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: 12

Answer

a. $ \frac{3}{5}$ b. $ \frac{4}{5}$ c. $ \frac{3}{4}$

Work Step by Step

Given $cot\theta=3$ and $\theta$ in quadrant III, form a right triangle with sides $1,3,\sqrt {10}$. We have $sin\theta=-\frac{\sqrt {10}}{10}$ and $cos\theta=-\frac{3\sqrt {10}}{10}$ a. $sin2\theta=2sin\theta cos\theta=2(-\frac{\sqrt {10}}{10})(-\frac{3\sqrt {10}}{10})=\frac{3}{5}$ b. $cos2\theta=1-2sin^2\theta=1-2(-\frac{\sqrt {10}}{10})^2=\frac{4}{5}$ c. $tan2\theta=\frac{sin2\theta}{cos2\theta}=\frac{3}{4}$ (or use the Double Angle Formula to get the same result.)
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