Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 30: 19


$\sin \theta= \frac{12}{13}$, $\tan \theta=\frac{12}{5}$.

Work Step by Step

$\cos^{2}\theta+\sin^{2}\theta=1$ $\implies \sin^{2}\theta= 1-\cos^{2}\theta=1-\frac{25}{169}=\frac{144}{169}$ $\implies \sin \theta= \frac{12}{13}$ ($\sin \theta$ is positive as $0\leq\theta\lt\frac{\pi}{2}$) $\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$.
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