cos^2 \theta\):
\[ \cos^2 \theta = \frac{289}{289} - \frac{256}{289} = \frac{33}{289} \]
Now, taking the square root:
\[ \cos \theta = \pm \sqrt{\frac{33}{289}} = \pm \frac{\sqrt{33}}{17} \]
Given the range \(\frac{\pi}{2} < \theta < \pi\), \(\cos \theta\) must be negative because cosine is negative in the second quadrant. Therefore:
\[ \cos \theta = -\frac{\sqrt{33}}{17} \]
The appropriate response, among the provided options, is:
\[ -\frac{\sqrt{33}}{17} \]