Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 680: 14

Answer

a. $ \frac{4\sqrt 5}{9}$ b. $ \frac{1}{9}$ c. $ 4\sqrt 5 $

Work Step by Step

Given $sin\theta=-\frac{2}{3}$ and $\theta$ in quadrant III, form a right triangle with sides $2, \sqrt 5, 3$. We have $cos\theta=-\frac{\sqrt 5}{3}$ and $tan\theta=\frac{2\sqrt 5}{5}$ a. $sin2\theta=2sin\theta cos\theta=2(-\frac{2}{3})(-\frac{\sqrt 5}{3})=\frac{4\sqrt 5}{9}$ b. $cos2\theta=1-2sin^2\theta=1-2(-\frac{2}{3})^2=\frac{1}{9}$ c. $tan2\theta=\frac{sin2\theta}{cos2\theta}=4\sqrt 5 $ (or use the Double Angle Formula to get the same result.)
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