Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 20

Answer

\[ - \frac{{2\sqrt 3 }}{3}\]

Work Step by Step

\[\begin{gathered} \sec \left( {\frac{{7\pi }}{6}} \right) \hfill \\ {\text{Write }}\frac{{7\pi }}{6}{\text{ as }}\pi + \frac{\pi }{6} \hfill \\ \sec \left( {\frac{{7\pi }}{6}} \right) = \sec \left( {\pi + \frac{\pi }{6}} \right) \hfill \\ {\text{Use the identity }}\sec \theta = \frac{1}{{\cos \theta }} \hfill \\ \sec \left( {\pi + \frac{\pi }{6}} \right) = \frac{1}{{\cos \left( {\pi + \frac{\pi }{6}} \right)}} \hfill \\ {\text{Where cos}}\left( {A + B} \right) = \cos A\cos B - \sin A\sin B \hfill \\ \sec \left( {\frac{{7\pi }}{6}} \right) = \frac{1}{{\cos \left( \pi \right)\cos \left( {\frac{\pi }{6}} \right) - \sin \left( \pi \right)\sin \left( {\frac{\pi }{6}} \right)}} \hfill \\ \sec \left( {\frac{{7\pi }}{6}} \right) = \frac{1}{{\left( { - 1} \right){\frac{{\sqrt 3 }}{2}} }} \hfill \\ \sec \left( {\frac{{7\pi }}{6}} \right) = - \frac{2}{{\sqrt 3 }} \hfill \\ \sec \left( {\frac{{7\pi }}{6}} \right) = - \frac{{2\sqrt 3 }}{3} \hfill \\ \end{gathered} \]
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