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: 105


The statement does not make sense.

Work Step by Step

We know that this statement does not make sense because to calculate the value of $\cos 100{}^\circ $ using the half angle formula, it is required to know the value of $\cos 200{}^\circ $. $\begin{align} & \cos \frac{\alpha }{2}=\pm \sqrt{\frac{1+\cos \alpha }{2}} \\ & \cos \frac{200{}^\circ }{2}=\pm \sqrt{\frac{1+\cos 200{}^\circ }{2}} \\ & \cos 100{}^\circ =\pm \sqrt{\frac{1+\cos 200{}^\circ }{2}} \end{align}$ Clearly, the value of $\cos 200{}^\circ $ can be computed with the help of the calculator but there is no known trigonometric value of $200{}^\circ $ angle.
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