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: 22



Work Step by Step

First, we use the division theorem to divide the absolute values and subtract the arguments: $\frac{16(cis 310^{\circ})}{8(cis70^{\circ})} \\=2cis(310^{\circ}-70^{\circ}) \\=2cis240^{\circ}$ Next, we change the expression into its equivalent form: $=2 cis 240^{\circ} \\=2(\cos240^{\circ}+i\sin240^{\circ})$ 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: $2 (\cos240^{\circ}+i\sin240^{\circ}) \\=2(-\frac{1}{2}-\frac{\sqrt{3}}{2}i) \\=-1-\sqrt{3}i$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.