Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - Chapter 5 Test Prep - Review Exercises - Page 559: 19

Answer

$\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2}, -1, -1, -\sqrt 2, \sqrt 2$

Work Step by Step

Given point $(-2\sqrt 2,2\sqrt 2)$ on the terminal side, we have: 1. $sin\theta=\frac{2\sqrt 2}{\sqrt {(-2\sqrt 2)^2+(2\sqrt 2)^2}}=\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}$ 2. $cos\theta=\frac{-2\sqrt 2}{\sqrt {(-2\sqrt 2)^2+(2\sqrt 2)^2}}=-\frac{1}{\sqrt 2}=-\frac{\sqrt 2}{2}$ 3. $tan\theta=\frac{2\sqrt 2}{-2\sqrt 2}=-1$ 4. $cot\theta=\frac{1}{tan\theta}=-1$ 5. $sec\theta=\frac{1}{cos\theta}=-\sqrt 2$ 6. $csc\theta=\frac{1}{sin\theta}=\sqrt 2$
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