Find the perpendicular distance from the origin to the line joining the two points $\left(a \cos \alpha, a \sin \alpha\right)$ and $\left(a \cos \beta, a \sin \beta\right)$.
Given points are: $\;$ $\left(x_1, y_1\right) = \left(a \cos \alpha, a \sin \alpha\right)$; $\;$ $\left(x_2, y_2\right) = \left(a \cos \beta, a \sin \beta\right)$
Slope of the line through the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is
$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha} = \dfrac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha}$
i.e. $\;$ $m = \dfrac{2 \cos \left(\dfrac{\beta + \alpha}{2}\right) \sin \left(\dfrac{\beta - \alpha}{2}\right)}{-2 \sin \left(\dfrac{\beta + \alpha}{2}\right) \sin \left(\dfrac{\beta - \alpha}{2}\right)} = \dfrac{- \cos \left(\dfrac{\alpha + \beta}{2}\right)}{\sin \left(\dfrac{\alpha + \beta}{2}\right)}$
$\therefore \;$ Equation of the line through the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is of the form
$y - y_1 = m \left(x - x_1\right)$
i.e. $\;$ $y - a \sin \alpha = \left[\dfrac{- \cos \left(\dfrac{\alpha + \beta}{2}\right)}{\sin \left(\dfrac{\alpha + \beta}{2}\right)}\right] \left(x - a \cos \alpha\right)$
i.e. $\;$ $y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \sin \alpha \sin \left(\dfrac{\alpha + \beta}{2}\right) = - x \cos \left(\dfrac{\alpha + \beta}{2}\right) + a \cos \alpha \cos \left(\dfrac{\alpha + \beta}{2}\right)$
i.e. $\;$ $x \cos \left(\dfrac{\alpha + \beta}{2}\right) + y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \left[\sin \alpha \sin \left(\dfrac{\alpha + \beta}{2}\right) + \cos \alpha \cos \left(\dfrac{\alpha + \beta}{2}\right)\right] = 0$
i.e. $\;$ $x \cos \left(\dfrac{\alpha + \beta}{2}\right) + y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \cos \left(\alpha - \dfrac{\alpha + \beta}{2}\right) = 0$
i.e. $\;$ $x \cos \left(\dfrac{\alpha + \beta}{2}\right) + y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \cos \left(\dfrac{\alpha - \beta}{2}\right) = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of a straight line $Ax + By + C = 0$ gives
$A = \cos \left(\dfrac{\alpha + \beta}{2}\right)$, $\;$ $B = \sin \left(\dfrac{\alpha + \beta}{2}\right)$, $\;$ $C = - a \cos \left(\dfrac{\alpha - \beta}{2}\right)$
Given point: $\left(x_3, y_3\right) = \left(0, 0\right)$
Perpendicular distance from the point $\left(x_3, y_3\right)$ to the line $Ax + By + C = 0$ is $= \dfrac{Ax_3 + By_3 + C}{\sqrt{A^2 + B^2}}$
$\therefore \;$ Required perpendicular distance is
$= \left|\dfrac{- a \cos \left(\dfrac{\alpha - \beta}{2}\right)}{\sqrt{\cos^2 \left(\dfrac{\alpha + \beta}{2}\right) + \sin^2 \left(\dfrac{\alpha + \beta}{2}\right)}}\right| = \left|- a \cos \left(\dfrac{\alpha - \beta}{2}\right)\right| = a \cos \left(\dfrac{\alpha - \beta}{2}\right)$