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: 32


$\sin\theta$ = $\frac{12}{13}$

Work Step by Step

We know from first Pythagorean identity that- $\sin\theta$ = ± $\sqrt (1-\cos^{2}\theta)$ As $\theta$ terminates in Q I, Therefore $\sin\theta$ will be positive- $\sin\theta$ = $\sqrt (1-\cos^{2}\theta)$ substitute the given value of $\cos\theta$- $\sin\theta$ = $\sqrt (1-(\frac{5}{13})^{2})$ $\sin\theta$ = $\sqrt (1-\frac{25}{169})$ $\sin\theta$ = $\sqrt (\frac{169 - 25}{169})$ = $\sqrt (\frac{144}{169})$ $\sin\theta$ = $\frac{12}{13}$
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