QUESTION 9 - 1 POINT
If $\sin (\theta)=\frac{7}{25}$, and $\theta$ is in Quadrant II, then what is $\sin \left(\frac{\theta}{2}\right)$ ? Type an exact answer, using radicals as needed. Simplify your answer completely and rationalize the denominator.

Provide your answer below:

Step 1 ：\( \sin^2(\theta) + \cos^2(\theta) = 1 \)

Step 2 ：\( \cos^2(\theta) = 1 - \sin^2(\theta) \)

Step 3 ：\( \cos^2(\theta) = 1 - \left(\frac{7}{25}\right)^2 \)

Step 4 ：\( \cos^2(\theta) = 1 - \frac{49}{625} \)

Step 5 ：\( \cos^2(\theta) = \frac{625}{625} - \frac{49}{625} \)

Step 6 ：\( \cos^2(\theta) = \frac{576}{625} \)

Step 7 ：\( \cos(\theta) = \pm \frac{24}{25} \)

Step 8 ：\( \cos(\theta) = -\frac{24}{25} \) (\text{since } \theta \text{ is in Quadrant II})

Step 9 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} \)

Step 10 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \left(-\frac{24}{25}\right)}{2}} \)

Step 11 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \frac{24}{25}}{2}} \)

Step 12 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{25}{25} + \frac{24}{25}} \)

Step 13 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{49}{25}} \)

Step 14 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{49}{25} \cdot \frac{2}{2}} \)

Step 15 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{98}{50}} \)

Step 16 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{49}{25}} \)

Step 17 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \frac{7}{5} \sqrt{\frac{1}{2}} \)

Step 18 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \frac{7}{5} \cdot \frac{\sqrt{2}}{2} \)

Step 19 ：\( \sin \left(\frac{\theta}{2}\right) = \pm \frac{7\sqrt{2}}{10} \)

Step 20 ：\( \sin \left(\frac{\theta}{2}\right) = \frac{7\sqrt{2}}{10} \) (\text{since } \frac{\theta}{2} \text{ will be in Quadrant I})

Step 21 ：\(\boxed{\sin \left(\frac{\theta}{2}\right) = \frac{7\sqrt{2}}{10}}\)