Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Test - Page 92: 5

Answer

$\displaystyle \sin 240^{o}= -\frac{\sqrt{3}}{2}$ $\displaystyle \cos 240^{o}= -\frac{1}{2}$ $\tan 240^{o}= \sqrt{3}$ $\displaystyle \csc 240^{o}= -\frac{2\sqrt{3}}{3}$ $\sec 240^{o}= -2$ $\displaystyle \cot 240^{o}= \frac{\sqrt{3}}{3}$

Work Step by Step

Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quadrant: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 0^{o}-\theta \end{array}\right]$ $240^{o}$ is in quadrant III, for which the reference angle is $\theta^{\prime}=\theta-180^{o}=240^{o}-180^{o}=60^{o}$ Signs: In quadrant III, tan and cot are positive, all the others are negative We use the table Function Values of Special Angles with the corresponding signs: $\displaystyle \sin 240^{o}=-\sin 60^{o}=-\frac{\sqrt{3}}{2}$ $\displaystyle \cos 240^{o}=-\cos 60^{o}=-\frac{1}{2}$ $\tan 240^{o}=\tan 60^{o}=\sqrt{3}$ $\displaystyle \csc 240^{o}=-\csc 60^{o}=-\frac{2\sqrt{3}}{3}$ $\sec 240^{o}=-\sec 60^{o}=-2$ $\displaystyle \cot 240^{o}=\cot 60^{o}=\frac{\sqrt{3}}{3}$
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