Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 458: 67



Work Step by Step

$g(f(\frac{12}{13}))=\cos(\sin^{-1}{\frac{12}{13}})$. Let $\theta=\sin^{-1}{\frac{12}{13}}$, hence we know that because of the definition of the sine function: $\sin{\theta}=\frac{12}{13}=\frac{\text{opposite}}{\text{hypotenuse}}$ The Pythagorean Theorem says that for a right triangle (if $z$ is the hypotenuse and $x,y$ are the other sides): $x^2+y^2=z^2$. Hence here: $\text{adjacent}=\sqrt{13^2-12^2}=\sqrt{25}=5.$ $\cos{\theta}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{5}{13}$.
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