Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - 5.2 Trigonometric Functions - 5.2 Exercises - Page 520: 146

Answer

$\cos\theta=\frac{\sqrt {5}}{8}$ $\sec\theta=\frac{8\sqrt {5}}{5}$ $ \sin\theta=-\frac{\sqrt{ 59}}{8}$ $\csc\theta=-\frac{8\sqrt {59}}{59}$ $\tan\theta=-\frac{\sqrt {295}}{5}$ $\cot\theta=-\frac{\sqrt {295}}{59}$

Work Step by Step

$\cos\theta=\frac{\sqrt {5}}{8}$ $\sec\theta=\frac{1}{\cos\theta}=\frac{8}{\sqrt 5}=\frac{8\sqrt {5}}{5}$ $\sin^{2}\theta=1-\cos^{2}\theta=1-(\frac{\sqrt {5}}{8})^{2}=\frac{59}{64}$ $\implies \sin\theta=\pm\sqrt {\frac{59}{64}}=-\frac{\sqrt{ 59}}{8}$ ($\sin\theta\lt0$ as $\tan\theta\lt0$ and $\cos\theta\gt0$) $\csc\theta=\frac{1}{\sin\theta}=-\frac{8}{\sqrt {59}}=-\frac{8\sqrt {59}}{59}$ $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-\frac{\sqrt {59}}{8}}{\frac{\sqrt 5}{8}}=-\frac{\sqrt {59}}{\sqrt {5}}=-\frac{\sqrt {295}}{5}$ $\cot\theta=\frac{1}{\tan\theta}=-\frac{\sqrt 5}{\sqrt {59}}=-\frac{\sqrt {295}}{59}$
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