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: 51


$\cos{\theta} =- \dfrac{\sqrt{5}}{3}$ $\tan{\theta}=-\dfrac{2 \sqrt{5}}{5}$ $\csc{\theta} =\dfrac{3 }{2}$ $\sec{\theta} =-\dfrac{3 \sqrt{5}}{5}$ $\cot{\theta} =-\dfrac{\sqrt{5}}{2}$

Work Step by Step

$\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} <0$ Since $\sin{\theta} >0$, then $\cos{\theta}$ must be negative. Thus, $\cos{\theta} = -\sqrt{1-\sin^2 {\theta}}$ $\cos{\theta} = -\sqrt{1-\left(\dfrac{2}{3} \right)^2}= - \dfrac{\sqrt{5}}{3}$ $\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\tan{\theta}= \dfrac{\dfrac{2}{3}}{- \dfrac{\sqrt{5}}{3}} = -\dfrac{2 \sqrt{5}}{5}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\csc{\theta} = \dfrac{1}{\dfrac{2}{3}}= \dfrac{3 }{2}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $\sec{\theta} = \dfrac{1}{-\dfrac{\sqrt{5}}{3}} = -\dfrac{3 \sqrt{5}}{5}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{-\dfrac{2 \sqrt{5}}{5}} = -\dfrac{\sqrt{5}}{2}$
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