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 682: 89

Answer

The required solution is $\underline{\tan \frac{\alpha }{2}=\frac{1-\cos \alpha }{\sin \alpha }}$

Work Step by Step

We know that the half-angle formula gives the relationship between angle $\alpha,$ and angle $\frac{\alpha }{2}.$ The described half-angle formula for tangent can be verified from the right-hand side. That means from $\frac{1-\cos \alpha }{\sin \alpha }$. So, the expression for $\cos \alpha $ and $\sin \alpha $ can be obtained from the double-angle formula and replacing angle $\theta $ with angle $\frac{\alpha }{2}.$ This means, we start with the double-angle formula, $\begin{align} & \cos 2\theta =1-2si{{n}^{2}}\theta \\ & \sin 2\theta =2\sin \theta \cos \theta \end{align}$ Then, replace angle $\theta $ with angle $\frac{\alpha }{2}.\text{ Then,}$ $\begin{align} & \cos \alpha =1-2si{{n}^{2}}\frac{\alpha }{2} \\ & \sin \alpha =2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} \end{align}$ Now, substitute these values in $\frac{1-\cos \alpha }{\sin \alpha }$ $\begin{align} & \frac{1-\cos \alpha }{\sin \alpha }=\frac{1-\left( 1-2{{\sin }^{2}}\frac{\alpha }{2} \right)}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}} \\ & =\frac{2{{\sin }^{2}}\frac{\alpha }{2}}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}} \\ & =\frac{\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}} \\ & =\tan \frac{\alpha }{2} \end{align}$ Thus, the half-angle formula for tangent is verified.
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