Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.1 - Definition II: Right Triangle Trigonometry - 2.1 Problem Set - Page 63: 49

Answer

- $ \frac{3\sqrt 3}{2} $ or $ - \frac{3}{2} \sqrt 3 $

Work Step by Step

Given expression = -3 $\sin2x$ = -3 $\sin2 (30^{\circ})$ ( replacing $ x$ with $ 30^{\circ}$) = -3 $\sin60^{\circ}$ = -3 $ \times (\frac{\sqrt 3}{2})$ ( substituting exact value of $\sin 60^{\circ}$) = - $ \frac{3\sqrt 3}{2} $ or $ - \frac{3}{2} \sqrt 3 $
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