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: 57

Answer

$\sin{\theta} = \dfrac{21}{29}$ $\cos{\theta} = -\dfrac{20}{29}$ $\tan{\theta} = -\dfrac{21}{20}$ $\sec{\theta} = -\dfrac{29}{20}$ $\csc{\theta} = \dfrac{29}{21}$ $\cot{\theta} = -\dfrac{20}{21}$

Work Step by Step

$\because \cos{\theta} = \dfrac{x}{r} = \dfrac{-20}{29}$ $\therefore x = -20 \hspace{10pt} \& \hspace{10pt} r = 29$ $$r^2 = x^2+y^2 \\ y^2 = r^2-x^2 \\ y = \pm \sqrt{r^2-x^2} =\pm \sqrt{29^2-(-20)^2} = \pm 21$$ As $\theta$ terminates in $QII$, any point on its terminal side will have a positive $y$ coordinate. $\therefore y = 21$ $\sin{\theta} = \dfrac{y}{r} = \dfrac{21}{29}$ $\tan{\theta} = \dfrac{y}{x} = -\dfrac{21}{20}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = -\dfrac{29}{20}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}} = \dfrac{29}{21}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}} = -\dfrac{20}{21}$
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