Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.1 Fundamental Identities - 7.1 Exercises - Page 658: 11


$\Rightarrow \sin \theta =\dfrac {\sqrt {7}}{4}$

Work Step by Step

$\sin \theta =\pm \sqrt {1-\cos ^{2}\theta }=\pm \sqrt {1-\left( \dfrac {3}{4}\right) ^{2}}=\pm \dfrac {\sqrt {7}}{4}$ in quadrant 1 sine is positive so $\Rightarrow \sin \theta =\dfrac {\sqrt {7}}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.