Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.3 The Product and Quotient Theorems - 8.3 Exercises - Page 376: 21



Work Step by Step

First, we use the division theorem to divide the absolute values and subtract the arguments: $\frac{3(cis 305^{\circ})}{9(cis65^{\circ})} \\=\frac{1}{3}cis(305^{\circ}-65^{\circ}) \\=\frac{1}{3}cis240^{\circ}$ Next, we change the expression into its equivalent form: $=\frac{1}{3} cis 240^{\circ} \\=\frac{1}{3}(\cos240^{\circ}+i\sin240^{\circ})$ Since we know that $\cos240^{\circ}=-\frac{1}{2}$ and $\sin240^{\circ}=-\frac{\sqrt{3}}{2}$, we can substitute these values in the expression and simplify: $\frac{1}{3} (\cos240^{\circ}+i\sin240^{\circ}) \\=\frac{1}{3}(-\frac{1}{2}-\frac{\sqrt{3}}{2}i) \\=-\frac{1}{6}-\frac{\sqrt{3}}{6}i$
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