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: 49

Answer

The exact value of the trigonometric function $\tan \frac{\theta }{2}$ is $\frac{1}{3}$.

Work Step by Step

Calculate the value of $\tan \frac{\theta }{2}$. Recall the half angle formula. $\begin{align} & \tan \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }} \\ & =\sqrt{\frac{1-\left( \frac{\text{base}}{\text{hypotenuse}} \right)}{1-\left( \frac{\text{base}}{\text{hypotenuse}} \right)}} \end{align}$ Substitute $4$ for the base and $5$ for the hypotenuse. $\begin{align} & \tan \frac{\theta }{2}=\sqrt{\frac{1-\left( \frac{\text{4}}{\text{5}} \right)}{1+\left( \frac{\text{4}}{\text{5}} \right)}} \\ & =\sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}} \\ & =\sqrt{\frac{1}{9}} \\ & =\frac{1}{3} \end{align}$ Therefore, the exact value of the trigonometric function $\tan \frac{\theta }{2}$ is $\frac{1}{3}$.
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