Answer:
[tex]\displaystyle \sin{\frac{\theta}{2}} = \frac{5\sqrt{26}}{26}\approx 0.196[/tex].
Step-by-step explanation:
[tex]\displaystyle \theta\in \left(\pi, \frac{3\pi}{2}\right)[/tex],
such that
[tex]\displaystyle \frac{\theta}{2} \in \left(\frac{\pi}{2}, \frac{3\pi}{4}\right)[/tex].
As a result,
- [tex]\displaystyle 0 < \sin{\frac{\theta}{2}} <1[/tex], and
- [tex]\displaystyle -1 < \cos{\frac{\theta}{2}} <0[/tex].
[tex]\displaystyle \tan{\frac{\theta}{2}} = \frac{\sin{\displaystyle \frac{\theta}{2}}}{\displaystyle \cos{\frac{\theta}{2}}}[/tex],
such that
- [tex]\displaystyle \tan{\frac{\theta}{2}} <0[/tex].
Let
[tex]\displaystyle t = \tan{\frac{\theta}{2}}[/tex].
[tex]t < 0[/tex].
By the double angle identity for tangents.
[tex]\displaystyle \frac{\displaystyle 2\tan{\frac{\theta}{2}}}{1-\displaystyle \left(\tan{\frac{\theta}{2}}\right)^{2}} = \tan{\theta}[/tex].
[tex]\displaystyle \frac{2t}{1 - t^{2}} = \frac{5}{12}[/tex].
[tex]24t = 5 - 5t^{2}[/tex].
Solve this quadratic equation for [tex]t[/tex]:
- [tex]\displaystyle t_1 = \frac{1}{5}[/tex], and
- [tex]t_2 = -5[/tex].
Discard [tex]t_1[/tex] for it is not smaller than zero.
Let [tex]\displaystyle s = \sin{\frac{\theta}{2}}[/tex].
[tex]0 < s <1[/tex].
By the definition of tangents:
[tex]\displaystyle \tan{\frac{\theta}{2}} = \frac{\displaystyle \sin{\frac{\theta}{2}}}{\displaystyle \cos{\frac{\theta}{2}}}[/tex].
Apply the Pythagorean Algorithm to express the cosine of [tex]\displaystyle \frac{\theta}{2}[/tex] in terms of [tex]s[/tex]. Note that [tex]\displaystyle \cos{\frac{\theta}{2}}[/tex] is expected to be smaller than zero.
[tex]\displaystyle \cos{\frac{\theta}{2}} = -\sqrt{1 - \left(\sin{\frac{\theta}{2}}\right)^{2}}= - \sqrt{1 - s^{2}}[/tex].
Solve for [tex]s[/tex]:
[tex]\displaystyle \frac{s}{- \sqrt{1 - s^{2}}} = -5[/tex].
[tex]s^{2} =25(1 - s^{2})[/tex].
[tex]\displaystyle s = \sqrt{\frac{25}{26}} = \frac{5\sqrt{26}}{26}[/tex].
Therefore
[tex]\displaystyle \sin{\frac{\theta}{2}} = \frac{5\sqrt{26}}{26}\approx 0.196[/tex].
