Precalculus (6th Edition) Blitzer

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

Chapter 4 - Review Exercises - Page 645: 49


$\sin \theta =\dfrac{y}{r}=-\dfrac{\sqrt{21}}{5}$ $\tan \theta =\dfrac{y}{x}=-\dfrac{\sqrt{21}}{2}$ $\csc \theta =\dfrac{1}{\sin \theta}=-\dfrac{5}{\sqrt{21}}=-\dfrac{5\sqrt{21}}{21}$ $\sec \theta =\dfrac{1}{\cos \theta}=\dfrac{5}{2}$ $\cot \theta = \dfrac{1}{\tan \theta}=-\dfrac{2}{\sqrt{21}}=-\dfrac{2\sqrt{21}}{21}$

Work Step by Step

Here, $\cos \theta =\dfrac{2}{5}=\dfrac{x}{r}$ We solve for $y$: $x^2+y^2=r^2$ $2^2+y^2=5^2$ $y=-\sqrt{21}$ $y$ is negative because we are in quadrant IV (cosine is positive and sine is negative). The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=-\dfrac{\sqrt{21}}{5}$ $\tan \theta =\dfrac{y}{x}=-\dfrac{\sqrt{21}}{2}$ $\csc \theta =\dfrac{1}{\sin \theta}=-\dfrac{5}{\sqrt{21}}=-\dfrac{5\sqrt{21}}{21}$ $\sec \theta =\dfrac{1}{\cos \theta}=\dfrac{5}{2}$ $\cot \theta = \dfrac{1}{\tan \theta}=-\dfrac{2}{\sqrt{21}}=-\dfrac{2\sqrt{21}}{21}$
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