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: 82


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

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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