Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 33

Answer

$\sin \theta =\dfrac{y}{r}=\dfrac{-2 \sqrt 2}{3} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-1}{3} \\ \tan \theta =\dfrac{y}{x}=2 \sqrt 2$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-3 \sqrt 2}{4} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{3}{2\sqrt 2}=-\dfrac{3\sqrt 2}{4} \\ \cot \theta =\dfrac{x}{y}=\dfrac{\sqrt 2}{4}$

Work Step by Step

Here, $ x=-1; r=3$ and $ r=\sqrt {x^2+y^2}$ This gives: $ y=-\sqrt {(3)^2-(-1)^2}=-2 \sqrt 2$; Because $\theta $ lies in Quadrant-III. The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=\dfrac{-2 \sqrt 2}{3} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-1}{3} \\ \tan \theta =\dfrac{y}{x}=2 \sqrt 2$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-3 \sqrt 2}{4} \\ \sec \theta =\dfrac{r}{x}=-\dfrac{3}{2\sqrt 2}=-\dfrac{3\sqrt 2}{4} \\ \cot \theta =\dfrac{x}{y}=\dfrac{\sqrt 2}{4}$
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