Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 5 - Section 5.3 - Double-Angle Formulas - 5.3 Problem Set - Page 298: 68

Answer

$2\sin^{-1} \dfrac{x}{2} - \dfrac{2x \sqrt{4-x^2}}{4-2x^2}$

Work Step by Step

$\sin{\theta} =\dfrac{x}{2}$ $\cos{\theta} =\sqrt{1-\sin^2{\theta}} = \dfrac{1}{2} \sqrt{4-x^2}$ $\tan{\theta} =\dfrac{\sin{\theta}}{\cos{\theta}} = \dfrac{x}{\sqrt{4-x^2}}$ $\tan{2\theta} =\dfrac{2\theta}{1-\tan^2{\theta}} = \dfrac{2x \sqrt{4-x^2}}{4-2x^2}$ $2\theta - \tan{2\theta} = 2\sin^{-1} \dfrac{x}{2} - \dfrac{2x \sqrt{4-x^2}}{4-2x^2}$
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