Question

Find perpendicular distance from the origin to the line joining the points $(\cos \theta ,\sin \theta )$ and $(\cos \varphi ,\sin \varphi )$

Answer 1

The equation of the line joining the points $(\cos \theta ,\sin \theta )$ and $(\cos \varphi ,\sin \varphi )$ is given by,

$y-\sin \theta =\frac{\sin \varphi -\sin \theta }{\cos \varphi -\cos \theta }(x-\cos \theta )$

$x(\sin \varphi -\sin \theta )+y(\cos \varphi -\cos \theta )+\cos \theta \sin \varphi -\cos \theta \sin \theta -\sin \theta \cos \varphi +\sin \theta \cos \theta =0$

$x(\sin \theta -\sin \varphi )+y(\cos \varphi -\cos \theta )+\sin (\varphi -\theta )=0$

Therefore, the perpendicular distance $\left(d\right)$ of the given line from point $(0,0)$ is

$d=\frac{\left|\right(0\left)\right(\sin \theta -\sin \varphi )+(0\left)\right(\cos \varphi -\cos \theta )+\sin (\varphi -\theta \left)\right|}{\sqrt{(\sin \theta -\sin \varphi {)}^2+(\cos \varphi -\cos \theta {)}^2}}$

$=\frac{\left|\sin \right(\varphi -\theta \left)\right|}{\sqrt{{\sin }^2\theta +{\sin }^2\varphi -2\sin \theta \sin \varphi +{\cos }^2\varphi +{\cos }^2\theta -2\cos \varphi \cos \theta }}$

$=\frac{\left|\sin \right(\varphi -\theta \left)\right|}{\sqrt{({\sin }^2\theta +{\cos }^2\theta )+({\sin }^2\varphi +{\cos }^2\varphi )-2(\sin \theta \sin \varphi +\cos \theta \cos \varphi )}}$

$=\frac{\left|\sin \right(\varphi -\theta \left)\right|}{\sqrt{2(1-\cos (\varphi -\theta \left)\right)}}$

$=\frac{\left|\sin \right(\varphi -\theta \left)\right|}{\sqrt{2\left\{2{\sin }^2\left\{\frac{\varphi -\theta }{2}\right\}\right\}}}$

$=\frac{\left|\sin \right(\varphi -\theta \left)\right|}{\left|2\sin \left\{\frac{\varphi -\theta }{2}\right\}\right|}$