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: 47

Answer

$$2 - \sqrt 3 $$

Work Step by Step

$$\eqalign{ & \tan \frac{\pi }{{12}} \cr & {\text{Write }}\frac{\pi }{{12}}{\text{ as }}\frac{\pi }{4} - \frac{\pi }{6} \cr & \tan \frac{\pi }{{12}} = \tan \left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) \cr & {\text{Use Tangent difference identity}} \cr & \tan \frac{\pi }{{12}} = \frac{{\tan \left( {\pi /4} \right) - \tan \left( {\pi /6} \right)}}{{1 + \tan \left( {\pi /4} \right)\tan \left( {\pi /6} \right)}} \cr & {\text{Substitute known values}} \cr & \tan \frac{\pi }{{12}} = \frac{{1 - \sqrt 3 /3}}{{1 + \sqrt 3 /3}} \cr & \tan \frac{\pi }{{12}} = \frac{{3 - \sqrt 3 }}{{3 + \sqrt 3 }} \cr & {\text{Rationalizing}} \cr & \tan \frac{\pi }{{12}} = \frac{{3 - \sqrt 3 }}{{3 + \sqrt 3 }} \times \frac{{3 - \sqrt 3 }}{{3 - \sqrt 3 }} \cr & \tan \frac{\pi }{{12}} = \frac{{12 - 6\sqrt 3 }}{6} \cr & \tan \frac{\pi }{{12}} = 2 - \sqrt 3 \cr} $$
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