Answer :
To find the various trigonometric functions for \(\alpha\) given \(\cos \left(\frac{\alpha}{2}\right)=\frac{1}{8}\) and \(\frac{\pi}{4}<\frac{\alpha}{2}<\frac{3 \pi}{8}\), we will use trigonometric identities and properties.
1. Calculate \(\sin \left(\frac{\alpha}{2}\right)\):
Using the Pythagorean identity for trigonometric functions:
[tex]\[ \sin^2 \left( \frac{\alpha}{2} \right) + \cos^2 \left( \frac{\alpha}{2} \right) = 1 \][/tex]
Given \(\cos \left( \frac{\alpha}{2} \right) = \frac{1}{8}\):
[tex]\[ \cos^2 \left( \frac{\alpha}{2} \right) = \left(\frac{1}{8}\right)^2 = \frac{1}{64} \][/tex]
Thus,
[tex]\[ \sin^2 \left( \frac{\alpha}{2} \right) = 1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{63}{64} \][/tex]
Therefore,
[tex]\[ \sin \left( \frac{\alpha}{2} \right) = \sqrt{\frac{63}{64}} = \frac{\sqrt{63}}{8} \][/tex]
2. Calculate \(\sin \alpha\):
Using the double-angle identity for sine:
[tex]\[ \sin(\alpha) = 2 \sin \left( \frac{\alpha}{2} \right) \cos \left( \frac{\alpha}{2} \right) \][/tex]
Substituting the known values:
[tex]\[ \sin(\alpha) = 2 \cdot \frac{\sqrt{63}}{8} \cdot \frac{1}{8} = \frac{2\sqrt{63}}{64} = \frac{\sqrt{63}}{32} \][/tex]
3. Calculate \(\cos \alpha\):
Using the double-angle identity for cosine:
[tex]\[ \cos(\alpha) = 2 \cos^2 \left( \frac{\alpha}{2} \right) - 1 \][/tex]
Substituting the known value:
[tex]\[ \cos(\alpha) = 2 \left(\frac{1}{8}\right)^2 - 1 = 2 \cdot \frac{1}{64} - 1 = \frac{2}{64} - 1 = \frac{1}{32} - 1 = -\frac{31}{32} \][/tex]
4. Calculate \(\tan \alpha\):
Using the identity for tangent:
[tex]\[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} \][/tex]
Substituting the computed values:
[tex]\[ \tan(\alpha) = \frac{\frac{\sqrt{63}}{32}}{-\frac{31}{32}} = -\frac{\sqrt{63}}{31} \][/tex]
5. Calculate \(\csc \alpha\):
Using the identity for cosecant:
[tex]\[ \csc(\alpha) = \frac{1}{\sin(\alpha)} \][/tex]
Substituting the computed value for \(\sin(\alpha)\):
[tex]\[ \csc(\alpha) = \frac{1}{\frac{\sqrt{63}}{32}} = \frac{32}{\sqrt{63}} \][/tex]
Rationalizing the denominator:
[tex]\[ \csc(\alpha) = \frac{32}{\sqrt{63}} \cdot \frac{\sqrt{63}}{\sqrt{63}} = \frac{32\sqrt{63}}{63} \][/tex]
6. Calculate \(\sec \alpha\):
Using the identity for secant:
[tex]\[ \sec(\alpha) = \frac{1}{\cos(\alpha)} \][/tex]
Substituting the computed value for \(\cos(\alpha)\):
[tex]\[ \sec(\alpha) = \frac{1}{-\frac{31}{32}} = -\frac{32}{31} \][/tex]
7. Calculate \(\cot \alpha\):
Using the identity for cotangent:
[tex]\[ \cot(\alpha) = \frac{1}{\tan(\alpha)} \][/tex]
Substituting the computed value for \(\tan(\alpha)\):
[tex]\[ \cot(\alpha) = \frac{1}{-\frac{\sqrt{63}}{31}} = -\frac{31}{\sqrt{63}} \][/tex]
Rationalizing the denominator:
[tex]\[ \cot(\alpha) = -\frac{31}{\sqrt{63}} \cdot \frac{\sqrt{63}}{\sqrt{63}} = -\frac{31\sqrt{63}}{63} = -\frac{\sqrt{63}}{2} \][/tex]
Summarizing the results:
- \(\sin \alpha = \frac{\sqrt{63}}{32}\)
- \(\cos \alpha = -\frac{31}{32}\)
- \(\tan \alpha = -\frac{\sqrt{63}}{31}\)
