Trigonometry (10th Edition)

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

Chapter 6 - Test - Page 287: 15

Answer

$\theta = 90^{\circ}$

Work Step by Step

$sin^2~\theta = -cos~2\theta$ $sin^2~\theta = -(cos^2~\theta-sin^2~\theta)$ $cos^2~\theta = 0$ $cos~\theta = 0$ $\theta = 90^{\circ}$
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