Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 693: 49



Work Step by Step

Use the half-angle formula, $\sin\frac{A}{2}=\pm\sqrt{\frac{1-\cos A}{2}}$. Note that $67.5^\circ$ is in Quadrant I, where sine is positive, so we take the positive square root. $\sin 67.5^\circ$ $=\sin \frac{135^\circ}{2}$ $=\sqrt{\frac{1-\cos 135^\circ}{2}}$ $=\sqrt{\frac{1-\left(-\frac{\sqrt{2}}{2}\right)}{2}}$ $=\sqrt{\frac{\left(1+\frac{\sqrt{2}}{2}\right)*2}{2*2}}$ $=\sqrt{\frac{2+\sqrt{2}}{4}}$ $=\frac{\sqrt{2+\sqrt{2}}}{\sqrt{4}}$ $=\frac{\sqrt{2+\sqrt{2}}}{2}$
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