What is the distance between the points P `( m cos 2 alpha , m sin 2 alpha)` and `Q( m cos 2 beta , m sin 2 beta) `?

Options

A. `|2m sin (alpha - beta)|`
B. `|2 m cos (alpha - beta)|`
C. `|m sin (2 alpha - 2 beta)|`
D. `|m sin (2 alpha - 2 beta)|`

Answer

To find the distance between the points \( P(m \cos 2\alpha, m \sin 2\alpha) \) and \( Q(m \cos 2\beta, m \sin 2\beta) \), 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 coordinates of points P and Q:**
- Point \( P \) has coordinates \( (m \cos 2\alpha, m \sin 2\alpha) \)
- Point \( Q \) has coordinates \( (m \cos 2\beta, m \sin 2\beta) \)

2. **Substitute the coordinates into the distance formula:**
- Here, \( x_1 = m \cos 2\alpha \), \( y_1 = m \sin 2\alpha \), \( x_2 = m \cos 2\beta \), and \( y_2 = m \sin 2\beta \).
- Thus, we have:
\[ d = \sqrt{(m \cos 2\beta - m \cos 2\alpha)^2 + (m \sin 2\beta - m \sin 2\alpha)^2} \]

3. **Factor out \( m \) from the expression:**
- Since \( m \) is common in both terms, we can factor it out:
\[ d = \sqrt{m^2 \left((\cos 2\beta - \cos 2\alpha)^2 + (\sin 2\beta - \sin 2\alpha)^2\right)} \]
- This simplifies to:
\[ d = m \sqrt{(\cos 2\beta - \cos 2\alpha)^2 + (\sin 2\beta - \sin 2\alpha)^2} \]

4. **Use the identity for the difference of angles:**
- We can use the identity \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \) and \( \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \) to simplify further.
- Let \( A = 2\beta \) and \( B = 2\alpha \):
\[ d = m \sqrt{(-2 \sin(\beta + \alpha) \sin(\beta - \alpha))^2 + (2 \cos(\beta + \alpha) \sin(\beta - \alpha))^2} \]

5. **Combine the terms:**
- This can be simplified to:
\[ d = m \sqrt{4 \sin^2(\beta - \alpha) \left(\sin^2(\beta + \alpha) + \cos^2(\beta + \alpha)\right)} \]
- Since \( \sin^2 x + \cos^2 x = 1 \):
\[ d = m \sqrt{4 \sin^2(\beta - \alpha)} = 2m |\sin(\beta - \alpha)| \]

### Final Answer:
Thus, the distance between the points \( P \) and \( Q \) is:
\[ d = 2m |\sin(\beta - \alpha)| \]