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: 12

Answer

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

Work Step by Step

Use the figure given in the exercise, we have: 1. $sin\theta=\frac{-3}{\sqrt {(-3)^2+(-3)^2}}=-\frac{3}{3\sqrt 2}=-\frac{\sqrt 2}{2}$ 2. $cos\theta=\frac{-3}{\sqrt {(-3)^2+(-3)^2}}=-\frac{3}{3\sqrt 2}=-\frac{\sqrt 2}{2}$ 3. $tan\theta=\frac{-3}{-3}=1$ 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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