If $$\sin \left( {\pi \cos x} \right) = \cos \left( {\pi \sin x} \right),$$      then what is one of the values of $$\sin 2x \,?$$

Solution :
$$\eqalign{ & {\text{Given that}}:\sin \left( {\pi \cos x} \right) = \cos \left( {\pi \sin x} \right) \cr & {\text{So}},\,\,\cos \left( {\frac{\pi }{2} - \pi \cos x} \right) = \cos \left( {\pi \sin x} \right) \cr & \Rightarrow \frac{\pi }{2} - \pi \cos x = \pi \sin x \cr & \Rightarrow \sin x + \cos x = \frac{1}{2} \cr} $$
Squaring both sides, we get
$$\eqalign{ & {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = \frac{1}{4} \cr & \Rightarrow \sin 2x = \frac{1}{4} - 1 = - \frac{3}{4} \cr} $$