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: 59


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

Work Step by Step

We know that the reference angle for $\theta =\frac{7\pi }{4}$ is: $\begin{align} & \theta '=2\pi -\frac{7\pi }{4} \\ & =\frac{8\pi -7\pi }{4} \\ & =\frac{\pi }{4} \end{align}$ Now, the function value for the reference angle is: $\sec \frac{\pi }{4}=\sqrt{2}$ Thus, we see that the angle $\frac{7\pi }{4}$ lies in quadrant IV, so the sec function is positive: $\begin{align} & \sec \frac{7\pi }{4}=\sec \frac{\pi }{4} \\ & =\sqrt{2} \end{align}$
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