Answer
The exact value of the trigonometric function $2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}$ is $\frac{7}{25}$.
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 $2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}$.
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 $\frac{7}{25}$.
