Answer :
Given that \(\tan \alpha = \frac{15}{8}\) and \(0^\circ < \alpha < 90^\circ\), let's find the exact values of \(\sin \frac{\alpha}{2}\), \(\cos \frac{\alpha}{2}\), and \(\tan \frac{\alpha}{2}\).
### Step-by-Step Solution
Step 1: Calculate \(\sec \alpha\)
From the identity \(1 + \tan^2 \alpha = \sec^2 \alpha\), we have:
[tex]\[ \sec \alpha = \sqrt{1 + \tan^2 \alpha} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \left(\frac{15}{8}\right)^2} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \frac{225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{64 + 225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{289}{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{\sqrt{289}}{\sqrt{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{17}{8} \][/tex]
Step 2: Calculate \(\cos \alpha\)
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), we have:
[tex]\[ \cos \alpha = \frac{1}{\sec \alpha} \][/tex]
[tex]\[ \cos \alpha = \frac{1}{\frac{17}{8}} \][/tex]
[tex]\[ \cos \alpha = \frac{8}{17} \][/tex]
Step 3: Calculate \(\sin \alpha\)
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we have:
[tex]\[ \sin^2 \alpha = 1 - \cos^2 \alpha \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \left(\frac{8}{17}\right)^2 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{289 - 64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{225}{289} \][/tex]
[tex]\[ \sin \alpha = \sqrt{\frac{225}{289}} \][/tex]
[tex]\[ \sin \alpha = \frac{15}{17} \][/tex]
Step 4: Calculate \(\sin \frac{\alpha}{2}\)
Using the half-angle formula for sine, \(\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\):
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{17 - 8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{9}{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}} \][/tex]
Step 5: Calculate \(\cos \frac{\alpha}{2}\)
Using the half-angle formula for cosine, \(\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\):
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{17 + 8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{25}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{25}{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{\sqrt{25}}{\sqrt{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}} \][/tex]
Step 6: Calculate \(\tan \frac{\alpha}{2}\)
Using the half-angle formula for tangent, \(\tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}\):
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{1 + \frac{8}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{9}{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{3}{5} \][/tex]
### Answers
a. \(\sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}}\)
b. \(\cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}}\)
c. [tex]\(\tan \frac{\alpha}{2} = \frac{3}{5}\)[/tex]
### Step-by-Step Solution
Step 1: Calculate \(\sec \alpha\)
From the identity \(1 + \tan^2 \alpha = \sec^2 \alpha\), we have:
[tex]\[ \sec \alpha = \sqrt{1 + \tan^2 \alpha} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \left(\frac{15}{8}\right)^2} \][/tex]
[tex]\[ \sec \alpha = \sqrt{1 + \frac{225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{64 + 225}{64}} \][/tex]
[tex]\[ \sec \alpha = \sqrt{\frac{289}{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{\sqrt{289}}{\sqrt{64}} \][/tex]
[tex]\[ \sec \alpha = \frac{17}{8} \][/tex]
Step 2: Calculate \(\cos \alpha\)
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), we have:
[tex]\[ \cos \alpha = \frac{1}{\sec \alpha} \][/tex]
[tex]\[ \cos \alpha = \frac{1}{\frac{17}{8}} \][/tex]
[tex]\[ \cos \alpha = \frac{8}{17} \][/tex]
Step 3: Calculate \(\sin \alpha\)
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we have:
[tex]\[ \sin^2 \alpha = 1 - \cos^2 \alpha \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \left(\frac{8}{17}\right)^2 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{289 - 64}{289} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{225}{289} \][/tex]
[tex]\[ \sin \alpha = \sqrt{\frac{225}{289}} \][/tex]
[tex]\[ \sin \alpha = \frac{15}{17} \][/tex]
Step 4: Calculate \(\sin \frac{\alpha}{2}\)
Using the half-angle formula for sine, \(\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\):
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{17 - 8}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{2}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \sqrt{\frac{9}{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{34}} \][/tex]
[tex]\[ \sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}} \][/tex]
Step 5: Calculate \(\cos \frac{\alpha}{2}\)
Using the half-angle formula for cosine, \(\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\):
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{17 + 8}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{\frac{25}{17}}{2}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \sqrt{\frac{25}{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{\sqrt{25}}{\sqrt{34}} \][/tex]
[tex]\[ \cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}} \][/tex]
Step 6: Calculate \(\tan \frac{\alpha}{2}\)
Using the half-angle formula for tangent, \(\tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}\):
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{8}{17}}{1 + \frac{8}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \sqrt{\frac{9}{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{25}} \][/tex]
[tex]\[ \tan \frac{\alpha}{2} = \frac{3}{5} \][/tex]
### Answers
a. \(\sin \frac{\alpha}{2} = \frac{3}{\sqrt{34}}\)
b. \(\cos \frac{\alpha}{2} = \frac{5}{\sqrt{34}}\)
c. [tex]\(\tan \frac{\alpha}{2} = \frac{3}{5}\)[/tex]
