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: 70

Answer

$\sin{\theta} =-\dfrac{2\sqrt{5}}{5}$ $\cos{\theta} = -\dfrac{\sqrt{5}}{5}$

Work Step by Step

$y = 2x \\ \therefore \dfrac{y}{x} = 2 \\ \because \tan{\theta} = \dfrac{y}{x} \\ \therefore \tan{\theta} = 2$ $\because$ The terminal side of $\theta$ lies in $QIII%$, then $x$ and $y$ are both negative. $\therefore $ Let $y=-2$ and $x =-1$ $r = \sqrt{x^2+y^2} = \sqrt{(-1)^2+(-2)^2} = \sqrt{5}$ $\sin{\theta} = \dfrac{y}{r} = \boxed{-\dfrac{2\sqrt{5}}{5}}$ $\cos{\theta} = \dfrac{x}{r} = \boxed{-\dfrac{\sqrt{5}}{5}}$
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