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To find the exact values of \(\sin \alpha\), \(\cos \alpha\), \(\tan \alpha\), \(\csc \alpha\), \(\sec \alpha\), and \(\cot \alpha\) given that \(\cos 2\alpha = \frac{2}{3}\) and \(0^\circ < 2\alpha < 90^\circ\), follow these steps:
### Step 1: Use the Double Angle Formula for Cosine
Given:
[tex]\[ \cos 2\alpha = \frac{2}{3} \][/tex]
We know the double angle formula for cosine:
[tex]\[ \cos 2\alpha = 2\cos^2 \alpha - 1 \][/tex]
Set \(\cos 2\alpha = \frac{2}{3}\):
[tex]\[ 2\cos^2 \alpha - 1 = \frac{2}{3} \][/tex]
### Step 2: Solve for \(\cos^2 \alpha\)
Rearrange the equation to isolate \(\cos^2 \alpha\):
[tex]\[ 2\cos^2 \alpha - 1 = \frac{2}{3} \][/tex]
[tex]\[ 2\cos^2 \alpha = \frac{2}{3} + 1 \][/tex]
[tex]\[ 2\cos^2 \alpha = \frac{2}{3} + \frac{3}{3} \][/tex]
[tex]\[ 2\cos^2 \alpha = \frac{5}{3} \][/tex]
Divide both sides by 2:
[tex]\[ \cos^2 \alpha = \frac{5}{6} \][/tex]
### Step 3: Find \(\cos \alpha\)
Take the positive square root (since \(0^\circ < \alpha < 45^\circ\) implies \(\cos \alpha\) is positive):
[tex]\[ \cos \alpha = \sqrt{\frac{5}{6}} \approx 0.9128709291752769 \][/tex]
### Step 4: Use the Pythagorean Identity to Find \(\sin \alpha\)
Using the identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Substitute \(\cos^2 \alpha = \frac{5}{6}\):
[tex]\[ \sin^2 \alpha + \frac{5}{6} = 1 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{5}{6} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{6}{6} - \frac{5}{6} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{1}{6} \][/tex]
Take the positive square root (since \(0^\circ < \alpha < 45^\circ\) implies \(\sin \alpha\) is positive):
[tex]\[ \sin \alpha = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \approx 0.40824829046386296 \][/tex]
### Step 5: Calculate Other Trigonometric Functions
#### \(\tan \alpha\)
Using:
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \][/tex]
[tex]\[ \tan \alpha = \frac{\frac{\sqrt{6}}{6}}{\sqrt{\frac{5}{6}}} \][/tex]
[tex]\[ \tan \alpha = \frac{\frac{\sqrt{6}}{6}}{\frac{\sqrt{30}}{6}} = \frac{\sqrt{6}}{\sqrt{30}} = \frac{1}{\sqrt{5}} \approx 0.44721359549995787 \][/tex]
#### \(\csc \alpha\)
Using:
[tex]\[ \csc \alpha = \frac{1}{\sin \alpha} \][/tex]
[tex]\[ \csc \alpha = \frac{1}{\frac{\sqrt{6}}{6}} = \frac{6}{\sqrt{6}} = \sqrt{6} \approx 2.4494897427831783 \][/tex]
#### \(\sec \alpha\)
Using:
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \][/tex]
[tex]\[ \sec \alpha = \frac{1}{\sqrt{\frac{5}{6}}} = \frac{\sqrt{6}}{\sqrt{5}} = \frac{\sqrt{30}}{5} \approx 1.0954451150103321 \][/tex]
#### \(\cot \alpha\)
Using:
[tex]\[ \cot \alpha = \frac{1}{\tan \alpha} \][/tex]
[tex]\[ \cot \alpha = \sqrt{5} \approx 2.2360679774997902 \][/tex]
Thus, the exact values are:
[tex]\[ \begin{align*} \sin \alpha &\approx 0.40824829046386296, \\ \cos \alpha &\approx 0.9128709291752769, \\ \tan \alpha &\approx 0.44721359549995787, \\ \csc \alpha &\approx 2.4494897427831783, \\ \sec \alpha &\approx 1.0954451150103321, \\ \cot \alpha &\approx 2.2360679774997902. \end{align*} \][/tex]
