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

Equation of the line joining points $(cos\theta ,sin\theta )$ and $(cos\varphi ,sin\varphi )$.

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

$\Rightarrow (cos\varphi -cos\theta )y-sin\theta cos\varphi +sin\theta cos\theta$

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

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

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

Now perpendicular distance from (0, 0) to the given line is

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

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

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

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

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