Trigonometry 7th Edition

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

Chapter 5 - Section 5.4 - Half-Angle Formulas - 5.4 Problem Set - Page 305: 66

Answer

$c$

Work Step by Step

$270^{\circ} < A< 360^{\circ}$ $\text{Dividing by }2\text{:}$ $135^{\circ} < \dfrac{A}{2} <180^{\circ}$ $\therefore \dfrac{A}{2} \text{ terminates in } QII$ $\cos{\dfrac{A}{2}} <0$ $\text{Using the half angle formula:}$ $$\cos{\dfrac{A}{2}} = -\sqrt{\dfrac{1+\cos{A}}{2}}$$ $\cos{\dfrac{A}{2}} = -\sqrt{\dfrac{1+\dfrac{2}{5}}{2} }= -\dfrac{\sqrt{70}}{10}$
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