Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 33


$\frac{\sqrt {2}-\sqrt {6}}{4}$

Work Step by Step

Using the addition formula, we have $\cos(\frac{\pi}{3}+\frac{\pi}{4})=\cos\frac{\pi}{3}\times\cos \frac{\pi}{4}-\sin\frac{\pi}{3}\times\sin\frac{\pi}{4}$ $=\frac{1}{2}\times\frac{1}{\sqrt 2}-\frac{\sqrt 3}{2}\times\frac{1}{\sqrt 2}=\frac{1-\sqrt 3}{2\sqrt 2}=\frac{\sqrt {2}-\sqrt {6}}{4}$
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