Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 645: 63


The exact value of the provided expression is $-\frac{\sqrt{3}}{2}$.

Work Step by Step

We know that $\theta =-\frac{\pi }{3}$ lies in quadrant IV, so the reference angle for $\theta =-\frac{\pi }{3}$ is: $\theta '=\frac{\pi }{3}$ Now, the function value for the reference angle is: $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ Thus, the angle $-\frac{\pi }{3}$ lies in quadrant IV, so the sine function is negative: $\begin{align} & \sin \left( -\frac{\pi }{3} \right)=-\sin \frac{\pi }{3} \\ & =-\frac{\sqrt{3}}{2} \end{align}$
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.