Two vectors $\overline A$ and $\overline B$ are orthogonal to each other, if their scalar product is zero i.e. $\overline A$.$\overline B$ = 0.

Here, $\overline A = \cos \omega t\widehat i + \sin \omega t\widehat j$

and $\overline B = \cos {{\omega t} \over 2}\widehat i + \sin {{\omega t} \over 2}\widehat j$

$\therefore \overline A .\overline B = \left( {\cos \omega t\widehat i + \sin \omega t\widehat j} \right)\left( {\cos {{\omega t} \over 2}\widehat i + \sin {{\omega t} \over 2}\widehat j} \right)$

= $\cos \omega t\cos {{\omega t} \over 2} + \sin \omega t\sin {{\omega t} \over 2}$
    $\left( \because {\widehat i.\widehat i = \widehat j.\widehat j = 1\,and\,\widehat i.\widehat j = \widehat j.\widehat i = 0} \right)$

= $\cos \left( {\omega t - {{\omega t} \over 2}} \right)$
    ($\because \cos (A - B) = cosAcosB + sinAsinB$)

But $\overline A .\overline B = 0$ (as $\overline A$ and $\overline B$ are orthogonal to each other)

$\therefore \cos \left( {\omega t - {{\omega t} \over 2}} \right) = 0$

$\cos \left( {\omega t - {{\omega t} \over 2}} \right) = \cos {\pi \over 2}\,\,or\,\omega t - {{\omega t} \over 2} = {\pi \over 2}$

${{\omega t} \over 2} = {\pi \over 2}\,or\,t = {\pi \over \omega }$