Trigonometry (11th Edition) Clone

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 281: 64


The least possible value of $t$ is $\frac{1}{1440}$ seconds.

Work Step by Step

$$i=I_{max}\sin(2\pi ft)$$ For $i=50,I_{max}=100,f=120$, we have $$50=100\sin(2\pi\times120t)$$ $$50=100\sin(240\pi t)$$ $$\sin(240\pi t)=\frac{1}{2}$$ $t$ refers to time, so as a rule, $t\in[0,\infty)$ Also $t$ is minimum when $240\pi t$ is minimum, as $240\pi$ is a constant. Therefore, we would be able to find the least of $t$ as we find the first value of $240\pi t$ that would have $\sin(240\pi t)=\frac{1}{2}$ over the interval $[0,2\pi)$. Such a value of $240\pi t$ can be found using the inverse function for sine: $$240\pi t=\sin^{-1}\frac{1}{2}$$ $$240\pi t=\frac{\pi}{6}$$ $$t=\frac{\pi}{6\times240\pi}$$ $$t=\frac{1}{1440}(seconds)$$ Therefore, the least value of $t$ satisfying the given data is $\frac{1}{1440}$ seconds.
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