## Precalculus (6th Edition) Blitzer

The exact value of the trigonometric function $2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}$ is $\frac{7}{25}$.
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 $2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}$. Recall the trigonometric half angle formula. \begin{align} & 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}=2\cdot \sqrt{\frac{1-\cos \alpha }{2}}\cdot \sqrt{\frac{1+\cos \alpha }{2}} \\ & =2\cdot \sqrt{\frac{1-\left( \frac{\text{base}}{\text{hypotenuse}} \right)}{2}}\cdot \sqrt{\frac{1+\left( \frac{\text{base}}{\text{hypotenuse}} \right)}{2}} \end{align} Substitute $24$ for the base and $25$ for the hypotenuse. \begin{align} & 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=2\cdot \sqrt{\frac{1-\left( \frac{\text{24}}{\text{25}} \right)}{2}}\cdot \sqrt{\frac{1+\left( \frac{\text{24}}{\text{25}} \right)}{2}} \\ & =2\cdot \left( \sqrt{\frac{1}{2\times 25}} \right)\cdot \left( \sqrt{\frac{49}{2\times 25}} \right) \\ & =2\cdot \left( \frac{7}{50} \right) \\ & =\frac{7}{25} \end{align} Therefore, the exact value of the trigonometric function $2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}$ is $\frac{7}{25}$.