Precalculus (6th Edition) Blitzer

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

Chapter 6 - Cumulative Review Exercises - Page 801: 10


The value is $0$.

Work Step by Step

As we know: $\sin \left( \frac{\pi }{3} \right)=\frac{\sqrt{3}}{2}$ And $\tan \left( \frac{\pi }{6} \right)=\frac{1}{\sqrt{3}}$ Therefore, on solving equation (1), we get, $\begin{align} & 2\sin \frac{\pi }{3}-3\tan \frac{\pi }{6}=2\left( \frac{\sqrt{3}}{2} \right)-3\left( \frac{1}{\sqrt{3}} \right) \\ & =\sqrt{3}-\frac{3}{\sqrt{3}} \\ & =\sqrt{3}-\sqrt{3} \\ & =0 \end{align}$
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