Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.3 - Trigonometric Functions - Exercises 1.3 - Page 27: 8


$\cos{x}=\dfrac{\sqrt{5}}{5}$ $\sin{x} = \dfrac{2\sqrt{5}}{5}$

Work Step by Step

Since $x$ lies in the first quadrant, $\sin{x}$ and $\cos{x}$ are positive. Using the formula: $$\sec{x}=\sqrt{1+\tan^2{x}}$$ $$\sec{x}=\sqrt{1+(2)^2}$$ $$\sec{x}=\sqrt{5}$$ $\because \cos{x}=\dfrac{1}{\sec{x}}$ $\therefore \cos{x}=\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$ Using the formula: $$\sin{x}=\sqrt{1-\cos^2{x}}$$ $$\sin{x}=\sqrt{1-\left( \dfrac{\sqrt{5}}{5}\right)^2}$$ $$\sin{x} = \dfrac{2\sqrt{5}}{5}$$
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