University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 27: 9


$$\sin x=-\frac{2\sqrt2}{3}$$ $$\tan x=-2\sqrt2$$

Work Step by Step

$$\cos x=\frac{1}{3}, x\in\Big[-\frac{\pi}{2},0\Big]$$ 1) As $x\in\Big[-\frac{\pi}{2},0\Big]$, it means angle $x$ is in the fourth quadrant, so $\sin x\lt0$ and $\tan x\lt0$. 2) To find $\sin x$, we employ the formula: $$\sin^2 x+\cos^2x=1$$ $$\sin^2x=1-\cos^2x=1-\Big(\frac{1}{3}\Big)^2=1-\frac{1}{9}=\frac{8}{9}$$ $$|\sin x|=\frac{2\sqrt2}{3}$$ Since $\sin x\lt0$, $$\sin x=-\frac{2\sqrt2}{3}$$ 3) To find $\tan x$, we employ the formula: $$\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{2\sqrt2}{3}}{\frac{1}{3}}=-2\sqrt2$$
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