## Precalculus (6th Edition) Blitzer

The exact value of the provided expression is $\frac{\sqrt{3}}{2}$.
We know that the reference angle for $\theta =\frac{11\pi }{6}$ is: \begin{align} & \theta '=2\pi -\frac{11\pi }{6} \\ & =\frac{12\pi -11\pi }{6} \\ & =\frac{\pi }{6} \end{align} Now, the function value for the reference angle is: $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ Thus, we see that the angle $\frac{11\pi }{6}$ lies in quadrant IV, so the cos function is positive: \begin{align} & \cos \frac{11\pi }{6}=\cos \frac{\pi }{6} \\ & =\frac{\sqrt{3}}{2} \end{align} .