Precalculus (6th Edition)

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

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises - Page 532: 43

Answer

$\dfrac {\sqrt {3}}{2}$

Work Step by Step

$\sin 60^{0}=\sqrt {1-\cos ^{2}60}=\sqrt {1-\left( \dfrac {1}{2}\right) ^{2}}=\dfrac {\sqrt {3}}{2}$
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.