Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises - Page 679: 60



Work Step by Step

$\sin \dfrac {5\pi }{9}\cos \dfrac {\pi }{18}-\cos \dfrac {5\pi }{9}\sin \dfrac {\pi }{18}=\sin \left( \dfrac {5\pi }{9}-\dfrac {\pi }{18}\right) =\sin \dfrac {\pi }{2}=1$
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.