Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set - Page 561: 49


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

Work Step by Step

Since, $\cos \dfrac{\pi}{6}=\dfrac{\sqrt 3}{2}$ Now, we have $ f(\dfrac{\pi}{6})= 2 \cos \dfrac{\pi}{6}-\cos (2 \times \dfrac{\pi}{6}) $ or, $=2 \times \dfrac{\sqrt 3}{2}-\dfrac{1}{2}$ After simplifications, we have or, $=\dfrac{2\sqrt 3-1}{2}$
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