\[ \tan \theta \cdot \cos 60^\circ = \frac{\sqrt{3}}{2} \Rightarrow \tan \theta \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \Rightarrow \tan \theta = \frac{\sqrt{3}}{2} \cdot 2 = \sqrt{3} \Rightarrow \tan \theta = \tan 60^\circ \Rightarrow \theta = 60^\circ \] Therefore, \[ \sin(\theta - 15^\circ) = \sin(60^\circ - 15^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}} \]