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: 25


$\cos\theta= - \frac{\sqrt {21}}{5}$ and $\tan\theta=-\frac{2\sqrt {21}}{21}$

Work Step by Step

$\sin\theta=0.4=\frac{2}{5}$ $\cos^{2}\theta=1-\sin^{2}\theta=1-\frac{4}{25}=\frac{21}{25}$ $\cos\theta=± \frac{\sqrt {21}}{5}$ Since $\theta$ lies in second quadrant, $\cos\theta$ is negative. Therefore $\cos\theta= - \frac{\sqrt {21}}{5}$ $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{2}{5}}{ - \frac{\sqrt {21}}{5}}=-\frac{2}{\sqrt {21}}=-\frac{2\sqrt {21}}{21}$
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