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: 40


$sin~\frac{7\pi}{12}=\frac{\sqrt {2+\sqrt 3}}{2}$ $cos~\frac{7\pi}{12}=-\frac{\sqrt {2-\sqrt 3}}{2}$ $tan~\frac{7\pi}{12}=-(2+\sqrt 3)$

Work Step by Step

$\frac{7\pi}{12}$ lies in the second quadrant. $sin~\frac{7\pi}{12}=sin\frac{\frac{7\pi}{6}}{2}=+\sqrt {\frac{1-cos~\frac{7\pi}{6}}{2}}=\sqrt {\frac{1-(-\frac{\sqrt 3}{2})}{2}}=\sqrt {\frac{2+\sqrt 3}{4}}=\frac{\sqrt {2+\sqrt 3}}{2}$ $cos~\frac{7\pi}{12}=cos\frac{\frac{7\pi}{6}}{2}=-\sqrt {\frac{1+cos~\frac{7\pi}{6}}{2}}=-\sqrt {\frac{1-\frac{\sqrt 3}{2}}{2}}=-\sqrt {\frac{2-\sqrt 3}{4}}=-\frac{\sqrt {2-\sqrt 3}}{2}$ $tan~\frac{7\pi}{12}=tan\frac{\frac{7\pi}{6}}{2}=\frac{1-cos~\frac{7\pi}{6}}{sin~\frac{7\pi}{6}}=\frac{1-(-\frac{\sqrt 3}{2})}{-\frac{1}{2}}=-2-\sqrt 3=-(2+\sqrt 3)$
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