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: 11


Option B

Work Step by Step

Double -angle Theorem can be defined as: $\tan \dfrac{\theta}{2}=\pm \sqrt {\dfrac{1- \cos \theta}{1+ \cos \theta}}$ $\tan (15^{\circ})=\tan \dfrac{(30^{\circ})}{2}$ or, $=\sqrt {\dfrac{1- \cos (30^{\circ})}{1+ \cos (30^{\circ})}}$ or, $=\sqrt {\dfrac{1- \cos (\sqrt 3/2)}{1+ \cos (\sqrt 3/2)}}$ or, $=\sqrt {\dfrac{2-\sqrt 3}{2+\sqrt 3}}$ or, $=\sqrt {\dfrac{2-\sqrt 3}{2+\sqrt 3}} \times \sqrt {\dfrac{2-\sqrt 3}{2-\sqrt 3}}$ or, $=2-\sqrt 3$ Hence, we can see that only option $B$ satisfies the equation.
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