Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 17

Answer

$\frac{1}{2}\sqrt{2-\sqrt{3}}$

Work Step by Step

Use the half-angle formula, $\sin\frac{u}{2}=\pm\sqrt{\frac{1-\cos u}{2}}$. Note that $15^\circ$ is in Quadrant I, where sine is positive, so we take the positive square root. $\sin 15^\circ$ $=\sin \frac{30^\circ}{2}$ $=\sqrt{\frac{1-\cos 30^\circ}{2}}$ $=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$ $=\sqrt{\frac{(1-\frac{\sqrt{3}}{2})*2}{2*2}}$ $=\sqrt{\frac{2-\sqrt{3}}{4}}$ $=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{4}}$ $=\frac{1}{2}\sqrt{2-\sqrt{3}}$
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