Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.7 Apply Double-Angle and Half-Angle Formulas - 14.7 Exercises - Skill Practice - Page 959: 6


$\dfrac{\sqrt{2-\sqrt 3}}{2}$

Work Step by Step

Double -angle Theorem can be defined as: $\cos \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1+ \cos \theta}{2}}$ $\cos (-75 ^{\circ})=\cos \dfrac{(-150^{\circ})}{2}$ Thus, $\sqrt {\dfrac{1+ \cos (-150^{\circ})}{2}}= \sqrt {\dfrac{1+ \cos (150^{\circ})}{2}}$ and $\sqrt {\dfrac{1- (\sqrt 3/2)}{2}}=\sqrt {\dfrac{2-\sqrt 3}{4}}$ Hence, $\cos(-75 ^{\circ})=\dfrac{\sqrt{2-\sqrt 3}}{2}$
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