Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.4 - Sum and Difference Formulas - 7.4 Exercises - Page 540: 96

Answer

$E=\sqrt {12.5} \sin \theta+\sqrt {12.5} \cos \theta$

Work Step by Step

We have: $\tan \dfrac{\pi}{4}=1$ When $a=b$; $\sqrt {a^2+b^2} = 5$ $ \implies 5=\sqrt {2a^2}$ or, $a=b=\sqrt {12.5} $ Now, $E=\sqrt {12.5} \sin \theta+\sqrt {12.5} \cos \theta$
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.