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 6.2 Trigonometric Equations I - 6.2 Exercises - Page 274: 68b


To produce voltage $E=10\sqrt3$, it takes at least $\frac{10}{3}$ seconds.

Work Step by Step

$$E=20\sin(\frac{\pi t}{4}-\frac{\pi}{2})$$ Since $t$ is time in seconds, the range of $t$ is $[0,\pi)$ For $E=10\sqrt3$, we get the equation $$20\sin(\frac{\pi t}{4}-\frac{\pi}{2})=10\sqrt3$$ $$\sin(\frac{\pi t}{4}-\frac{\pi}{2})=\frac{10\sqrt3}{20}=\frac{\sqrt3}{2}$$ Over the interval $[0,2\pi)$, $\frac{\pi t}{4}-\frac{\pi}{2}=\frac{\pi}{3}$ is the smallest value which we would have $\sin(\frac{\pi t}{4}-\frac{\pi}{2})=\frac{\sqrt3}{2}$, which would also lead to least positive value of $t$. $$\frac{\pi t}{4}-\frac{\pi}{2}=\frac{\pi}{3}$$ $$\frac{\pi t}{4}=\frac{5\pi}{6}$$ $$t=\frac{5\pi}{6}\times\frac{4}{\pi}$$ $$t=\frac{10}{3}(seconds)$$ So to produce voltage $E=10\sqrt3$, it takes at least $\frac{10}{3}$ seconds.
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