- \(\csc \alpha = \frac{32\sqrt{63}}{63}\)
- \(\sec \alpha = -\frac{32}{31}\)
- [tex]\(\cot \alpha = -\frac{\sqrt{63}}{2}\)[/tex]
1. Calculate \(\sin \left(\frac{\alpha}{2}\right)\):
Using the Pythagorean identity for trigonometric functions:
[tex]\[ \sin^2 \left( \frac{\alpha}{2} \right) + \cos^2 \left( \frac{\alpha}{2} \right) = 1 \][/tex]
Given \(\cos \left( \frac{\alpha}{2} \right) = \frac{1}{8}\):
[tex]\[ \cos^2 \left( \frac{\alpha}{2} \right) = \left(\frac{1}{8}\right)^2 = \frac{1}{64} \][/tex]
Thus,
[tex]\[ \sin^2 \left( \frac{\alpha}{2} \right) = 1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{63}{64} \][/tex]
Therefore,
[tex]\[ \sin \left( \frac{\alpha}{2} \right) = \sqrt{\frac{63}{64}} = \frac{\sqrt{63}}{8} \][/tex]
2. Calculate \(\sin \alpha\):
Using the double-angle identity for sine:
[tex]\[ \sin(\alpha) = 2 \sin \left( \frac{\alpha}{2} \right) \cos \left( \frac{\alpha}{2} \right) \][/tex]
Substituting the known values:
[tex]\[ \sin(\alpha) = 2 \cdot \frac{\sqrt{63}}{8} \cdot \frac{1}{8} = \frac{2\sqrt{63}}{64} = \frac{\sqrt{63}}{32} \][/tex]
3. Calculate \(\cos \alpha\):
Using the double-angle identity for cosine:
[tex]\[ \cos(\alpha) = 2 \cos^2 \left( \frac{\alpha}{2} \right) - 1 \][/tex]
Substituting the known value:
[tex]\[ \cos(\alpha) = 2 \left(\frac{1}{8}\right)^2 - 1 = 2 \cdot \frac{1}{64} - 1 = \frac{2}{64} - 1 = \frac{1}{32} - 1 = -\frac{31}{32} \][/tex]
4. Calculate \(\tan \alpha\):
Using the identity for tangent:
[tex]\[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} \][/tex]
Substituting the computed values:
[tex]\[ \tan(\alpha) = \frac{\frac{\sqrt{63}}{32}}{-\frac{31}{32}} = -\frac{\sqrt{63}}{31} \][/tex]
5. Calculate \(\csc \alpha\):
Using the identity for cosecant:
[tex]\[ \csc(\alpha) = \frac{1}{\sin(\alpha)} \][/tex]
Substituting the computed value for \(\sin(\alpha)\):
[tex]\[ \csc(\alpha) = \frac{1}{\frac{\sqrt{63}}{32}} = \frac{32}{\sqrt{63}} \][/tex]
Rationalizing the denominator:
[tex]\[ \csc(\alpha) = \frac{32}{\sqrt{63}} \cdot \frac{\sqrt{63}}{\sqrt{63}} = \frac{32\sqrt{63}}{63} \][/tex]
6. Calculate \(\sec \alpha\):
Using the identity for secant:
[tex]\[ \sec(\alpha) = \frac{1}{\cos(\alpha)} \][/tex]
Substituting the computed value for \(\cos(\alpha)\):
[tex]\[ \sec(\alpha) = \frac{1}{-\frac{31}{32}} = -\frac{32}{31} \][/tex]
7. Calculate \(\cot \alpha\):
Using the identity for cotangent:
[tex]\[ \cot(\alpha) = \frac{1}{\tan(\alpha)} \][/tex]
Substituting the computed value for \(\tan(\alpha)\):
[tex]\[ \cot(\alpha) = \frac{1}{-\frac{\sqrt{63}}{31}} = -\frac{31}{\sqrt{63}} \][/tex]
Rationalizing the denominator:
[tex]\[ \cot(\alpha) = -\frac{31}{\sqrt{63}} \cdot \frac{\sqrt{63}}{\sqrt{63}} = -\frac{31\sqrt{63}}{63} = -\frac{\sqrt{63}}{2} \][/tex]
Summarizing the results:
- \(\sin \alpha = \frac{\sqrt{63}}{32}\)
- \(\cos \alpha = -\frac{31}{32}\)
- \(\tan \alpha = -\frac{\sqrt{63}}{31}\)
- \(\csc \alpha = \frac{32\sqrt{63}}{63}\)
- \(\sec \alpha = -\frac{32}{31}\)
- [tex]\(\cot \alpha = -\frac{\sqrt{63}}{2}\)[/tex]
