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: 23


$\sin \theta =\dfrac{y}{r}=\dfrac{-4}{5} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-3}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{4}{3}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-5}{4} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{3}{4}$

Work Step by Step

Here, $ x= -3; r=5$ $ r=\sqrt {x^2+y^2}$ This gives: $ y=-\sqrt {(5)^2-(-3)^2}=-4$; Because $\theta $ lies in Quadrant-III. The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=\dfrac{-4}{5} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-3}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{4}{3}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-5}{4} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{3}{4}$
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