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: 44

Answer

$2\sqrt {3}+1$

Work Step by Step

RECALL: (1) $\cot{\theta}=\dfrac{\cos{\theta}}{\sin{\theta}}$ (2) $\csc{\theta}=\dfrac{1}{\sin{\theta}}$ With $\sin{\frac{\pi}{3}}=\frac{\sqrt3}{2}$ and $\cos{\frac{\pi}{4}}=\sin{\frac{\pi}{4}}=\frac{\sqrt2}{2}$, then $3\csc\dfrac {\pi}{3}+\cot \dfrac {\pi }{4} \\=3\times \left(\dfrac {1}{\sin \dfrac {\pi }{3}}\right)+\dfrac {\cos \dfrac {\pi }{4}}{\sin \dfrac {\pi }{4}} \\=3\times \left(\dfrac {1}{\dfrac {\sqrt {3}}{2}}\right)+\dfrac {\dfrac {\sqrt {2}}{2}}{\dfrac {\sqrt {2}}{2}} \\=(3 \times \frac{2}{\sqrt3})+1 \\=2\sqrt {3}+1$
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