Answer
$\sin 528 \pi t $
Work Step by Step
Amplitude, $ A=|1|=1$
Now, $ f=\dfrac{\omega}{2 \pi} \implies \omega =2 \pi f $
and period, $ P=\dfrac{2 \pi}{\omega} $
So, $\omega =2 \pi (264)=528 \pi $
Therefore, $ d=A \sin \omega t= \sin 528 \pi t $
