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: 12


$$\cos x=-\frac{\sqrt3}{2}$$ $$\tan x=\frac{\sqrt3}{3}$$

Work Step by Step

$$\sin x=-\frac{1}{2}, x\in\Big[\pi,\frac{3\pi}{2}\Big]$$ 1) As $x\in\Big[\pi,\frac{3\pi}{2}\Big]$, it means angle $x$ is in the third quadrant, so $\cos x\lt0$ and $\tan x\gt0$. 2) To find $\cos x$, we employ the formula: $$\cos^2x=1-\sin^2x=1-\Big(-\frac{1}{2}\Big)^2=1-\frac{1}{4}=\frac{3}{4}$$ $$|\cos x|=\frac{\sqrt3}{2}$$ Since $\cos x\lt0$, $$\cos x=-\frac{\sqrt3}{2}$$ 3) To find $\tan x$, we employ the formula: $$\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$$
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