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

Options

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

Answer

To solve the problem where \( \sin \beta \) is the geometric mean between \( \sin \alpha \) and \( \cos \alpha \), we need to find \( \cos 2\beta \). Here’s a step-by-step solution:

### Step 1: Set up the equation for the geometric mean
Given that \( \sin \beta \) is the geometric mean of \( \sin \alpha \) and \( \cos \alpha \), we can express this mathematically as:
\[ \sin \beta = \sqrt{\sin \alpha \cdot \cos \alpha} \]

### Step 2: Square both sides
To eliminate the square root, we square both sides:
\[ \sin^2 \beta = \sin \alpha \cdot \cos \alpha \]

### Step 3: Use the double angle identity
We know from trigonometric identities that:
\[ \sin \alpha \cdot \cos \alpha = \frac{1}{2} \sin(2\alpha) \]
Thus, we can rewrite our equation as:
\[ \sin^2 \beta = \frac{1}{2} \sin(2\alpha) \]

### Step 4: Relate \( \sin^2 \beta \) to \( \cos 2\beta \)
Using the identity \( \cos 2\beta = 1 - 2\sin^2 \beta \), we can substitute \( \sin^2 \beta \):
\[ \cos 2\beta = 1 - 2\left(\frac{1}{2} \sin(2\alpha)\right) \]
This simplifies to:
\[ \cos 2\beta = 1 - \sin(2\alpha) \]

### Step 5: Final expression
Thus, we have:
\[ \cos 2\beta = 1 - \sin(2\alpha) \]

### Conclusion
The final answer for \( \cos 2\beta \) is:
\[ \cos 2\beta = 1 - \sin(2\alpha) \]