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: 68

Answer

$\sin{t} = \dfrac{1}{5}$ $\cos{t} =-\dfrac{2\sqrt{6}}{5}$ $\tan{t} =\dfrac{\sqrt{6}}{12}$ $\sec{t} =- \dfrac{5\sqrt{6}}{12}$ $\cot{t} = 2\sqrt{6}$

Work Step by Step

$\because \csc{t} = 5 \hspace{20pt} \therefore \sin{t} = \dfrac{1}{5}$ $\because \cos{t} <0 \\ \therefore \cos{t} = -\sqrt{1-\sin^2{t}} \\ = -\sqrt{1-\left(\dfrac{}1{5}\right)^2} = -\dfrac{2\sqrt{6}}{5}$ $\tan{t} = \dfrac{\sin{t}}{\cos{t}} = \dfrac{\sqrt{6}}{12}$ $\sec{t} = \dfrac{1}{\cos{t}} = - \dfrac{5\sqrt{6}}{12}$ $\cot{t} = \dfrac{1}{\tan{t}} = 2\sqrt{6}$
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