Question

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

Answer 1

Step 1: Determine the equation of the Line

Let $A=(\cos \theta ,\sin \theta )$ and $B=(\cos \varphi ,\sin \varphi )$ be the given points.

Then equation of $AB$ is of the form,

$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

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

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

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

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

$x(\sin \varphi -\sin \theta )-y(\cos \varphi -\cos \theta )+\sin (\theta -\varphi )=0$ $\left[\because \sin \left(\theta -\varphi \right)=\sin \theta \cos \varphi -\cos \theta \sin \varphi \right]$

Step 2: Determine the distance

We know that distance of this line from the origin,

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

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

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

$=\left|\frac{\sin (\theta -\varphi )}{\sqrt{2}·\sqrt{2{\sin }^2\left(\frac{\theta -\varphi }{2}\right)}}\right|$ $\left[\because 1-\cos x=2{\sin }^2\frac{x}{2},\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}\right]$

$\begin{array}{l}=\left|\frac{2\sin \left(\frac{\theta -\varphi }{2}\right)·\cos \left(\frac{\theta -\varphi }{2}\right)}{\sqrt{2}\sqrt{2}·\sin \left(\frac{\theta -\varphi }{2}\right)}\right|\\ =\left|\cos \left(\frac{\theta -\varphi }{2}\right)\right|\end{array}$

$\begin{array}{l}=\left|\cos \left(\frac{\theta -\varphi }{2}\right)\right|\end{array}$

Hence, the required perpendicular distance is$\left|\cos \left(\frac{\theta -\varphi }{2}\right)\right|$.