Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 7 - Review - Exercises - Page 579: 65

Answer

$-\frac{12\sqrt {10}}{31}$

Work Step by Step

Step 1. Let $u=cos^{-1}\frac{3}{7}$, we have $cos(u)=\frac{3}{7}$ and $sin(u)=\sqrt {1-cos^2u}=\frac{2\sqrt {10}}{7}, tan(u)=sin(u)/cos(u)=\frac{2\sqrt {10}}{3}$ Step 2. Use the Double-Angle Formula $tan(2cos^{-1}\frac{3}{7})=tan(2u)=\frac{2tan(u)}{1-tan^2u}=\frac{4\sqrt {10}}{3(1-40/9)}=-\frac{12\sqrt {10}}{31}$
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