Trigonometry 7th Edition

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

Chapter 1 - Test - Page 51: 20

Answer

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

Work Step by Step

Given $\sin\theta$ = $\frac{1}{2}$ From first Pythagorean identity- $\cos\theta$ = ± $\sqrt {1-\sin^{2}\theta}$ As $\theta$ terminates in QII, therefore $\cos\theta$ is negative- $\cos\theta$ = - $\sqrt {1-\sin^{2}\theta}$ = - $\sqrt {1-(\frac{1}{2})^{2}}$ = - $\sqrt {1-\frac{1}{4}}$ = - $\sqrt {\frac{4-1}{4}}$ = - $\sqrt {\frac{3}{4}}$ $\cos\theta$ = - $\frac{\sqrt 3}{2}$ From ratio identity- $\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$ = $\frac{1/2}{-\sqrt 3/2}$ $\tan\theta$ = -$\frac{1}{\sqrt 3}$ Using reciprocal identities- $\csc\theta$ = $\frac{1}{\sin\theta}$ $\csc\theta$ = $\frac{1}{1/2}$ = 2 $\sec\theta$ = $\frac{1}{\cos\theta}$ $\sec\theta$ = $\frac{1}{-\sqrt 3/2}$ = -$\frac{2}{\sqrt 3}$ $\cot\theta$ = $\frac{1}{\tan\theta}$ $\cot\theta$ = $\frac{1}{-1/\sqrt 3}$ = - $\sqrt 3$
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