Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 402: 16

Answer

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

Work Step by Step

With $P= \left(-\dfrac{1}{5},\dfrac{2\sqrt{6}}{5} \right) = (x,y)$, then $x = -\dfrac{1}{5} \text{ and } \hspace{15pt} y =\dfrac{2\sqrt{6}}{5}$. Thus, $\sin{t} = y$ $ \sin{t} = \dfrac{2\sqrt{6}}{5}$ $\cos{t} = x$ $ \cos{t} = -\dfrac{1}{5}$ $\tan{t} = \dfrac{y}{x}$ $\tan{t} = \dfrac{\dfrac{2\sqrt{6}}{5}}{ -\dfrac{1}{5} } = -2 \sqrt{6}$ $\csc{t} = \dfrac{1}{y}$ $\csc{t} = \dfrac{1}{\dfrac{2\sqrt{6}}{5}} = \dfrac{5 \sqrt{6}}{12}$ $\sec{t} = \dfrac{1}{x}$ $\sec{t} = \dfrac{1}{-\dfrac{1}{5}}= -5$ $\cot{t} = \dfrac{x}{y}$ $\cot{t} = \dfrac{-\dfrac{1}{5} }{\dfrac{2\sqrt{6}}{5}} = -\dfrac{\sqrt{6}}{12}$
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.