Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.2 Trigonometric Functions: Unit Circle Approach - 6.2 Assess Your Understanding: 41

Answer

$0$

Work Step by Step

RECALL: $\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}$ With $\sin{\frac{\pi}{3}} = \frac{\sqrt3}{2}$, $\sin{\frac{\pi}{6}}=\frac{1}{2}$, and $\cos{\frac{\pi}{6}}=\frac{\sqrt3}{2}$, then $2\sin \dfrac {\pi }{3}-3\tan \dfrac {\pi }{6} \\=2\times \left(\dfrac {\sqrt {3}}{2}\right)-3\left(\dfrac {\sin \dfrac {\pi }{6}}{\cos \dfrac {\pi }{6}}\right) \\=\sqrt {3}-3\times \left(\dfrac {\dfrac {1}{2}}{\dfrac {\sqrt {3}}{2}}\right) \\=\sqrt {3}-\sqrt {3} \\=0$
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