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: 20



Work Step by Step

First, we use the division theorem to divide the absolute values and subtract the arguments: $\frac{12(\cos23^{\circ}+i\sin23^{\circ})}{6(\cos293^{\circ}+i\sin 293^{\circ})} \\=2(\cos (23^{\circ}-293^{\circ})+i\sin(23^{\circ}-293^{\circ})) \\=2(\cos-270^{\circ}+i\sin-270^{\circ})$ Since we know that $\cos-270^{\circ}=\cos270^{\circ}=0$ and $\sin-270^{\circ}=-\sin270^{\circ}=1$, we can substitute these values in the expression and simplify: $2(\cos-270^{\circ}+i\sin-270^{\circ}) \\=2(i) \\=2i$
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