Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

Given that [tex]\cos 2 \alpha = \frac{2}{3}[/tex] and [tex]0^{\circ} \ \textless \ 2 \alpha \ \textless \ 90^{\circ}[/tex], determine the exact values of [tex]\sin \alpha[/tex], [tex]\cos \alpha[/tex], [tex]\tan \alpha[/tex], [tex]\csc \alpha[/tex], [tex]\sec \alpha[/tex], and [tex]\cot \alpha[/tex].

Sagot :

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]