Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.3 - Trigonometric Functions - Exercises 1.3 - Page 28: 49

Answer

$\frac{1}{4}(2-\sqrt 3)$

Work Step by Step

$ sin^2\frac{\pi}{12}=\frac{1}{2}(1-cos\frac{\pi}{6})=\frac{1}{2}(1-\frac{\sqrt 3}{2})=\frac{1}{4}(2-\sqrt 3)$
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