## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 27: 11

#### Answer

$$\cos x=-\frac{2\sqrt5}{5}$$ $$\sin x=-\frac{\sqrt5}{5}$$

#### Work Step by Step

$$\tan x=\frac{1}{2}, x\in\Big[\pi,\frac{3\pi}{2}\Big]$$ 1) As $x\in\Big[\pi,\frac{3\pi}{2}\Big]$, it means angle $x$ is in the third quadrant, so $\sin x\lt0$ and $\cos x\lt0$. 2) To find $\cos x$, we employ the formula: $$1+\tan^2x=\frac{1}{\cos^2x}$$ $$\frac{1}{\cos^2x}=1+\Big(\frac{1}{2}\Big)^2=1+\frac{1}{4}=\frac{5}{4}$$ $$\cos^2 x=\frac{4}{5}$$ $$|\cos x|=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}$$ Since $\cos x\lt0$, $$\cos x=-\frac{2\sqrt5}{5}$$ 3) To find $\sin x$, we employ the formula: $$\tan x=\frac{\sin x}{\cos x}$$ $$\sin x=\tan x\times\cos x=-\frac{2\sqrt5}{5}\times\frac{1}{2}=-\frac{\sqrt5}{5}$$

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