If \(\tan \theta \cos 60^\circ = \cfrac{\sqrt{3}}{2}\), then the value of \(\sin (\theta - 15^\circ)\) is \(\cfrac{1}{\sqrt{2}}\). ___ \[ \tan \theta \cos 60^\circ = \cfrac{\sqrt{3}}{2} \Rightarrow \tan \theta \times \cfrac{1}{2} = \cfrac{\sqrt{3}}{2} \Rightarrow \tan \theta = \cfrac{\sqrt{3}}{2} \times \cfrac{2}{1} = \sqrt{3} \Rightarrow \tan \theta = \tan 60^\circ \Rightarrow \theta = 60^\circ \] \[ \therefore \sin(\theta - 15^\circ) = \sin(60^\circ - 15^\circ) = \sin 45^\circ = \cfrac{1}{\sqrt{2}} \quad \text{(Answer)} \]