Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.4 - Introduction to Identities - 1.4 Problem Set - Page 41: 51

Answer

$\sin{\theta} =-\dfrac{3\sqrt{13}}{13}$ $\cos{\theta} =\dfrac{2\sqrt{13}}{13}$ $\tan{\theta} =-\dfrac{3}{2}$ $\csc{\theta} =-\dfrac{\sqrt{13}}{13}$ $\sec{\theta} =\dfrac{\sqrt{13}}{2}$ $\cot{\theta} =-\dfrac{2}{3} $

Work Step by Step

$\cos{\theta} = \dfrac{2\sqrt{13}}{13}$ $\because \theta \in QIV \hspace{20pt} \therefore \sin{\theta}$ is negative. $\sin{\theta} = - \sqrt{1-\cos^2{\theta}} = - \sqrt{1-\left(\dfrac{2\sqrt{13}}{13}\right)^2} = -\dfrac{3\sqrt{13}}{13}$ $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} = -\dfrac{3}{2}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}} = -\dfrac{\sqrt{13}}{13}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{\sqrt{13}}{2}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}} = -\dfrac{2}{3} $
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