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: 11



Work Step by Step

First, we use the product theorem to multiply the absolute values and add the arguments: $(5 cis 90^{\circ})(3 cis (45^{\circ})) \\=5(3) cis (90^{\circ}+45^{\circ}) \\=15 cis (135^{\circ})$ Next, we change the expression into its equivalent form: $=15 cis (135^{\circ}) \\=15(\cos135^{\circ}+i\sin135^{\circ})$ Since we know that $\cos135^{\circ}=-\frac{1}{\sqrt{2}}$ and $\sin135^{\circ}=\frac{1}{\sqrt{2}}$, we substitute these values in the expression and simplify: $15(\cos90^{\circ}+i\sin90^{\circ}) \\=15(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i) \\=-\frac{15}{\sqrt{2}}+\frac{15}{\sqrt{2}}i$
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.