Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 238: 71

Answer

$AE = \sqrt{6}-\sqrt{2}$ $sin~15^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}$

Work Step by Step

We can use the Pythagorean theorem to find the length of $AE$: $(AE)^2 = (DE)^2-(AD)^2$ $AE = \sqrt{(DE)^2-(AD)^2}$ $AE = \sqrt{(4)^2-(\sqrt{6}+\sqrt{2})^2}$ $AE = \sqrt{16-(6+2\sqrt{12}+2)}$ $AE = \sqrt{6-2\sqrt{12}+2}$ $AE = \sqrt{(\sqrt{6}-\sqrt{2})~(\sqrt{6}-\sqrt{2})}$ $AE = \sqrt{6}-\sqrt{2}$ We can find $sin~15^{\circ}$: $sin~15^{\circ} = \frac{opposite}{hypotenuse}$ $sin~15^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}$
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