Trigonometry (10th Edition)

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

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.4 De Moivre's Theorem: Powers and Roots of Complex Numbers - 8.4 Exercises - Page 377: 48

Answer

$sin~2\theta = 2~cos~\theta ~sin~\theta$

Work Step by Step

From Exercise 45: $(cos~\theta+i~sin~\theta)^2 = cos~2\theta + i~sin~2\theta$ From Exercise 46: $(cos~\theta+i~sin~\theta)^2 = cos^2~\theta - sin^2~\theta + 2i~cos~\theta ~sin~\theta$ We can equate the imaginary part of each expression: $i~sin~2\theta = 2i~cos~\theta ~sin~\theta$ We can cancel $i$ form each side of the equation: $sin~2\theta = 2~cos~\theta ~sin~\theta$
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