If sin `beta` is the GM between `sin alpha` and `cos alpha`, then cos `2beta` is equal to

Options

A. `2sin^(2)(pi/4-alpha)`
B. `2cos^(2)(pi/4-alpha)`
C. `2cos^(2)((3pi)/4+2alpha)`
D. `2sin^(2)(pi/4+alpha)`

Answer

To solve the problem, we need to find the value of \( \cos 2\beta \) given that \( \sin \beta \) is the geometric mean (GM) of \( \sin \alpha \) and \( \cos \alpha \).

### Step-by-Step Solution:

1. **Understanding the Geometric Mean**:
The geometric mean of two numbers \( a \) and \( b \) is given by \( \sqrt{ab} \). Here, we have:
\[ \sin \beta = \sqrt{\sin \alpha \cdot \cos \alpha} \]

2. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ \sin^2 \beta = \sin \alpha \cdot \cos \alpha \]

3. **Using the Double Angle Identity**:
We know from trigonometric identities that:
\[ \sin^2 \beta = \frac{1 - \cos 2\beta}{2} \]
Therefore, we can set up the equation:
\[ \frac{1 - \cos 2\beta}{2} = \sin \alpha \cdot \cos \alpha \]

4. **Multiplying by 2**:
To eliminate the fraction, multiply both sides by 2:
\[ 1 - \cos 2\beta = 2 \sin \alpha \cdot \cos \alpha \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ \cos 2\beta = 1 - 2 \sin \alpha \cdot \cos \alpha \]

6. **Using the Identity for \( \sin \alpha \cdot \cos \alpha \)**:
Recall that \( \sin \alpha \cdot \cos \alpha = \frac{1}{2} \sin 2\alpha \). Thus, we can substitute:
\[ \cos 2\beta = 1 - \sin 2\alpha \]

### Final Result:
The value of \( \cos 2\beta \) is:
\[ \cos 2\beta = 1 - \sin 2\alpha \]