Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 68: 68

Answer

a) $\frac{-\pi}{4}$ b) $\frac{119}{169}$

Work Step by Step

a) $\arcsin(\sin{\frac{5\pi}{4}})$ Using the unit circle: $\arcsin(-\frac{\sqrt{2}}{2})$ Sine of what gives us $-\frac{\sqrt{2}}{2}$? If we look at the unit circle, we see that it is $\frac{-\pi}{4}$. b) Draw a right triangle with a leg length of $5$ and a hypotenuse of $13$. The angle $\theta$ should be across the $5$. Using the Pythagorean Theorem ($a^2+b^2=c^2$), we find that the length of the missing leg is $12$. $\cos(2\sin^{-1}(\frac{5}{13}))$ According to our triangle, $\sin^{-1}(\frac{5}{13})$ is our $\theta$, so now, $\cos(2\sin^{-1}(\frac{5}{13}))=\cos(2\theta)$. Using double angle identities ($\cos(2\theta)=\cos^2(x)-sin^2(x)$): $\cos(2\theta)=\cos^2(x)-sin^2(x)$ According to our triangle, $\cos(x)=\frac{12}{13}$ and $\sin(x)=\frac{5}{13}$ $\cos^2(x)-sin^2(x)=(\frac{12}{13})^2-(\frac{5}{13})^2=\frac{144}{169}-\frac{25}{169}=\frac{119}{169}$
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