Trigonometry (11th Edition) Clone

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

Chapter 3 - Radian Measure and the Unit Circle - Section 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 125: 88



Work Step by Step

With $y=0.35154709$ and $x=-0.9361702$, then \begin{align*} \tan{\theta}&=\dfrac{0.35154709}{-0.9361702}\\\\ \theta&=\tan^{-1}\left(\dfrac{0.35154709}{-0.9361702}\right)\\\ \theta&=-0.3592231654 \end{align*} This angle is in Quadrant IV. Note that the given point is in Quadrant II. Since $\tan{\theta}=\tan{(\theta+\pi)}$, then the value of $\theta$ in Quadrant II must be: \begin{align*} \theta&=-0.3592231654+\pi\\ &=2.782369488 \end{align*} Solve for the arc length using the formula $S=r\theta$ with $r=1$ and $\theta=2.782369488$ to obtain: \begin{align*} S&=r\theta\\ &=1(2.782369488)\\ &=2.782369488 \end{align*} Therefore, the length of the shortest arc from $(1, 0)$ to the given point is $2.782369488$.
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