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: 8


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

Work Step by Step

$$\tan x = 2, x\in\Big[0,\frac{\pi}{2}\Big]$$ 1) As $x\in\Big[0,\frac{\pi}{2}\Big]$, it means angle $x$ is in the first quadrant, so $\sin x\gt0$ and $\cos x\gt0$. 2) To find $\cos x$, we employ the formula: $$\tan^2x+1=\frac{1}{\cos^2x}$$ $$\frac{1}{\cos^2x}=2^2+1=5$$ $$\cos^2x=\frac{1}{5}$$ $$|\cos x|=\frac{\sqrt5}{5}$$ Since $\cos x\gt0$, $$\cos x=\frac{\sqrt5}{5}$$ 3) To find $\sin x$, we employ the formula: $$\tan x=\frac{\sin x}{\cos x}$$ $$\sin x=\tan x\cos x=2\times\frac{\sqrt5}{5}=\frac{2\sqrt5}{5}$$
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