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 - Page 547: 4

Answer

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

Work Step by Step

(-$\frac{\sqrt 2}{2}$ , $\frac{\sqrt 2}{2}$) sin t corresponds to y value So sin t = $\frac{\sqrt 2}{2}$ cos t corresponds to x value cos t = -$\frac{\sqrt 2}{2}$ tan t is computed by $\frac{y}{x}$ So tan t = $\frac{\frac{\sqrt 2}{2}}{\frac{-\sqrt 2}{2}}$ = -1 csc t = $\frac{1}{sin t}$ So, csc t = $\frac{2}{\sqrt 2}$ sec t = $\frac{1}{cos t}$ so sec t = -$\frac{2}{\sqrt 2}$ cot t = $\frac{1}{tan t}$ so cot t =-1
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