Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.3 - Solving Right Triangles - 2.3 Problem Set - Page 83: 76

Answer

$\sin\theta=-\frac{1}{2}$ $\cos\theta=-\frac{\sqrt 3}{2}$ $\sec\theta=-\frac{2\sqrt 3}{3}$ $\tan\theta=\frac{\sqrt 3}{3}$ $\cot\theta=\sqrt 3$

Work Step by Step

$\csc\theta=-2$ $\sin\theta=\frac{1}{\csc\theta}=\frac{1}{-2}=-\frac{1}{2}$ $\sin^{2}\theta+\cos^{2}\theta=1\implies\cos^{2}\theta=1-\sin^{2}\theta$ $=1-(-\frac{1}{2})^{2}=\frac{3}{4}$ $\cos\theta$ is negative in the third quadrant. Therefore, $\cos\theta=-\sqrt {\frac{3}{4}}=-\frac{\sqrt 3}{2}$ $\sec\theta=\frac{1}{\cos\theta}=-\frac{2}{\sqrt 3}=-\frac{2\sqrt 3}{3}$ $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-\frac{1}{2}}{-\frac{\sqrt 3}{2}}=\frac{1}{\sqrt 3}=\frac{\sqrt 3}{3}$ $\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{1}{\sqrt 3}}=\sqrt 3$
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