Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Test - Page 244: 1

Answer

$cos~\theta = \frac{24}{25}$ $sin~\theta = -\frac{7}{25}$ $tan~\theta = -\frac{7}{24}$ $sec~\theta = \frac{25}{24}$ $csc~\theta = -\frac{25}{7}$ $cot~\theta = -\frac{24}{7}$

Work Step by Step

Since $\theta$ is in quadrant IV, $sin ~\theta$ is negative, $cos~\theta$ is positive, and $tan~\theta$ is negative. $cos~\theta = \frac{24}{25}$ $sin^2~\theta+cos^2~\theta = 1$ $sin^2~\theta = 1 - cos^2~\theta$ $sin^2~\theta = 1 - (\frac{24}{25})^2$ $sin~\theta = -\sqrt{1 - (\frac{24}{25})^2}$ $sin~\theta = -\frac{7}{25}$ $tan~\theta = \frac{sin~\theta}{cos~\theta}$ $tan~\theta = \frac{-\frac{7}{25}}{\frac{24}{25}}$ $tan~\theta = -\frac{7}{24}$ $sec~\theta = \frac{1}{cos~\theta}$ $sec~\theta = \frac{1}{(24/25)}$ $sec~\theta = \frac{25}{24}$ $csc~\theta = \frac{1}{sin~\theta}$ $csc~\theta = \frac{1}{(-7/25)}$ $csc~\theta = -\frac{25}{7}$ $cot~\theta = \frac{1}{tan~\theta}$ $cot~\theta = \frac{1}{(-7/24)}$ $cot~\theta = -\frac{24}{7}$
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