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: 63

Answer

$ \sin{\theta} = \dfrac{5}{13}$ $\cos{\theta} =-\dfrac{12}{13} $ $\tan{\theta} =-\dfrac{5}{12}$ $ \csc{\theta} =\dfrac{13}{5}$ $\sec{\theta} = -\dfrac{13}{12}$ $\cot{\theta} =-\dfrac{12}{5}$

Work Step by Step

$ \csc{\theta} = \dfrac{r}{y} = \dfrac{13}{5}$ $\therefore y = 5 \hspace{20pt} r = 13$ $\cos{\theta} = \dfrac{x}{r} \hspace{20pt} \because \cos{\theta} <0 \hspace{20pt}\therefore x $ is negative. $x = -\sqrt{r^2-y^2} = -\sqrt{(13)^2-(5)^2} = -12$ $\sin{\theta} = \dfrac{y}{r} = \dfrac{5}{13}$ $\cos{\theta} = \dfrac{x}{r} = -\dfrac{12}{13} $ $\tan{\theta} = \dfrac{y}{x} = -\dfrac{5}{12}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = -\dfrac{13}{12}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}} = -\dfrac{12}{5}$
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