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: 7


$-3 \sqrt 2-2 \sqrt 3$

Work Step by Step

We know these trigonometric values $\cos \dfrac{3\pi}{4} =\dfrac{-\sqrt 2}{2}$ and $\tan \dfrac{-\pi}{3}=-\sqrt 3$ We evaluate the given expression to obtain: $6 \cos ( \dfrac{3\pi}{4}) +2 \tan (\dfrac{-\pi}{3})=6 \times \dfrac{-\sqrt 2}{2} +2 \times (-\sqrt 3)$ or, $=-3 \sqrt 2-2 \sqrt 3$
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