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

Published by Pearson

# Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 44

#### Answer

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

#### Work Step by Step

Since $\theta \in QIV$, then $\sin{\theta} <0$. Thus, $\sin{\theta} = -\sqrt{1-\cos^2 {\theta}}$ $\sin{\theta} = -\sqrt{1-\left(\dfrac{3}{5} \right)^2} = -\sqrt{\dfrac{16}{25}}=-\dfrac{4}{5}$ $\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\tan{\theta}= \dfrac{-\dfrac{4}{5}}{\dfrac{3}{5}} = -\dfrac{4}{3}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\csc{\theta} = \dfrac{1}{-\dfrac{4}{5}}= -\dfrac{5}{4}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $\sec{\theta} = \dfrac{1}{\dfrac{3}{5}} = \dfrac{5}{3}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{-\dfrac{4}{3}} = -\dfrac{3}{4}$

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