### Step 1: Use the Double Angle Formula for Cosine
Given:
[tex]\[ \cos 2\alpha = \frac{2}{3} \][/tex]
We know the double angle formula for cosine:
[tex]\[ \cos 2\alpha = 2\cos^2 \alpha - 1 \][/tex]
Set \(\cos 2\alpha = \frac{2}{3}\):
[tex]\[ 2\cos^2 \alpha - 1 = \frac{2}{3} \][/tex]
### Step 2: Solve for \(\cos^2 \alpha\)
Rearrange the equation to isolate \(\cos^2 \alpha\):
[tex]\[ 2\cos^2 \alpha - 1 = \frac{2}{3} \][/tex]
[tex]\[ 2\cos^2 \alpha = \frac{2}{3} + 1 \][/tex]
[tex]\[ 2\cos^2 \alpha = \frac{2}{3} + \frac{3}{3} \][/tex]
[tex]\[ 2\cos^2 \alpha = \frac{5}{3} \][/tex]
Divide both sides by 2:
[tex]\[ \cos^2 \alpha = \frac{5}{6} \][/tex]
### Step 3: Find \(\cos \alpha\)
Take the positive square root (since \(0^\circ < \alpha < 45^\circ\) implies \(\cos \alpha\) is positive):
[tex]\[ \cos \alpha = \sqrt{\frac{5}{6}} \approx 0.9128709291752769 \][/tex]
### Step 4: Use the Pythagorean Identity to Find \(\sin \alpha\)
Using the identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Substitute \(\cos^2 \alpha = \frac{5}{6}\):
[tex]\[ \sin^2 \alpha + \frac{5}{6} = 1 \][/tex]
[tex]\[ \sin^2 \alpha = 1 - \frac{5}{6} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{6}{6} - \frac{5}{6} \][/tex]
[tex]\[ \sin^2 \alpha = \frac{1}{6} \][/tex]
Take the positive square root (since \(0^\circ < \alpha < 45^\circ\) implies \(\sin \alpha\) is positive):
[tex]\[ \sin \alpha = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \approx 0.40824829046386296 \][/tex]
### Step 5: Calculate Other Trigonometric Functions
#### \(\tan \alpha\)
Using:
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \][/tex]
[tex]\[ \tan \alpha = \frac{\frac{\sqrt{6}}{6}}{\sqrt{\frac{5}{6}}} \][/tex]
[tex]\[ \tan \alpha = \frac{\frac{\sqrt{6}}{6}}{\frac{\sqrt{30}}{6}} = \frac{\sqrt{6}}{\sqrt{30}} = \frac{1}{\sqrt{5}} \approx 0.44721359549995787 \][/tex]
#### \(\csc \alpha\)
Using:
[tex]\[ \csc \alpha = \frac{1}{\sin \alpha} \][/tex]
[tex]\[ \csc \alpha = \frac{1}{\frac{\sqrt{6}}{6}} = \frac{6}{\sqrt{6}} = \sqrt{6} \approx 2.4494897427831783 \][/tex]
#### \(\sec \alpha\)
Using:
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \][/tex]
[tex]\[ \sec \alpha = \frac{1}{\sqrt{\frac{5}{6}}} = \frac{\sqrt{6}}{\sqrt{5}} = \frac{\sqrt{30}}{5} \approx 1.0954451150103321 \][/tex]
#### \(\cot \alpha\)
Using:
[tex]\[ \cot \alpha = \frac{1}{\tan \alpha} \][/tex]
[tex]\[ \cot \alpha = \sqrt{5} \approx 2.2360679774997902 \][/tex]
Thus, the exact values are:
[tex]\[ \begin{align*} \sin \alpha &\approx 0.40824829046386296, \\ \cos \alpha &\approx 0.9128709291752769, \\ \tan \alpha &\approx 0.44721359549995787, \\ \csc \alpha &\approx 2.4494897427831783, \\ \sec \alpha &\approx 1.0954451150103321, \\ \cot \alpha &\approx 2.2360679774997902. \end{align*} \][/tex]
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