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: 50

Answer

$\dfrac {\sqrt {2}}{4}\left( \sqrt {3}-1\right) $

Work Step by Step

$\sin \dfrac {\pi }{12}=\sin \left( \dfrac {\pi }{3}-\dfrac {\pi }{4}\right) =\sin \dfrac {\pi }{3}\cos \dfrac {\pi }{4}-\cos \dfrac {\pi }{3}\sin \dfrac {\pi }{4}=\dfrac {\sqrt {3}}{2}\dfrac {\sqrt {2}}{2}-\dfrac {1}{2}\dfrac {\sqrt {2}}{2}=\dfrac {\sqrt {2}}{4}\left( \sqrt {3}-1\right) $
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.