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: 6

Answer

$20+20\sqrt 3i$

Work Step by Step

Step 1: $[8(\cos300^{\circ}+i\sin300^{\circ})][5(\cos120^{\circ}+i\sin120^{\circ}]$ Step 2: Using the product theorem, the expression becomes $8(5)[\cos(300^{\circ}+120^{\circ})+i\sin(300^{\circ}+120^{\circ})]$ Step 3: Simplifying, $40(\cos420^{\circ}+i\sin420^{\circ})$ Step 4: Since $60^{\circ}$ and $420^{\circ}$ are coterminal angles, the expression becomes $40(\cos60^{\circ}+i\sin60^{\circ})$ Step 5: We know that $\cos60^{\circ}=\frac{1}{2}$ and $\sin60^{\circ}=\frac{\sqrt 3}{2}$ Step 6: Substituting these values in the expression, $40(\frac{1}{2}+\frac{\sqrt 3}{2}i)$ Step 7: Therefore, the final answer is $20+20\sqrt 3i$
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