Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set: 27

Answer

tan t = $\frac{\sqrt 2}{4}$ csc t = 3 sec t = $\frac{3\sqrt 2}{4}$ cot t = 2$\sqrt 2$

Work Step by Step

tan t is computed by $\frac{sin t}{cos t}$ tan t = $\frac{sin t}{cos t}$ =$\frac{\frac{1}{3}}{\frac{2\sqrt 2}{3}}$ =$\frac{1}{2\sqrt 2}$*$\frac{\sqrt 2}{\sqrt 2}$ = $\frac{\sqrt 2}{4}$ csc t = $\frac{1}{sin t}$ = $\frac{1}{\frac{1}{3}}$ =3 sec t = $\frac{1}{cos t}$ = $\frac{1}{\frac{2\sqrt 2}{3}}$ = $\frac{3}{2\sqrt 2}$ * $\frac{\sqrt 2}{\sqrt 2}$ = $\frac{3\sqrt 2}{4}$ cot t = $\frac{1}{tan t}$ = $\frac{1}{\frac{1}{2\sqrt 2}}$ = 2$\sqrt 2$
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