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: 24

Answer

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

Work Step by Step

Use the half-angle formula, $\cos\frac{u}{2}=\pm\sqrt{\frac{1+\cos u}{2}}$. Note that $\frac{3\pi}{8}$ is in Quadrant I, where cosine is positive, so we take the positive square root. $\cos \frac{3\pi}{8}$ $=\cos \frac{\frac{3\pi}{4}}{2}$ $=\sqrt{\frac{1+\cos \frac{3\pi}{4}}{2}}$ $=\sqrt{\frac{1+(-\frac{\sqrt{2}}{2})}{2}}$ $=\sqrt{\frac{(1-\frac{\sqrt{2}}{2})*2}{2*2}}$ $=\sqrt{\frac{2-\sqrt{2}}{4}}$ $=\frac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}$ $=\frac{1}{2}\sqrt{2-\sqrt{2}}$
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