Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises - Page 679: 48


$2+\sqrt {3}$

Work Step by Step

$\tan \dfrac {5\pi }{12}=\tan \left( \dfrac {\pi }{4}+\dfrac {\pi }{6}\right) =\dfrac {\tan \dfrac {\pi }{4}+\tan \dfrac {\pi }{6}}{1-\tan \dfrac {\pi }{4}\tan \dfrac {\pi }{6}}=\dfrac {1+\dfrac {\sqrt {3}}{3}}{1-\dfrac {\sqrt {3}}{3}}=\dfrac {3+\sqrt {3}}{3-\sqrt {3}}=$ $=\dfrac {\left( 3+\sqrt {3}\right) \left( 3+\sqrt {3}\right) }{\left( 3-\sqrt {3}\right) \left( 3+\sqrt {3}\right) }=\dfrac {3^{2}+2\times 3\times \sqrt {3}+\sqrt {3}^{3}}{3^{2}-\sqrt {3}^{2}}=\dfrac {12+6\sqrt {3}}{6}=2+\sqrt {3}$
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.