Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.2 - Trigonometric Functions of Real Numbers - 5.2 Exercises - Page 418: 65

Answer

$\sin{t} =-\dfrac{2\sqrt{2}}{3}$ $\cos{t} = \dfrac{1}{3}$ $\tan{t} =-2\sqrt{2}$ $\csc{t} =-\dfrac{3\sqrt{2}}{4}$ $\cot{t} = -\dfrac{\sqrt{2}}{4}$

Work Step by Step

$\because\sec{t} =3 \hspace{20pt} \therefore \cos{t} = \dfrac{1}{3}$ $\because t$ terminates in $ QIV \hspace{20pt} \therefore \sin{t} $ is negative $\sin^2{t} = 1-\cos^2{t} \\ \sin{t} = -\sqrt{1-\cos^2{t}} \\ = -\sqrt{1-\left(\dfrac{1}{3}\right)^2} = -\dfrac{2\sqrt{2}}{3}$ $\tan{t} = \dfrac{\sin{t}}{\cos{t}} = -2\sqrt{2}$ $\csc{t} =\dfrac{1}{\sin{t}} = -\dfrac{3\sqrt{2}}{4}$ $\cot{t} = \dfrac{1}{\tan{t}}-\dfrac{\sqrt{2}}{4}$
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