Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 497: 27

Answer

$(2-\sqrt 2)\sqrt {2+\sqrt 2}$

Work Step by Step

Step 1. $cos(\frac{15\pi}{8})=\sqrt {\frac{1+cos(\frac{15\pi}{4})}{2}}=\sqrt {\frac{1+\frac{\sqrt 2}{2}}{2}}=\frac{\sqrt {2+\sqrt 2}}{2}$ Step 2. $sec(\frac{15\pi}{8})=\frac{1}{cos(\frac{15\pi}{8})}=\frac{2}{\sqrt {2+\sqrt 2}}=\frac{2\sqrt {2+\sqrt 2}}{2+\sqrt 2}=(2-\sqrt 2)\sqrt {2+\sqrt 2}$
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