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: 17

Answer

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

Work Step by Step

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