Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 645: 60


The exact value of the provided expression is $\frac{\sqrt{3}}{2}$.

Work Step by Step

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: $\begin{align} & \cos \frac{11\pi }{6}=\cos \frac{\pi }{6} \\ & =\frac{\sqrt{3}}{2} \end{align}$ .
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