The distance between the points `(cos theta, sin theta)` and `(sin theta, -cos theta)` is 

Options

A. `sqrt(3)`
B. `sqrt(2)`
C. 1
D. 0

Answer

To find the distance between the points \((\cos \theta, \sin \theta)\) and \((\sin \theta, -\cos \theta)\), we will use the distance formula. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step-by-Step Solution:

1. **Identify the Points**:
Let the first point \(A\) be \((x_1, y_1) = (\cos \theta, \sin \theta)\) and the second point \(B\) be \((x_2, y_2) = (\sin \theta, -\cos \theta)\).

2. **Substitute the Coordinates into the Distance Formula**:
Using the distance formula, we substitute the coordinates:
\[ d = \sqrt{(\sin \theta - \cos \theta)^2 + (-\cos \theta - \sin \theta)^2} \]

3. **Simplify the Expressions**:
First, simplify \((- \cos \theta - \sin \theta)\):
\[ -\cos \theta - \sin \theta = -(\sin \theta + \cos \theta) \]
Now, we can rewrite the distance formula:
\[ d = \sqrt{(\sin \theta - \cos \theta)^2 + (-(\sin \theta + \cos \theta))^2} \]

4. **Expand the Squares**:
Expand both squares:
\[ d = \sqrt{(\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta) + (\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta)} \]

5. **Combine Like Terms**:
Combine the terms inside the square root:
\[ d = \sqrt{(\sin^2 \theta + \cos^2 \theta) + (\sin^2 \theta + \cos^2 \theta)} \]
Since \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ d = \sqrt{1 + 1} = \sqrt{2} \]

6. **Final Answer**:
Therefore, the distance between the points \((\cos \theta, \sin \theta)\) and \((\sin \theta, -\cos \theta)\) is:
\[ d = \sqrt{2} \]