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: 23

Answer

$\dfrac{\sqrt{x^2+2x-8}}{x+1}$

Work Step by Step

$\theta = \sec^{-1} \dfrac{x+1}{3}$ $\sec{\theta} = \dfrac{x+1}{3}$ $\cos{\theta} = \dfrac{3}{x+1}$ $\sin^2{\theta} = 1-\cos^2{\theta} = 1-\dfrac{9}{(x+1)^2}$ $\sin^2{\theta} = \dfrac{(x+1)^2-9}{(x+1)^2} = \dfrac{x^2+2x-8}{(x+1)^2}$ $\sin{\theta} = \dfrac{\sqrt{x^2+2x-8}}{x+1}$
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