Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 8 - Section 8.2 - Trigonometric Form for Complex Numbers - 8.2 Problem Set - Page 433: 76

Answer

$\color{blue}{33/65}$

Work Step by Step

Using a Sum Formula and substituting the given $\sin A = 3/5, \sin B = 5/13$ gives: $$\begin{align*} \cos(A+B) &= \cos A\cos B - \sin A\sin B \\ \cos(A+B) &= \color{green}{\cos A}\color{magenta}{\cos B} - (3/5)(5/13). \qquad (\text{Eq. 1}) \end{align*}$$ We need $\color{magenta}{\cos A}$ and $\color{green}{\cos B}$ to complete the answer. Since $\cos^2 A + \sin^2 A =1$ and $A\in QI$ (so that $\cos A \gt 0$), $$\begin{align*} \color{green}{\cos A}\ &= \sqrt{1 - \sin^2A} \\ &= \sqrt{1 - (3/5)^2} \\ &= \sqrt{1 - 9/25} \\ &= \sqrt{16/25} \\ \color{green}{\cos A}\ &\color{green}{= 4/5} \end{align*}$$ Similarly, for $\cos B,\ B \in QI$, $$\begin{align*} \color{magenta}{\cos B}\ &= \sqrt{1 - \sin^2B} \\ &= \sqrt{1 - (5/13)^2} \\ &= \sqrt{1 - 25/169} \\ &= \sqrt{144/169} \\ \color{magenta}{\cos B}\ &\color{magenta}{= 12/13} \end{align*}$$ Substituting the values obtained for $\cos A$ and $\cos B$ into (Eq. 1) above gives $$\begin{align*} \cos(A+B) &= \color{green}{(4/5)}\color{magenta}{(12/13)} - (3/5)(5/13) \\ &= 48/65 - 15/65 \\ \color{blue}{\cos(A+B)} &\color{blue}{= 33/65} \end{align*}$$
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