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: 83


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

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 $sin~15^{\circ}$: $sin~15^{\circ} = \frac{opposite}{hypotenuse}$ $sin~15^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}$
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.