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: 20


$\cos \theta = \frac{4}{5}$ and $\tan \theta= \frac{3}{4}$

Work Step by Step

$\cos^{2}\theta+\sin^{2}\theta=1$ $\implies \cos^{2}\theta= 1-\sin^{2}\theta=1-\frac{9}{25}=\frac{16}{25}$ $\implies \cos \theta= \frac{4}{5}$ ($\cos \theta$ is positive as $0\leq\theta\lt\frac{\pi}{2}$) $\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$.
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