Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Chapter Review - Review Exercises - Page 458: 6


$\dfrac{3 \sqrt 2}{2} - \dfrac{ 4\sqrt 3}{3}$

Work Step by Step

We know these trigonometric values $\sin 45^{\circ}=\sin \dfrac{\pi}{4} =\dfrac{ \sqrt 2}{2}$ and $\tan \dfrac{\pi}{6}=\dfrac{\sqrt 3}{3}$ We evaluate the given expression to obtain: $3 \sin 45^{\circ}-4 \tan \dfrac{\pi}{6} =3 \sin \dfrac{\pi}{4}-4 \tan \dfrac{\pi}{6}$ or, $=\dfrac{3 \sqrt 2}{2} - \dfrac{ 4\sqrt 3}{3}$
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