## Precalculus (6th Edition) Blitzer

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