Trigonometry 7th Edition

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

Chapter 5 - Section 5.5 - Additional Identities - 5.5 Problem Set - Page 311: 13

Answer

$-\dfrac{5}{13}$

Work Step by Step

$\alpha = \tan^{-1} {\dfrac{3}{2}} $ $\tan{\alpha} = \dfrac{3}{2}$ $\sec{\alpha} =\sqrt{1+\tan^2{\alpha}} = \dfrac{\sqrt{13}}{2}$ $\cos{\alpha} = \dfrac{1}{\sec{\alpha}} = \dfrac{2}{\sqrt{13}}$ $\cos{2\alpha} =2\cos^2{\alpha} -1 = -\dfrac{5}{13}$
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