## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Quiz (Sections 6.1-6.3) - Page 282: 7a

#### Answer

The least possible value of $t$ such that $V=0$ is $0$ seconds.

#### Work Step by Step

$$V=\cos2\pi t$$ with $0\le t\le\frac{1}{2}$ For $V=1$, we have $$\cos2\pi t=1$$ Here the condition states that $0\le t\le\frac{1}{2}$. In other words, $t\in[0,\frac{1}{2}]$ Also $t$ is minimum when $2\pi t$ is minimum, as $2\pi$ is a constant. Therefore, we would be able to find the least value of $t$ as we find the first value of $2\pi t$ that would have $\cos(2\pi t)=1$ over the interval $[0,2\pi)$. Such a value of $2\pi t$ can be found using the inverse function for cosine: $$2\pi t=\cos^{-1}1$$ $$2\pi t=0$$ $$t=0(seconds)$$ $0\in[0,\frac{1}{2}]$ so this solution is accepted. In conclusion, the least possible value of $t$ such that $V=0$ is $0$ seconds.

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