The points `A(0, 0), B(cos alpha, sin alpha) and C(cos beta, sin beta)` are the vertices of a right angled triangle if :

Options

A. (a) `sin((alpha-beta)/(2))=(1)/(sqrt(2))`
B. (b) `cos((alpha-beta)/(2))=-(1)/(sqrt(2))`
C. (c) `cos((alpha-beta)/(2))=(1)/(sqrt(2))`
D. (d) `sin((alpha-beta)/(2))=-(1)/(sqrt(2))`

Answer

To determine the conditions under which the points \( A(0, 0) \), \( B(\cos \alpha, \sin \alpha) \), and \( C(\cos \beta, \sin \beta) \) form the vertices of a right-angled triangle, we will use the distance formula and the Pythagorean theorem.

### Step 1: Calculate the distances between the points

1. **Distance \( AB \)**:
\[ AB = \sqrt{(\cos \alpha - 0)^2 + (\sin \alpha - 0)^2} = \sqrt{\cos^2 \alpha + \sin^2 \alpha} = \sqrt{1} = 1 \]

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

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

### Step 2: Apply the Pythagorean theorem

For triangle \( ABC \) to be a right triangle, one of the sides must be the hypotenuse. We will assume \( BC \) is the hypotenuse. According to the Pythagorean theorem:
\[ BC^2 = AB^2 + AC^2 \]

Substituting the values we calculated:
\[ BC^2 = 1^2 + 1^2 = 2 \]
Thus,
\[ BC = \sqrt{2} \]

### Step 3: Set up the equation for \( BC \)

Now we need to equate the expression for \( BC \) to \( \sqrt{2} \):
\[ \sqrt{(\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2} = \sqrt{2} \]

### Step 4: Square both sides

Squaring both sides gives:
\[ (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 = 2 \]

### Step 5: Expand and simplify

Expanding the left-hand side:
\[ \cos^2 \alpha - 2\cos \alpha \cos \beta + \cos^2 \beta + \sin^2 \alpha - 2\sin \alpha \sin \beta + \sin^2 \beta = 2 \]

Using the identity \( \cos^2 x + \sin^2 x = 1 \):
\[ 1 - 2\cos \alpha \cos \beta + 1 - 2\sin \alpha \sin \beta = 2 \]
This simplifies to:
\[ 2 - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = 2 \]

### Step 6: Rearranging the equation

Rearranging gives:
\[ -2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = 0 \]
Thus,
\[ \cos \alpha \cos \beta + \sin \alpha \sin \beta = 0 \]

### Step 7: Use the cosine of angle difference identity

This can be rewritten using the cosine of the difference of angles:
\[ \cos(\alpha - \beta) = 0 \]

### Conclusion

The points \( A(0, 0) \), \( B(\cos \alpha, \sin \alpha) \), and \( C(\cos \beta, \sin \beta) \) form the vertices of a right-angled triangle if:
\[ \alpha - \beta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]