Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - Review Questions - Page 501: 48

Answer

$cos~\theta=-\frac{\sqrt {55}}{8}$ $tan~\theta=-\frac{3\sqrt {55}}{55}$ $cot~\theta=-\frac{\sqrt {55}}{3}$ $sec~\theta=-\frac{8\sqrt {55}}{55}$ $csc~\theta=\frac{8}{3}$

Work Step by Step

$sin^\theta+cos^2\theta=1$ $(\frac{3}{8})^2+cos^2\theta=1$ $cos^2\theta=1-\frac{9}{64}=\frac{55}{64}$ $cos~\theta=-\frac{\sqrt {55}}{8}$ $tan~\theta=\frac{sin~\theta}{cos~\theta}=\frac{\frac{3}{8}}{-\frac{\sqrt {55}}{8}}=-\frac{3\sqrt {55}}{55}$ $cot~\theta=\frac{cos~\theta}{sin~\theta}=\frac{-\frac{\sqrt {55}}{8}}{\frac{3}{8}}=-\frac{\sqrt {55}}{3}$ $sec~\theta=\frac{1}{cos~\theta}=-\frac{8}{\sqrt {55}}=-\frac{8\sqrt {55}}{55}$ $csc~\theta=\frac{1}{sin~\theta}=\frac{8}{3}$
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