If the points `A(0,0),B(cosalpha,sinalpha)`, and `C(cosbeta,sinbeta)` are the vertices of a right- angled triangle, then

Options

A. `sin.(alpha-beta)/(2)=(1)/(sqrt2)`
B. `cos.(alpha-beta)/(2)=(1)/(sqrt2)`
C. `cos.(alpha-beta)/(2)=-(1)/(sqrt2)`
D. `sin.(alpha-beta)/(2)=-(1)/(sqrt2)`

Explanation

To prove that the points \( A(0,0) \), \( B(\cos \alpha, \sin \alpha) \), and \( C(\cos \beta, \sin \beta) \) form a right-angled triangle, we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle
1. **Length of AB**:
\[ AB = \sqrt{(\cos \alpha - 0)^2 + (\sin \alpha - 0)^2} = \sqrt{\cos^2 \alpha + \sin^2 \alpha} = \sqrt{1} = 1 \]

2. **Length of AC**:
\[ AC = \sqrt{(\cos \beta - 0)^2 + (\sin \beta - 0)^2} = \sqrt{\cos^2 \beta + \sin^2 \beta} = \sqrt{1} = 1 \]

3. **Length of BC**:
\[ BC = \sqrt{(\cos \beta - \cos \alpha)^2 + (\sin \beta - \sin \alpha)^2} \]

### Step 2: Use the Pythagorean theorem
Since \( AB = AC = 1 \), we can assume that \( A \) is the right angle. Therefore, we need to check if:
\[ AB^2 + AC^2 = BC^2 \]
Substituting the lengths we found:
\[ 1^2 + 1^2 = BC^2 \Rightarrow 2 = BC^2 \Rightarrow BC = \sqrt{2} \]

### Step 3: Check the slopes of the lines
1. **Slope of AB**:
\[ \text{slope of } AB = \frac{\sin \alpha - 0}{\cos \alpha - 0} = \tan \alpha \]

2. **Slope of AC**:
\[ \text{slope of } AC = \frac{\sin \beta - 0}{\cos \beta - 0} = \tan \beta \]

### Step 4: Check if the lines are perpendicular
Two lines are perpendicular if the product of their slopes is -1:
\[ \tan \alpha \cdot \tan \beta = -1 \]
This implies:
\[ \tan \alpha = -\cot \beta \]

### Step 5: Relate angles
Using the identity \( \tan \alpha = \tan\left(-\frac{\pi}{2} - \beta\right) \), we can deduce:
\[ \alpha - \beta = -\frac{\pi}{2} \]

### Step 6: Conclusion
Thus, we have shown that \( A(0,0) \), \( B(\cos \alpha, \sin \alpha) \), and \( C(\cos \beta, \sin \beta) \) form a right-angled triangle at point \( A \).