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 8.3 The Product and Quotient Theorems - 8.3 Exercises - Page 369: 10



Work Step by Step

First, we use the product theorem to multiply the absolute values and add the arguments: $(\sqrt{6} cis 120^{\circ})(\sqrt{6} cis (-30^{\circ})) \\=\sqrt{6}\sqrt{6} cis (120^{\circ}-30^{\circ}) \\=6 cis (90^{\circ})$ Next, we change the expression into its equivalent form: $=6 cis (90^{\circ}) \\6(\cos90^{\circ}+i\sin90^{\circ})$ Since we know that $\cos90^{\circ}=0$ and $\sin90^{\circ}=1$, we substitute these values in the expression and simplify: $6(\cos90^{\circ}+i\sin90^{\circ}) \\=6(i) \\=6i$
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