To find the value of \(\cos \theta\) given that \(\sin \theta = \frac{1}{6}\) and \(\frac{\pi}{2} < \theta < \pi\), we can use the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
First, we can find \(\sin^2 \theta\):
\[ \sin^2 \theta = \left(\frac{1}{6}\right)^2 = \frac{1}{36} \]
Next, substitute \(\sin^2 \theta\) into the Pythagorean identity:
\[ \frac{1}{36} + \cos^2 \theta = 1 \]
Now, subtract \(\frac{1}{36}\) from both sides:
\[ \cos^2 \theta = 1 - \frac{1}{36} \]
To combine the fractions, we find a common denominator (which is 36):
\[ 1 = \frac{36}{36} \] \[ \cos^2 \theta = \frac{36}{36} - \frac{1}{36} = \frac{35}{36} \]
Now, take the square root of both sides:
\[ \cos \theta = \pm \sqrt{\frac{35}{36}} = \pm \frac{\sqrt{35}}{6} \]
Since \(\frac{\pi}{2} < \theta < \pi\), \(\theta\) is in the second quadrant, where the cosine value is negative. Therefore:
\[ \cos \theta = -\frac{\sqrt{35}}{6} \]
Thus, the final answer is:
\[ \cos \theta = -\frac{\sqrt{35}}{6} \]