Finding the Distance Between Two Points Using Coordinates

The problem asks us to find the distance between two points, P and Q, given their coordinates in terms of trigonometric functions and a constant 'm'. The points are given as \(P(m \cos 2\alpha, m \sin 2\alpha)\) and \(Q(m \cos 2\beta, m \sin 2\beta)\).

To solve this, we will use the standard distance formula for two points \((x_1, y_1)\) and \((x_2, y_2)\) in a Cartesian coordinate system, which is:

\(\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

In this case, we have:

• \(x_1 = m \cos 2\alpha\)
• \(y_1 = m \sin 2\alpha\)
• \(x_2 = m \cos 2\beta\)
• \(y_2 = m \sin 2\beta\)

Now, let's substitute these values into the distance formula:

\(PQ = \sqrt{(m \cos 2\beta - m \cos 2\alpha)^2 + (m \sin 2\beta - m \sin 2\alpha)^2}\)

We can factor out 'm' from both terms inside the square root:

\(PQ = \sqrt{[m (\cos 2\beta - \cos 2\alpha)]^2 + [m (\sin 2\beta - \sin 2\alpha)]^2}\)

\(PQ = \sqrt{m^2 (\cos 2\beta - \cos 2\alpha)^2 + m^2 (\sin 2\beta - \sin 2\alpha)^2}\)

Factor out \(m^2\) from under the square root:

\(PQ = \sqrt{m^2 [(\cos 2\beta - \cos 2\alpha)^2 + (\sin 2\beta - \sin 2\alpha)^2]}\)

\(PQ = |m| \sqrt{(\cos 2\beta - \cos 2\alpha)^2 + (\sin 2\beta - \sin 2\alpha)^2}\)

Now, let's expand the squared terms inside the square root:

\((\cos 2\beta - \cos 2\alpha)^2 = \cos^2 2\beta - 2 \cos 2\beta \cos 2\alpha + \cos^2 2\alpha\)

\((\sin 2\beta - \sin 2\alpha)^2 = \sin^2 2\beta - 2 \sin 2\beta \sin 2\alpha + \sin^2 2\alpha\)

Summing these expanded terms:

\((\cos^2 2\beta - 2 \cos 2\beta \cos 2\alpha + \cos^2 2\alpha) + (\sin^2 2\beta - 2 \sin 2\beta \sin 2\alpha + \sin^2 2\alpha)\)

Rearrange the terms:

\((\cos^2 2\beta + \sin^2 2\beta) + (\cos^2 2\alpha + \sin^2 2\alpha) - 2 (\cos 2\beta \cos 2\alpha + \sin 2\beta \sin 2\alpha)\)

Using the fundamental trigonometric identity \(\cos^2 \theta + \sin^2 \theta = 1\):

\(1 + 1 - 2 (\cos 2\beta \cos 2\alpha + \sin 2\beta \sin 2\alpha)\)

Using the angle subtraction formula for cosine, \(\cos(A - B) = \cos A \cos B + \sin A \sin B\), where \(A = 2\alpha\) and \(B = 2\beta\):

\(2 - 2 \cos (2\alpha - 2\beta)\)

So, the expression under the square root becomes:

\(PQ = |m| \sqrt{2 - 2 \cos (2(\alpha - \beta))}\)

Factor out 2 from the term inside the square root:

\(PQ = |m| \sqrt{2 (1 - \cos (2(\alpha - \beta)))}\)

Now, we use the half-angle identity \(1 - \cos 2\theta = 2 \sin^2 \theta\), where \(\theta = \alpha - \beta\):

\(1 - \cos (2(\alpha - \beta)) = 2 \sin^2 (\alpha - \beta)\)

Substitute this back into the distance formula:

\(PQ = |m| \sqrt{2 (2 \sin^2 (\alpha - \beta))}\)

\(PQ = |m| \sqrt{4 \sin^2 (\alpha - \beta)}\)

Take the square root:

\(PQ = |m| \times |2 \sin (\alpha - \beta)|\)

Since |m| is a magnitude and distance is non-negative, we combine the absolute values:

\(PQ = |2m \sin (\alpha - \beta)|\)

This is the distance between the points P and Q.

Revision Table: Key Concepts for Distance Calculation

Additional Information: Points on a Circle

The points given, \(P(m \cos 2\alpha, m \sin 2\alpha)\) and \(Q(m \cos 2\beta, m \sin 2\beta)\), have a specific form. If we consider a point \((x, y) = (m \cos \theta, m \sin \theta)\), the square of its distance from the origin (0,0) is \(x^2 + y^2 = (m \cos \theta)^2 + (m \sin \theta)^2 = m^2 \cos^2 \theta + m^2 \sin^2 \theta = m^2 (\cos^2 \theta + \sin^2 \theta) = m^2(1) = m^2\). Thus, the distance from the origin is \( \sqrt{m^2} = |m| \). This means that any point of the form \((m \cos \theta, m \sin \theta)\) lies on a circle centered at the origin (0,0) with a radius of \(|m|\).

The points P and Q are both on a circle centered at the origin with radius \(|m|\). The distance we calculated is the chord length connecting these two points on the circle. The angles \(2\alpha\) and \(2\beta\) represent the angles made by the position vectors of points P and Q with the positive x-axis, respectively.