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: 61

Answer

$\sin{\theta} =-\dfrac{3}{5}$ $\cos{\theta} =-\dfrac{4}{5} $ $ \tan{\theta} =\dfrac{3}{4}$ $\csc{\theta} = -\dfrac{5}{3}$ $\sec{\theta} =-\dfrac{5}{4}$ $\cot{\theta} =\dfrac{4}{3}$

Work Step by Step

$ \tan{\theta} = \dfrac{y}{x} = \dfrac{3}{4}$ As $\theta$ terminates in $QIII$, any point on its terminal side will have negative $x$ and $y$ coordinates. $\therefore y = -3 \hspace{20pt} x =-4$ $$r^2 = x^2+y^2 \\ r = \sqrt{x^2+y^2} = \sqrt{(-4)^2+(-3)^2} = 5$$ $\sin{\theta} = \dfrac{y}{r} = -\dfrac{3}{5}$ $\cos{\theta} = \dfrac{x}{r} = -\dfrac{4}{5} $ $\csc{\theta} = \dfrac{1}{\sin{\theta}} = -\dfrac{5}{3}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = -\dfrac{5}{4}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}} = \dfrac{4}{3}$
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