Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Mid-Chapter Check Point - Page 577: 12


$$\sin \theta =-\frac{\sqrt{40}}{7} \\ \tan \theta = -\frac{\sqrt{40}}{3} \\ \sec \theta =\frac{7}{3} \\ \cot \theta = - \frac{3\sqrt{40}}{40} \\ \csc \theta = - \frac{7\sqrt{40}}{40}$$

Work Step by Step

By using the assumption that $\cos \theta =\frac{3}{7}, \quad \sin \theta \lt 0$ and applying appropriate trigonometric identities, we can find the value of the remaining trigonometric functions of $\theta$ as follows.$$\sin \theta = \sqrt{1- \cos^2 \theta }=\sqrt{1-(\frac{3}{7})^2}=-\frac{\sqrt{40}}{7} \\ \tan \theta = \frac{\sin \theta }{\cos \theta }= \frac{-\frac{\sqrt{40}}{7}}{\frac{3}{7}} =-\frac{\sqrt{40}}{3} \\ \sec \theta = \frac{1}{\cos \theta }=\frac{1}{\frac{3}{7}}=\frac{7}{3} \\ \cot \theta =\frac{1}{\tan \theta }= \frac{1}{-\frac{\sqrt{40}}{3}}=-\frac{3}{\sqrt{40}}=-\frac{3\sqrt{40}}{40} \\ \csc \theta = \frac{1}{\sin \theta }= \frac{1}{-\frac{\sqrt{40}}{7}}=-\frac{7}{\sqrt{40}}=-\frac{7\sqrt{40}}{40}$$
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