Trigonometry (11th Edition) Clone

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

Chapter 6 - Quiz (Sections 6.1-6.3) - Page 282: 2c



Work Step by Step

$$y=\sec^{-1}(-\frac{2\sqrt3}{3})$$ *Remember: $\sec^{-1}A=x$ means that $\sec x=A$ Therefore, to find $y$, we need to recall which value of $y$ would have $\sec y=-\frac{2\sqrt3}{3}$. Now, if you find it hard to relate to secant function, use the following identity to switch it to cosine: $$\cos y=\frac{1}{\sec y}=\frac{1}{-\frac{2\sqrt3}{3}}=-\frac{3}{2\sqrt3}=-\frac{\sqrt3}{2}$$ So an equivalent task is to find $y$ so that $\cos y=-\frac{\sqrt3}{2}$. $\cos y\lt0$ shows that we can find a value of $y$ in either quadrant II or III of the unit circle. The choice is arbitrary. I would choose quadrant II here. In quadrant II, $\frac{5\pi}{6}$ is the value of the angle with which $\cos(\frac{5\pi}{6})=-\frac{\sqrt3}{2}$, or similarly, $\sec(\frac{5\pi}{6})=-\frac{2\sqrt3}{3}$ Therefore, $$y=\frac{5\pi}{6}$$
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