Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.5 - Multiple-Angle and Product-to-Sum Formulas - 7.5 Exercises - Page 548: 39


$sin~\frac{\pi}{8}=\frac{\sqrt {2-\sqrt 2}}{2}$ $cos~\frac{\pi}{8}=\frac{\sqrt {2+\sqrt 2}}{2}$ $tan~\frac{\pi}{8}=\sqrt 2-1$

Work Step by Step

$\frac{\pi}{8}$ lies in the first quadrant. $sin~\frac{\pi}{8}=sin\frac{\frac{\pi}{4}}{2}=+\sqrt {\frac{1-cos~\frac{\pi}{4}}{2}}=\sqrt {\frac{1-\frac{\sqrt 2}{2}}{2}}=\sqrt {\frac{2-\sqrt 2}{4}}=\frac{\sqrt {2-\sqrt 2}}{2}$ $cos~\frac{\pi}{8}=cos\frac{\frac{\pi}{4}}{2}=+\sqrt {\frac{1+cos~\frac{\pi}{4}}{2}}=\sqrt {\frac{1+\frac{\sqrt 2}{2}}{2}}=\sqrt {\frac{2+\sqrt 2}{4}}=\frac{\sqrt {2+\sqrt 2}}{2}$ $tan~\frac{\pi}{8}=tan\frac{\frac{\pi}{4}}{2}=\frac{1-cos~\frac{\pi}{4}}{sin~\frac{\pi}{4}}=\frac{1-\frac{\sqrt 2}{2}}{\frac{\sqrt 2}{2}}=\frac{2-\sqrt 2}{\sqrt 2}=\sqrt 2-1$
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