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: 34


$\sin(\frac{\pi}{3}-\frac{\pi}{4})=\frac{\sqrt {6}-\sqrt {2}}{4}$

Work Step by Step

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