Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.2 Trigonometric Functions: Unit Circle Approach - 6.2 Assess Your Understanding - Page 379: 64

Answer

$sin(t) =-\frac{1}{2}$, $cos(t) =\frac{\sqrt 3}{2}$, $tan(t) =-\frac{\sqrt 3}{3}$, $cot(t) =-\sqrt 3$, $sec(t) =\frac{2\sqrt 3}{3}$, $csc(t) =-2$.

Work Step by Step

Given $t=-\frac{13\pi}{6}=-2\pi-\frac{\pi}{6}$, the terminal side is in quadrant IV with $t_0=\frac{\pi}{6}$ to the $+x$-axis , we have: $sin(t)=-sin(t_0)=-\frac{1}{2}$, $cos(t)=cos(t_0)=\frac{\sqrt 3}{2}$, $tan(t)=-tan(t_0)=-\frac{\sqrt 3}{3}$, $cot(t)=-cot(t_0)=-\sqrt 3$, $sec(t)=sec(t_0)=\frac{2\sqrt 3}{3}$, $csc(t)=csc(t_0)=-2$.
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