Algebra and Trigonometry 10th Edition

Published by Cengage Learning

Chapter 7 - 7.4 - Sum and Difference Formulas - 7.4 Exercises - Page 540: 90

Answer

$b \cos \beta \theta+a \sin \beta \theta$

Work Step by Step

$\tan C=\dfrac{opposite}{adjacent}=\dfrac{b}{a}$ $hypotenuse =\sqrt {a^2+b^2}$ $\cos C=\dfrac{adjacent}{hypotenuse}=\dfrac{a}{\sqrt {a^2+b^2}}$ $\sin C=\dfrac{opposite}{hypotenuse}=\dfrac{b}{\sqrt {a^2+b^2}}$ Now, $\sqrt {a^2+b^2}(\cos \beta \theta \cos C +\sin \beta \theta \sin C)$ or, $\sqrt {a^2+b^2}[\cos \beta \theta (\dfrac{b}{\sqrt {a^2+b^2}}) +\sin \beta \theta (\dfrac{a}{\sqrt {a^2+b^2}})]$ or, $b \cos \beta \theta+a \sin \beta \theta$

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.