Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.5 - Vectors: A Geometric Approach - 2.5 Problem Set - Page 108: 53


$\sin \theta=\frac{2\sqrt 5}{5}$ $\cos \theta=\frac{\sqrt 5}{5}$

Work Step by Step

We can consider an arbitrary point that satisfies y=2x and lies in quadrant I. Let us consider the point (1,2). Then, $r=\sqrt {x^{2}+y^{2}}=\sqrt {1^{2}+2^{2}}=\sqrt 5$ $\sin \theta=\frac{y}{r}=\frac{2}{\sqrt 5}=\frac{2\sqrt 5}{5}$ $\cos \theta=\frac{x}{r}=\frac{1}{\sqrt 5}=\frac{\sqrt 5}{5}$
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