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: 24


$\sin \theta =\dfrac{y}{r}=\dfrac{-12}{13} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-5}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{12}{5}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-13}{12} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-13}{5} \\ \cot \theta =\dfrac{x}{y}=\dfrac{5}{12}$

Work Step by Step

Here, $ y=-12; r=13$ $ r=\sqrt {x^2+y^2}$ This gives: $ x=-\sqrt {(13)^2-(-12)^2}=-5$; Because $\theta $ lies in Quadrant-III. The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=\dfrac{-12}{13} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-5}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{12}{5}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-13}{12} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-13}{5} \\ \cot \theta =\dfrac{x}{y}=\dfrac{5}{12}$
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