Precalculus (6th Edition) Blitzer

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

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 550: 100


The given statement makes sense.

Work Step by Step

The statement talks about the periodicity of the sine function, so the period of the sine function is $2\pi $, or it can be written as $\sin \left( x+2\pi \right)=\sin x$ So, $\begin{align} & \sin \frac{7\pi }{3}=\sin \left( 2\pi +\frac{\pi }{3} \right) \\ & =\sin \frac{\pi }{3} \\ & =\frac{\sqrt{3}}{2} \end{align}$ So, by using the periodic properties of the sine function, the exact value of $\sin \frac{7\pi }{3}$ can be calculated. Hence, the statement is true.
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