Question

Find the distances between the following pairs of points.
$(a\cos \alpha ,a\sin \alpha )$ and $(a\cos \beta ,a\sin \beta )$.

Answer 1

As per distance formula:

Distance between points $(x_1,y_1)$ and $(x_2,y_2)$$=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

Let, $P(a\cos \alpha ,a\sin \alpha )$ and $Q(a\cos \beta ,a\sin \beta )$

Then distance between $P$ and $Q$ is

$PQ=\sqrt{{(a\cos \beta -a\cos \alpha )}^2+{(a\sin \beta -a\sin \alpha )}^2}PQ=\sqrt{a^2{\cos }^2\beta +a^2{\cos }^2\alpha -2a^2\cos \alpha \cos \beta +a^2{\sin }^2\beta +a^2{\sin }^2\alpha -2a^2\sin \alpha \sin \beta }PQ=\sqrt{a^2({\cos }^2\beta +{\sin }^2\beta )+a^2({\cos }^2\alpha +{\sin }^2\alpha )-2a^2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )}$

Using identity: $\cos (a-b)=\cos a\cos b-\sin a\sin b$

$PQ=\sqrt{2a^2-2a^2\cos (\alpha -\beta )}$

Using $\cos 2a=1-2{\sin }^{2}a$

$PQ=\sqrt{2a^2-2a^2(1-2{\sin }^2\left(\frac{\alpha -\beta }{2}\right))}PQ=\sqrt{2a^2-2a^2+4a^2{\sin }^2\left(\frac{\alpha -\beta }{2}\right)}PQ=\sqrt{4a^2{\sin }^2\left(\frac{\alpha -\beta }{2}\right)}PQ=2a\sin \left(\frac{\alpha -\beta }{2}\right)$