Step 1 :Since \(\cot \theta = -\frac{2}{7}\), we know that \(\cos \theta = -\frac{2}{\sqrt{2^2 + 7^2}} = -\frac{2}{\sqrt{53}}\) and \(\sin \theta = \frac{7}{\sqrt{2^2 + 7^2}} = \frac{7}{\sqrt{53}}\).
Step 2 :However, since \(\theta\) is in the fourth quadrant where sine is negative, we must take the negative of the above value for \(\sin \theta\).
Step 3 :So, \(\sin \theta = -\frac{7}{\sqrt{53}}\).
Step 4 :We can rationalize the denominator to get \(\sin \theta = -\frac{7\sqrt{53}}{53}\).
Step 5 :So, the final answer is \(\boxed{-\frac{7\sqrt{53}}{53}}\).