Trigonometry (11th Edition) Clone

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

Chapter 3 - Test - Page 138: 15


$\sin{(\frac{7\pi}{6})} = -\frac{1}{2} \\\cos{\frac{7\pi}{6}=-\frac{\sqrt3}{2}} \\\tan{\frac{7\pi}{6}}=\frac{\sqrt3}{3} \\\cot{\frac{7\pi}{6}}=\sqrt3 \\\sec{\frac{7\pi}{6}}=-\frac{2\sqrt3}{3} \\\csc{\frac{7\pi}{6}}=-2$

Work Step by Step

Recall: (1) An angle and its reference angle either have the same trigonometric function values or differ only in signs. (2) $\frac{\pi}{6}$ is a special angle whose sine value is known to be $\frac{1}{2}$ and whose cosine value is known to be $\frac{\sqrt3}{2}$. (3) $\tan{s} = \frac{\sin{s}}{\cos{s}}$ (4) $\cot{s}= \frac{\cos{s}}{\sin{s}}$ (5) $\sec{s} = \frac{1}{\cos{s}}$ (6) $\csc{s} = \frac{1}{\sin{s}}$ Note that the reference angle of $\frac{7\pi}{6}$ is $\frac{\pi}{6}$. However, since $\frac{7\pi}{6}$ is in Quadrant III, then its sine and cosine values are both negative . Therefore, $\sin{(\frac{7\pi}{6})} = -\frac{1}{2} \\\cos{\frac{7\pi}{6}=-\frac{\sqrt3}{2}} \\\tan{\frac{7\pi}{6}}=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\frac{\sqrt3}{3} \\\cot{\frac{7\pi}{6}}=\dfrac{-\frac{\sqrt3}{3}}{-\frac{1}{2}}=\sqrt3 \\\sec{\frac{7\pi}{6}}=\dfrac{1}{-\frac{\sqrt3}{2}}=-\frac{2\sqrt3}{3} \\\csc{\frac{7\pi}{6}}=\dfrac{1}{-\frac{1}{2}}=-2$
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