Answer :
We begin with the given information: \(\cos(2\alpha) = \frac{1}{3}\) and \(0^\circ < 2\alpha < 90^\circ\). We need to find the values of \(\sin \alpha\), \(\cos \alpha\), \(\tan \alpha\), \(\csc \alpha\), \(\sec \alpha\), and \(\cot \alpha\).
### Step 1: Determine \(\cos \alpha\)
We use the double-angle identity for cosine:
[tex]\[ \cos(2\alpha) = 2 \cos^2(\alpha) - 1 \][/tex]
Given that \(\cos(2\alpha) = \frac{1}{3}\), we can set up the equation:
[tex]\[ 2 \cos^2(\alpha) - 1 = \frac{1}{3} \][/tex]
Solving for \(\cos^2(\alpha)\):
[tex]\[ 2 \cos^2(\alpha) = \frac{1}{3} + 1 = \frac{4}{3} \][/tex]
[tex]\[ \cos^2(\alpha) = \frac{4}{3} \cdot \frac{1}{2} = \frac{2}{3} \][/tex]
Taking the positive square root (since \(0^\circ < 2\alpha < 90^\circ\) implies \(0^\circ < \alpha < 45^\circ\), where cosine is positive):
[tex]\[ \cos(\alpha) = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3} \approx 0.8165 \][/tex]
### Step 2: Determine \(\sin \alpha\)
Using the Pythagorean identity:
[tex]\[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \][/tex]
[tex]\[ \sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \frac{2}{3} = \frac{1}{3} \][/tex]
Taking the positive square root:
[tex]\[ \sin(\alpha) = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.5774 \][/tex]
### Step 3: Determine \(\tan \alpha\)
Using the definition of tangent:
[tex]\[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{6}}{3}} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.7071 \][/tex]
### Step 4: Determine \(\csc \alpha\)
Using the definition of cosecant:
[tex]\[ \csc(\alpha) = \frac{1}{\sin(\alpha)} = \frac{1}{\frac{\sqrt{3}}{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \approx 1.7321 \][/tex]
### Step 5: Determine \(\sec \alpha\)
Using the definition of secant:
[tex]\[ \sec(\alpha) = \frac{1}{\cos(\alpha)} = \frac{1}{\frac{\sqrt{6}}{3}} = \frac{3}{\sqrt{6}} = \frac{3 \cdot \sqrt{6}}{6} = \frac{\sqrt{6}}{2} \approx 1.2247 \][/tex]
### Step 6: Determine \(\cot \alpha\)
Using the definition of cotangent:
[tex]\[ \cot(\alpha) = \frac{1}{\tan(\alpha)} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.4142 \][/tex]
### Summary
The exact values are:
- \(\sin(\alpha) = \frac{\sqrt{3}}{3} \approx 0.5774\)
- \(\cos(\alpha) = \frac{\sqrt{6}}{3} \approx 0.8165\)
- \(\tan(\alpha) = \frac{\sqrt{2}}{2} \approx 0.7071\)
- \(\csc(\alpha) = \sqrt{3} \approx 1.7321\)
- \(\sec(\alpha) = \frac{\sqrt{6}}{2} \approx 1.2247\)
- [tex]\(\cot(\alpha) = \sqrt{2} \approx 1.4142\)[/tex]
### Step 1: Determine \(\cos \alpha\)
We use the double-angle identity for cosine:
[tex]\[ \cos(2\alpha) = 2 \cos^2(\alpha) - 1 \][/tex]
Given that \(\cos(2\alpha) = \frac{1}{3}\), we can set up the equation:
[tex]\[ 2 \cos^2(\alpha) - 1 = \frac{1}{3} \][/tex]
Solving for \(\cos^2(\alpha)\):
[tex]\[ 2 \cos^2(\alpha) = \frac{1}{3} + 1 = \frac{4}{3} \][/tex]
[tex]\[ \cos^2(\alpha) = \frac{4}{3} \cdot \frac{1}{2} = \frac{2}{3} \][/tex]
Taking the positive square root (since \(0^\circ < 2\alpha < 90^\circ\) implies \(0^\circ < \alpha < 45^\circ\), where cosine is positive):
[tex]\[ \cos(\alpha) = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3} \approx 0.8165 \][/tex]
### Step 2: Determine \(\sin \alpha\)
Using the Pythagorean identity:
[tex]\[ \sin^2(\alpha) + \cos^2(\alpha) = 1 \][/tex]
[tex]\[ \sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \frac{2}{3} = \frac{1}{3} \][/tex]
Taking the positive square root:
[tex]\[ \sin(\alpha) = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.5774 \][/tex]
### Step 3: Determine \(\tan \alpha\)
Using the definition of tangent:
[tex]\[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{6}}{3}} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.7071 \][/tex]
### Step 4: Determine \(\csc \alpha\)
Using the definition of cosecant:
[tex]\[ \csc(\alpha) = \frac{1}{\sin(\alpha)} = \frac{1}{\frac{\sqrt{3}}{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \approx 1.7321 \][/tex]
### Step 5: Determine \(\sec \alpha\)
Using the definition of secant:
[tex]\[ \sec(\alpha) = \frac{1}{\cos(\alpha)} = \frac{1}{\frac{\sqrt{6}}{3}} = \frac{3}{\sqrt{6}} = \frac{3 \cdot \sqrt{6}}{6} = \frac{\sqrt{6}}{2} \approx 1.2247 \][/tex]
### Step 6: Determine \(\cot \alpha\)
Using the definition of cotangent:
[tex]\[ \cot(\alpha) = \frac{1}{\tan(\alpha)} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.4142 \][/tex]
### Summary
The exact values are:
- \(\sin(\alpha) = \frac{\sqrt{3}}{3} \approx 0.5774\)
- \(\cos(\alpha) = \frac{\sqrt{6}}{3} \approx 0.8165\)
- \(\tan(\alpha) = \frac{\sqrt{2}}{2} \approx 0.7071\)
- \(\csc(\alpha) = \sqrt{3} \approx 1.7321\)
- \(\sec(\alpha) = \frac{\sqrt{6}}{2} \approx 1.2247\)
- [tex]\(\cot(\alpha) = \sqrt{2} \approx 1.4142\)[/tex]
