Show that the distance between the points `P(acosalpha, asinalpha)` and `Q(a cos 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 \( Q(a \cos \beta, a \sin \beta) \) is given by \( 2a \sin\left(\frac{\alpha - \beta}{2}\right) \), we will follow these steps:

### Step 1: Write the distance formula
The distance \( d \) between two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) in the coordinate plane is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For our points \( P(a \cos \alpha, a \sin \alpha) \) and \( Q(a \cos \beta, a \sin \beta) \), we have:
\[ d = \sqrt{(a \cos \beta - a \cos \alpha)^2 + (a \sin \beta - a \sin \alpha)^2} \]

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

### Step 3: Expand the terms inside the square root
Now we will expand the expression:
\[ (\cos \beta - \cos \alpha)^2 + (\sin \beta - \sin \alpha)^2 \]
Using the identity \( (x - y)^2 = x^2 - 2xy + y^2 \):
\[ = \cos^2 \beta - 2 \cos \beta \cos \alpha + \cos^2 \alpha + \sin^2 \beta - 2 \sin \beta \sin \alpha + \sin^2 \alpha \]
Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \):
\[ = (1 + 1) - 2(\cos \beta \cos \alpha + \sin \beta \sin \alpha) \]
This simplifies to:
\[ = 2 - 2(\cos \beta \cos \alpha + \sin \beta \sin \alpha) \]

### Step 4: Use the cosine of the difference formula
Using the cosine of the difference identity:
\[ \cos(\beta - \alpha) = \cos \beta \cos \alpha + \sin \beta \sin \alpha \]
We can rewrite the expression as:
\[ = 2 - 2 \cos(\beta - \alpha) \]
Thus, we have:
\[ d = a \sqrt{2 - 2 \cos(\beta - \alpha)} \]

### Step 5: Simplify using the sine double angle identity
Using the identity \( 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \):
\[ 2 - 2 \cos(\beta - \alpha) = 2(1 - \cos(\beta - \alpha)) = 2 \cdot 2 \sin^2\left(\frac{\beta - \alpha}{2}\right) = 4 \sin^2\left(\frac{\beta - \alpha}{2}\right) \]
So we have:
\[ d = a \sqrt{4 \sin^2\left(\frac{\beta - \alpha}{2}\right)} = a \cdot 2 \sin\left(\frac{\beta - \alpha}{2}\right) \]

### Final Result
Thus, the distance \( d \) between points \( P \) and \( Q \) is:
\[ d = 2a \sin\left(\frac{\beta - \alpha}{2}\right) \]