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: 19

Answer

$-2$

Work Step by Step

First, we use the division theorem to divide the absolute values and subtract the arguments: $\frac{10(\cos50^{\circ}+i\sin50^{\circ})}{5(\cos230^{\circ}+i\sin 230^{\circ})} \\=2(\cos (50^{\circ}-230^{\circ})+i\sin(50^{\circ}-230^{\circ})) \\=2(\cos-180^{\circ}+i\sin-180^{\circ})$ Since we know that $\cos-180^{\circ}=\cos180^{\circ}=-1$ and $\sin-180^{\circ}=-\sin180^{\circ}=0$, we can substitute these values in the expression and simplify: $2(\cos-180^{\circ}+i\sin-180^{\circ}) \\=2(-1) \\=-2$
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