If a particle is moving with angular velocity $=\omega$

Its angle of rotation is given by $\omega t$

Now displacement $=$ length of line $A B$

Position vector of a particle is given by

$\vec{R}=i a \cos \omega t+j a \sin \omega t$

$\vec{R}_{0}=a i$

displacement$, \vec{d}=\vec{R}-\vec{R}_{\mathrm{o}}$

$=a(\cos \omega t-1) i+a \sin \omega j$

$d=\sqrt{(a(\cos \omega t-1))^{2}+(a \sin \omega)^{2}}$

$=a \sqrt{2(1-\cos \omega t)}=a \sqrt{2 \times 2(\sin \omega t / 2)^{2}}=2 a \sin \omega t / 2$