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: 48

Answer

$\sin{\theta} =-\dfrac{3}{5}$ $\tan{\theta}=-\dfrac{3}{4}$ $\csc{\theta} =- \dfrac{5}{3}$ $\sec{\theta} =\dfrac{5}{4}$ $\cot{\theta} =-\dfrac{4}{3}$

Work Step by Step

Since $270^{\circ} < \theta <360^{\circ}$, then $\theta \in QIV$ and $\sin{\theta} <0$. Thus, $\sin{\theta} = -\sqrt{1-\cos^2 {\theta}}$ $\sin{\theta} = -\sqrt{1-\left(\dfrac{4}{5} \right)^2} = -\dfrac{3}{5}$ $\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\tan{\theta}= \dfrac{-\dfrac{3}{5}}{\dfrac{4}{5}} = -\dfrac{3}{4}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\csc{\theta} = \dfrac{1}{-\dfrac{3}{5}}=- \dfrac{5}{3}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $\sec{\theta} = \dfrac{1}{\dfrac{4}{5}} = \dfrac{5}{4}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{-\dfrac{3}{4}} = -\dfrac{4}{3}$
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