Trigonometry 7th Edition

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

Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 33: 65

Answer

$\sin{\theta} =\dfrac{2\sqrt{5}}{5}$ $\cos{\theta} =\dfrac{\sqrt{5}}{5} $ $\tan{\theta} =2$ $\csc{\theta} =\dfrac{\sqrt{5}}{2}$ $\sec{\theta} =\sqrt{5}$ $ \cot{\theta} =\dfrac{1}{2}$

Work Step by Step

$\cos{\theta} = \dfrac{x}{r} \hspace{20pt} \because \cos{\theta} > 0 \hspace{10pt} \therefore x$ is positive. $ \cot{\theta} = \dfrac{x}{y} = \dfrac{1}{2}$ $\therefore x = 1 \hspace{20pt} y = 2$ $r = \sqrt{x^2+y^2} = \sqrt{(1)^2+(2)^2} = \sqrt{5}$ $\sin{\theta} = \dfrac{y}{r} = \dfrac{2\sqrt{5}}{5}$ $\cos{\theta} = \dfrac{x}{r} = \dfrac{\sqrt{5}}{5} $ $\tan{\theta} = \dfrac{y}{x} = 2$ $\csc{\theta} = \dfrac{1}{\sin{\theta}} = \dfrac{\sqrt{5}}{2}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = \sqrt{5}$
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