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: 79

Answer

$sin~72^{\circ} = \sqrt{\frac{5+\sqrt{5}}{8}}$

Work Step by Step

$sin~72^{\circ} = sin~(90^{\circ}-18^{\circ}) = cos~18^{\circ}$ $cos^2~x+sin^2~x = 1$ $cos~x = \sqrt{1-sin^2~x}$ We can find the value of $cos~18^{\circ}$: $cos~18^{\circ} = \sqrt{1-(sin~18^{\circ})^2}$ $cos~18^{\circ} = \sqrt{1-(\frac{\sqrt{5}-1}{4})^2}$ $cos~18^{\circ} = \sqrt{1-(\frac{6-2\sqrt{5}}{16})}$ $cos~18^{\circ} = \sqrt{\frac{10+2\sqrt{5}}{16}}$ $cos~18^{\circ} = \sqrt{\frac{5+\sqrt{5}}{8}}$ Since $sin~72^{\circ} = cos~18^{\circ}$, then $sin~72^{\circ} = \sqrt{\frac{5+\sqrt{5}}{8}}$
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