Answer
$-\frac{\sqrt 2}{2}$
Work Step by Step
As the period of sine function is $360^{\circ}$,
$\sin(1305^{\circ})=\sin(1305^{\circ}-(360^{\circ}\times4))$
$=\sin (-135^{\circ})=-\sin 135^{\circ}$
$=-\sin(180^{\circ}-45^{\circ})=-\sin45^{\circ}$
( As $\sin(180^{\circ}-x)=\sin x$ )
$=-\frac{1}{\sqrt 2}=-\frac{\sqrt 2}{2}$
