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: 63

Answer

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

Work Step by Step

Given $t=-\frac{14\pi}{3}=-4\pi-\frac{2\pi}{3}$, the terminal side is in quadrant III with $t_0=\frac{2\pi}{3}$ to the $-x$-axis , we have: $sin(t)=-sin(t_0)=-\frac{\sqrt 3}{2}$, $cos(t)=-cos(t_0)=-\frac{1}{2}$, $tan(t)=tan(t_0)=\sqrt 3$, $cot(t)=cot(t_0)=\frac{\sqrt 3}{3}$, $sec(t)=-sec(t_0)=-2$, $csc(t)=-csc(t_0)=-\frac{2\sqrt 3}{3}$.
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