Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 250: 72a

Answer

$T = \frac{1}{\omega}$

Work Step by Step

$V = a~sin~(2\pi~\omega~t)$ We can find the period $T$: $2\pi~\omega~T = 2\pi$ $T = \frac{2\pi}{2\pi~\omega}$ $T = \frac{1}{\omega}$
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