Show that the distance between the points `P(acosalpha, asinalpha)` and `Q(a sin beta,a sinbeta)` is `2a sin(a-b)/(2)`

Explanation

To show that the distance between the points \( P(a \cos \alpha, a \sin \alpha) \) and \( Q(a \sin \beta, a \sin \beta) \) is \( 2a \sin \left( \frac{\alpha - \beta}{2} \right) \), we will follow these steps:

### Step 1: Write the Distance Formula
The distance \( d \) between two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
In our case, the coordinates of the points are:
- \( P(a \cos \alpha, a \sin \alpha) \)
- \( Q(a \sin \beta, a \sin \beta) \)

### Step 2: Substitute the Coordinates
Substituting the coordinates of points \( P \) and \( Q \) into the distance formula:
\[ d = \sqrt{(a \sin \beta - a \cos \alpha)^2 + (a \sin \beta - a \sin \alpha)^2} \]

### Step 3: Factor out \( a^2 \)
We can factor out \( a^2 \) from the square root:
\[ d = \sqrt{a^2 \left( (\sin \beta - \cos \alpha)^2 + (\sin \beta - \sin \alpha)^2 \right)} = a \sqrt{(\sin \beta - \cos \alpha)^2 + (\sin \beta - \sin \alpha)^2} \]

### Step 4: Expand the Squares
Now we will expand the squares inside the square root:
\[ d = a \sqrt{(\sin^2 \beta - 2 \sin \beta \cos \alpha + \cos^2 \alpha) + (\sin^2 \beta - 2 \sin \beta \sin \alpha + \sin^2 \alpha)} \]

### Step 5: Combine Like Terms
Combining the terms gives:
\[ d = a \sqrt{2 \sin^2 \beta - 2 \sin \beta (\cos \alpha + \sin \alpha) + (\cos^2 \alpha + \sin^2 \alpha)} \]
Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\[ d = a \sqrt{2 \sin^2 \beta - 2 \sin \beta (\cos \alpha + \sin \alpha) + 1} \]

### Step 6: Use the Cosine of Angle Difference
Now we will use the cosine of the angle difference formula:
\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]
This implies:
\[ 1 - \cos(\alpha - \beta) = 2 \sin^2\left(\frac{\alpha - \beta}{2}\right) \]
Thus, we can rewrite the distance as:
\[ d = 2a \sin\left(\frac{\alpha - \beta}{2}\right) \]

### Final Result
Thus, we have shown that:
\[ d = 2a \sin\left(\frac{\alpha - \beta}{2}\right) \]