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 299: 78

Answer

$d$

Work Step by Step

$\dfrac{2\tan{x}}{1+\tan^2{x}} = \dfrac{2\tan{x}}{\sec^2{x}}= 2 \tan{x}\cos^2{x}$ $= 2 \,\dfrac{\sin{x}}{\cos{x}} \, \cos^2{x} = 2 \sin{x} \cos{x} = \sin{2x}$
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