Answer :
Given \(\sec \alpha = -\frac{5}{3}\) and \(\frac{\pi}{2} < \alpha < \pi\):
First, find \(\cos \alpha\):
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \Rightarrow \cos \alpha = -\frac{3}{5} \][/tex]
Next, find \(\sin \alpha\) using the Pythagorean identity, \(\sin^2 \alpha + \cos^2 \alpha = 1\):
[tex]\[ \sin^2 \alpha + \left(-\frac{3}{5}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \alpha + \frac{9}{25} = 1 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \][/tex]
[tex]\[ \sin \alpha = \sqrt{\frac{16}{25}} = \pm \frac{4}{5} \][/tex]
Since \(\frac{\pi}{2} < \alpha < \pi\), \(\sin \alpha\) is positive:
[tex]\[ \sin \alpha = \frac{4}{5} \][/tex]
Now, use the half-angle identities to find \(\sin \frac{\alpha}{2}\), \(\cos \frac{\alpha}{2}\), and \(\tan \frac{\alpha}{2}\):
1. \(\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\):
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - (-\frac{3}{5})}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \sqrt{\frac{4}{5}} \approx 0.8944 \][/tex]
2. \(\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\):
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + (-\frac{3}{5})}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{2}{10}} = \sqrt{\frac{1}{5}} = \sqrt{\frac{1}{5}} \approx 0.4472 \][/tex]
3. \(\tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\):
[tex]\[ \tan \frac{\alpha}{2} = \frac{\sqrt{\frac{4}{5}}}{\sqrt{\frac{1}{5}}} = \sqrt{4} = 2 \][/tex]
So, we have the values as:
a. \(\sin \frac{\alpha}{2} = \sqrt{\frac{4}{5}} \approx 0.8944\)
b. \(\cos \frac{\alpha}{2} = \sqrt{\frac{1}{5}} \approx 0.4472\)
c. [tex]\(\tan \frac{\alpha}{2} = 2\)[/tex]
First, find \(\cos \alpha\):
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \Rightarrow \cos \alpha = -\frac{3}{5} \][/tex]
Next, find \(\sin \alpha\) using the Pythagorean identity, \(\sin^2 \alpha + \cos^2 \alpha = 1\):
[tex]\[ \sin^2 \alpha + \left(-\frac{3}{5}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \alpha + \frac{9}{25} = 1 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \][/tex]
[tex]\[ \sin \alpha = \sqrt{\frac{16}{25}} = \pm \frac{4}{5} \][/tex]
Since \(\frac{\pi}{2} < \alpha < \pi\), \(\sin \alpha\) is positive:
[tex]\[ \sin \alpha = \frac{4}{5} \][/tex]
Now, use the half-angle identities to find \(\sin \frac{\alpha}{2}\), \(\cos \frac{\alpha}{2}\), and \(\tan \frac{\alpha}{2}\):
1. \(\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\):
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - (-\frac{3}{5})}{2}} = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} = \sqrt{\frac{4}{5}} \approx 0.8944 \][/tex]
2. \(\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\):
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + (-\frac{3}{5})}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{2}{10}} = \sqrt{\frac{1}{5}} = \sqrt{\frac{1}{5}} \approx 0.4472 \][/tex]
3. \(\tan \frac{\alpha}{2} = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\):
[tex]\[ \tan \frac{\alpha}{2} = \frac{\sqrt{\frac{4}{5}}}{\sqrt{\frac{1}{5}}} = \sqrt{4} = 2 \][/tex]
So, we have the values as:
a. \(\sin \frac{\alpha}{2} = \sqrt{\frac{4}{5}} \approx 0.8944\)
b. \(\cos \frac{\alpha}{2} = \sqrt{\frac{1}{5}} \approx 0.4472\)
c. [tex]\(\tan \frac{\alpha}{2} = 2\)[/tex]
