Find the perpendicular distance from the orig | vedclass

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(C) The equation of the line passing through $(\cos 	heta, \sin 	heta)$ and $(\cos \phi, \sin \phi)$ is given by:$y - \sin 	heta = \frac{\sin \phi

Vedclass · Accepted Answer

$\left| \cos \left( \frac{	heta - \phi}{2} ight) ight|$

Vedclass · Answer

(C) The equation of the line passing through $(\cos 	heta, \sin 	heta)$ and $(\cos \phi, \sin \phi)$ is given by:$y - \sin 	heta = \frac{\sin \phi - \sin 	heta}{\cos \phi - \cos 	heta} (x - \cos 	heta)$Using the formula $\sin C - \sin D = 2 \sin \frac{C-D}{2} \cos \frac{C+D}{2}$ and $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$,the slope is:$m = \frac{2 \sin \frac{\phi-	heta}{2} \cos \frac{\phi+	heta}{2}}{-2 \sin \frac{\phi+	heta}{2} \sin \frac{\phi-	heta}{2}} = -\cot \left( \frac{\phi+	heta}{2} ight)$Thus,the equation is $y - \sin 	heta = -\cot \left( \frac{\phi+	heta}{2} ight) (x - \cos 	heta)$.Multiplying by $\sin \left( \frac{\phi+	heta}{2} ight)$:$y \sin \left( \frac{\phi+	heta}{2} ight) - \sin 	heta \sin \left( \frac{\phi+	heta}{2} ight) = -x \cos \left( \frac{\phi+	heta}{2} ight) + \cos 	heta \cos \left( \frac{\phi+	heta}{2} ight)$$x \cos \left( \frac{\phi+	heta}{2} ight) + y \sin \left( \frac{\phi+	heta}{2} ight) = \cos 	heta \cos \left( \frac{\phi+	heta}{2} ight) + \sin 	heta \sin \left( \frac{\phi+	heta}{2} ight)$Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$:$x \cos \left( \frac{\phi+	heta}{2} ight) + y \sin \left( \frac{\phi+	heta}{2} ight) - \cos \left( \frac{\phi-	heta}{2} ight) = 0$The perpendicular distance from $(0,0)$ is:$d = \frac{| -\cos \left( \frac{\phi-	heta}{2} ight) |}{\sqrt{\cos^2 \left( \frac{\phi+	heta}{2} ight) + \sin^2 \left( \frac{\phi+	heta}{2} ight)}} = \left| \cos \left( \frac{\phi-	heta}{2} ight) ight|$

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

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