Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Test - Page 250: 6


$sin(22.5^{\circ}) = \sqrt{{\frac{2-\sqrt{2}}{4}}}$

Work Step by Step

$sin~\frac{\theta}{2} = \sqrt{\frac{1-cos~\theta}{2}}$ $sin(22.5^{\circ}) = sin~\frac{45^{\circ}}{2}$ $sin(22.5^{\circ}) = \sqrt{\frac{1-cos~45^{\circ}}{2}}$ $sin(22.5^{\circ}) = \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}$ $sin(22.5^{\circ}) = \sqrt{\frac{\frac{2-\sqrt{2}}{2}}{2}}$ $sin(22.5^{\circ}) = \sqrt{{\frac{2-\sqrt{2}}{4}}}$
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