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: 62

Answer

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

Work Step by Step

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