Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 43


$\cos{\theta} =-\dfrac{5}{13}$ $\tan{\theta}=-\dfrac{12}{5}$ $\csc{\theta} =\dfrac{13}{12}$ $\sec{\theta} =-\dfrac{13}{5}$ $\cot{\theta} =-\dfrac{5}{12}$

Work Step by Step

Since $\theta \in QII$, then $\cos{\theta} <0$. $\cos{\theta} = -\sqrt{1-\sin^2 {\theta}}$ $\cos{\theta} = -\sqrt{1-\left(\dfrac{12}{13} \right)^2} = -\dfrac{5}{13}$ $\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\tan{\theta}= \dfrac{\dfrac{12}{13}}{-\dfrac{5}{13}} = -\dfrac{12}{5}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\csc{\theta} = \dfrac{1}{\dfrac{12}{13}}= \dfrac{13}{12}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $\sec{\theta} = \dfrac{1}{\dfrac{5}{13}} = -\dfrac{13}{5}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{\dfrac{12}{5}} = -\dfrac{5}{12}$
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