Precalculus (6th Edition) Blitzer

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

Chapter 4 - Test - Page 647: 4

Answer

$\sin \theta =\dfrac{y}{r}=\dfrac{5\sqrt {29}}{29} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-2\sqrt {29}}{29} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-5}{2}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{\sqrt {29}}{5} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-\sqrt {29}}{2} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-2}{5}$

Work Step by Step

Here, $ x=-2; y=5$ $ r=\sqrt {(-2)^2+(5)^2}=\sqrt {29}$ The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=\dfrac{5\sqrt {29}}{29} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-2\sqrt {29}}{29} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-5}{2}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{\sqrt {29}}{5} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-\sqrt {29}}{2} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-2}{5}$
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