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: 50



Work Step by Step

Use the half-angle formula, $\sin\frac{u}{2}=\pm\sqrt{\frac{1-\cos u}{2}}$. Note that $195^\circ$ is in Quadrant III, where sine is negative, so we take the positive square root. $\sin 195^\circ$ $=\sin \frac{390^\circ}{2}$ $=-\sqrt{\frac{1-\cos 390^\circ}{2}}$ $=-\sqrt{\frac{1-\cos 30^\circ}{2}}$ $=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$ $=-\sqrt{\frac{(1-\frac{\sqrt{3}}{2})*2}{2*2}}$ $=-\sqrt{\frac{2-\sqrt{3}}{4}}$ $=-\frac{\sqrt{2-\sqrt{3}}}{\sqrt{4}}$ $=-\frac{\sqrt{2-\sqrt{3}}}{2}$
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