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 560: 27

Answer

$\frac{2\sqrt 5}{5}, -\frac{\sqrt {5}}{5}, -2, -\frac{1}{2}, -\sqrt 5, \frac{\sqrt {5}}{2}$

Work Step by Step

1. Given $sec\theta=\frac{1}{cos\theta}=-\sqrt 5$ and $\theta$ in quadrant II, we have $cos\theta=-\frac{\sqrt 5}{5}$. Form a right triangle with sides $\sqrt {5^2-5}, \sqrt 5, 5$ or $2\sqrt {5}, \sqrt 5, 5$, we have: 2. $sin\theta=\frac{2\sqrt 5}{5}$ 3. $cos\theta=-\frac{\sqrt {5}}{5}$ 4. $tan\theta=-\frac{2\sqrt {5}}{\sqrt {5}}=-2$ 5. $cot\theta=\frac{1}{tan\theta}=-\frac{1}{2}$ 6. $sec\theta=\frac{1}{cos\theta}=-\sqrt 5$ 7. $csc\theta=\frac{1}{sin\theta}=\frac{\sqrt {5}}{2}$
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