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: 57


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

Work Step by Step

The reference angle for $\theta =240{}^\circ $ is: $\begin{align} & \theta '=240{}^\circ -180{}^\circ \\ & =60{}^\circ \end{align}$ Now, the function value for the reference angle is: $\sin 60{}^\circ =\frac{\sqrt{3}}{2}$ So, we see that the angle $240{}^\circ $ lies in quadrant III, so the sine function is negative: $\begin{align} & \sin 240{}^\circ =-\sin 60{}^\circ \\ & =-\frac{\sqrt{3}}{2} \end{align}$
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