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 239: 78

Answer

$cos~72^{\circ} = \frac{\sqrt{5}-1}{4}$

Work Step by Step

$cos~x = sin(90^{\circ}-x)$ We can find the value of $cos~72^{\circ}$: $cos~72^{\circ} = sin(90^{\circ}-72^{\circ})$ $cos~72^{\circ} = sin(18^{\circ})$ $cos~72^{\circ} = \frac{\sqrt{5}-1}{4}$
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