Given that \(\sin \theta = \frac{1}{6}\) and \( \frac{\pi}{2} < \theta < \pi\), we are tasked with finding \(\cos \theta\).
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First, since \(\theta\) is in the second quadrant (where sine is positive and cosine is negative), we can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \]
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Substitute \(\sin \theta\) into the equation: \[ \left(\frac{1}{6}\right)^2 + \cos^2 \theta = 1. \]
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Calculate \(\left(\frac{1}{6}\right)^2\): \[ \frac{1}{36} + \cos^2 \theta = 1. \]
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Rearranging the equation gives: \[ \cos^2 \theta = 1 - \frac{1}{36}. \]
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Find a common denominator: \[ 1 = \frac{36}{36} \Rightarrow \cos^2 \theta = \frac{36}{36} - \frac{1}{36} = \frac{35}{36}. \]
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Taking the square root: \[ \cos \theta = \pm \sqrt{\frac{35}{36}} = \pm \frac{\sqrt{35}}{6}. \]
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Since \(\theta\) is in the second quadrant, where cosine is negative, we choose the negative value: \[ \cos \theta = -\frac{\sqrt{35}}{6}. \]
Thus, the value of \(\cos \theta\) is: \[ \boxed{-\frac{\sqrt{35}}{6}}. \]