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: 70

Answer

$cos~15^{\circ} = \frac{\sqrt{6}+\sqrt{2}}{4} = 0.9659$

Work Step by Step

In the triangle $EAD$, the length of $AD$ is $\sqrt{6}+\sqrt{2}$ In the triangle $EAD$, the length of $ED$ is twice the radius, so the length of $ED$ is $4$ Note that triangle $EAD$ is a right triangle since the angle $EAD = 90^{\circ}$. We can use angle $ADB$ of triangle $EAD$ to find $cos~15^{\circ}$: $cos~15^{\circ} = \frac{adjacent}{hypotenuse}$ $cos~15^{\circ} = \frac{AD}{ED}$ $cos~15^{\circ} = \frac{\sqrt{6}+\sqrt{2}}{4} = 0.9659$
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