Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.6 Double-angle and Half-angle Formulas - 6.6 Assess Your Understanding - Page 518: 19

Answer

$\dfrac{\sqrt{{2-\sqrt2}}}{2}$

Work Step by Step

The half-angle Identity for sine can be expressed as: $\sin {(\dfrac{\theta}{2})}=\pm\sqrt{\dfrac{1-\cos{\theta}}{2}}$ Therefore, $\sin{(22.5^{\circ})}=\sqrt{\dfrac{1-\cos{(45^{\circ})}}{2}}\\=\sqrt{\dfrac{1-\dfrac{\sqrt2}{2}}{2}}\\=\sqrt{\dfrac{2-\sqrt2}{4}}\\ =\dfrac{\sqrt{{2-\sqrt2}}}{2}$
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