Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 244: 84


$tan~15^{\circ} = 2-\sqrt{3}$

Work Step by Step

Note that triangle $ACD$ is a right triangle since angle $ACD = 90^{\circ}$ We can use triangle $ACD$ to find $tan~15^{\circ}$: $tan~15^{\circ} = \frac{opposite}{adjacent}$ $tan~15^{\circ} = \frac{AC}{CD}$ $tan~15^{\circ} = \frac{1}{2+\sqrt{3}}$ $tan~15^{\circ} = \frac{1}{2+\sqrt{3}}~\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$ $tan~15^{\circ} = \frac{2-\sqrt{3}}{1}$ $tan~15^{\circ} = 2-\sqrt{3}$
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