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 683: 104


The statement does not make sense..

Work Step by Step

This, statement does not make sense. Consider an example. Let us suppose we want to find the value of $\sin \left( {{112.5}^{\circ }} \right)$. Because $112.5{}^\circ =\frac{225}{2}$, the half angle formula for $\sin \frac{\alpha }{2}$ with $\alpha =225{}^\circ $ can be used: $\sin \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{2}}$ The value can be computed as given below: $\begin{align} & \cos 225{}^\circ =\cos \left( 180{}^\circ +45{}^\circ \right) \\ & =-\cos \left( 45{}^\circ \right) \\ & =-\frac{1}{\sqrt{2}} \end{align}$ Therefore, by using the aforementioned formula: $\begin{align} & \sin \left( {{112.5}^{\circ }} \right)=\pm \sqrt{\frac{1-\left( -\frac{1}{\sqrt{2}} \right)}{2}} \\ & =\pm \frac{\sqrt{2-\sqrt{2}}}{2} \end{align}$ $112.5{}^\circ $ lies in the second quadrant, therefore, $\sin \left( 112.5{}^\circ \right)$ will be positive. $\sin \left( {{112.5}^{\circ }} \right)=\frac{\sqrt{2-\sqrt{2}}}{2}$ Thus, the sign of the half angle is determined by the quadrant where the half angle $\frac{\alpha }{2}$ lies. So, the angle and its half do not necessarily lie on the same quadrant.
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