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: 13

Answer

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

Work Step by Step

With $P= \left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2} \right) = (x,y)$, then $x = \dfrac{\sqrt{3}}{2}, \text{ and } \hspace{15pt} y = \dfrac{1}{2}$. Thus, $\sin{t} = y$ $ \sin{t} = \dfrac{1}{2}$ $\cos{t} = x$ $ \cos{t} = \dfrac{\sqrt{3}}{2}$ $\tan{t} = \dfrac{y}{x}$ $\tan{t} = \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} = \dfrac{\sqrt{3}}{3}$ $\csc{t} = \dfrac{1}{y}$ $\csc{t} = \dfrac{1}{\dfrac{1}{2}} = 2$ $\sec{t} = \dfrac{1}{x}$ $\sec{t} = \dfrac{1}{\dfrac{\sqrt{3}}{2}}= \dfrac{2\sqrt{3}}{3}$ $\cot{t} = \dfrac{x}{y}$ $\cot{t} = \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}= \sqrt{3}$
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.