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: 52

Answer

The exact value of the trigonometric function $\tan \frac{\alpha }{2}$ is $\frac{1}{7}$.

Work Step by Step

Calculate the value of the hypotenuse. For the right angle triangle. $\text{hypotenuse}=\sqrt{\text{perpendicula}{{\text{r}}^{2}}+\text{bas}{{\text{e}}^{2}}}$ Substitute $24$ for the base and $7$ for the perpendicular. $\begin{align} & \text{hypotenuse}=\sqrt{{{\text{7}}^{2}}+\text{2}{{\text{4}}^{2}}} \\ & =\sqrt{625} \\ & =25 \end{align}$ Calculate the value of $\tan \frac{\alpha }{2}$. Recall the half angle formula. $\begin{align} & \tan \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \\ & =\sqrt{\frac{1-\left( \frac{\text{base}}{\text{hypotenuse}} \right)}{1+\left( \frac{\text{base}}{\text{hypotenuse}} \right)}} \end{align}$ Substitute $24$ for the base and $25$ for the hypotenuse. $\begin{align} & \tan \frac{\alpha }{2}=\sqrt{\frac{1-\left( \frac{24}{25} \right)}{1+\left( \frac{24}{25} \right)}} \\ & =\sqrt{\frac{\left( \frac{1}{25} \right)}{\left( \frac{49}{25} \right)}} \\ & =\sqrt{\frac{1}{49}} \\ & =\frac{1}{7} \end{align}$ Therefore, the exact value of the trigonometric function $\tan \frac{\alpha }{2}$ is $\frac{1}{7}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.