Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.1 - The Unit Circle - 5.1 Exercises - Page 408: 45

Answer

a. $\quad \displaystyle \overline{t}= \frac{\pi}{3}$ b. $\displaystyle \quad P(-\frac{1}{2}, -\frac{\sqrt{3}}{2}$)

Work Step by Step

The reference number associated with the real number $t$ is the shortest distance along the unit circle between the terminal point determined by $t$ and the x-axis. For a given t, find its terminal point on the unit circle (positive=counterclockwise) and associate it with the terminal point of some t between 0 and $ 2\pi$ If the terminal point "lands" in quadrants II, III or IV, choose the symmetric terminal number in quadrant I$:$ t in Q.II $\Rightarrow \overline{t}=\pi-t$ t in Q.III$\Rightarrow \overline{t}=t-\pi$ t in Q.IV$\Rightarrow \overline{t}=2\pi-t$ ------------------- The terminal point of $-\displaystyle \frac{2\pi}{3}=-\pi+\frac{\pi}{3}$ is in Q.III (clockwise, $-\pi=- \displaystyle \frac{3\pi}{3}$), the same as the terminal point of $\displaystyle \pi+\frac{\pi}{3}=\frac{4\pi}{3},$ its reference number is $\displaystyle \frac{4\pi}{3}-\pi=\frac{\pi}{3} $ The terminal point for $\displaystyle \frac{\pi}{3}$ is $(\displaystyle \frac{1}{2}, \frac{\sqrt{3}}{2}$) ... Table 1, fig.6. In Q.III, both the x- and y-coordinates are negative, so by symmetry the terminal point for $-\displaystyle \frac{2\pi}{3}$ is $(-\displaystyle \frac{1}{2}, -\frac{\sqrt{3}}{2}$)
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