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: 70

Answer

$\sin{t} = \dfrac{4\sqrt{17}}{17}$ $\cos{t} = -\dfrac{\sqrt{17}}{17}$ $\csc{t} =\dfrac{\sqrt{17}}{4}$ $\sec{t}=-\sqrt{17}$ $\cot{t} =-\dfrac{1}{4}$

Work Step by Step

$\tan{t} =-4$ $\because \csc{t} > 0 \hspace{5pt} \& \tan{t} < 0 \hspace{20pt} \therefore \sec{t} $ is negative $\sec^2{t} = 1+\tan^2{t} \\ \sec{t} = -\sqrt{1+\tan^2{t}} \\ = -\sqrt{1+(-4)^2} = \sec{t} =-\sqrt{17}$ $\cos{t} = \dfrac{1}{\sec{t}} = -\dfrac{\sqrt{17}}{17}$ $\sin{t} = \cos{t} \times \tan{t} = \dfrac{4\sqrt{17}}{17}$ $\csc{t} = \dfrac{1}{\sin{t}} = \dfrac{\sqrt{17}}{4}$ $\cot{t} = \dfrac{1}{\tan{t}} =-\dfrac{1}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.