Trigonometry 7th Edition

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

Chapter 1 - Section 1.4 - Introduction to Identities - 1.4 Problem Set - Page 40: 47

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

We know from first Pythagorean identity that- $\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$ Given $\sin\theta$ is negative and $\theta$ is not in QIII, therefore $\theta$ terminates in Q IV. Hence $\cos\theta$ will be positive- OR $\cos\theta$ = $\sqrt (1-\sin^{2}\theta)$ substitute the given value of $\sin\theta$- $\cos\theta$ = $\sqrt (1-(\frac{-1}{2})^{2})$ $\cos\theta$ = $\sqrt (1-\frac{1}{4})$ $\cos\theta$ = $\sqrt (\frac{4-1}{4})$ = $\sqrt (\frac{3}{4})$ $\cos\theta$ = $\frac{\sqrt 3}{2}$ By ratio identity- $\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$ Substituting the values of $\sin\theta$ and $\cos\theta$- $\tan\theta$ = $\frac{-1/2}{\sqrt 3/2}$ = - $\frac{1}{\sqrt 3}$ From reciprocal identities- $\csc\theta$ = $\frac{1}{\sin\theta}$ = $\frac{1}{-1/2}$ = -2 $\sec\theta$ = $\frac{1}{\cos\theta}$ = $\frac{1}{\sqrt 3/2}$ = $\frac{2}{\sqrt 3}$ $\cot\theta$ = $\frac{1}{\tan\theta}$ = $\frac{1}{-1/\sqrt 3}$ = - $\sqrt 3$
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