If sin (π cos x) = cos (π sin x),

(c) $\pm \frac{1}{3}$

$\sin (\pi \cos x)=\cos (\pi \sin x)$

As we know that $\sin x=-\cos \left(\frac{\pi}{2}+x\right)$

$\Rightarrow-\cos \left(\frac{\pi}{2}+\pi \cos x\right)=\cos (\pi \sin x)$

$\Rightarrow \frac{-\pi}{2}-\pi \cos x=\pi \sin x$

$\Rightarrow \pi \sin x-\pi \cos x=\frac{\pi}{2}$

$\Rightarrow \sin x-\cos x=\frac{1}{2}$

Squaring both sides we get,

$\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=\frac{1}{4}$

$\Rightarrow 1-\sin 2 x=\frac{1}{4}$

$\Rightarrow \sin 2 x=\frac{1}{3}$

And we know that $\sin x=\cos \left(\frac{\pi}{2}-x\right)$

$\Rightarrow \cos \left(\frac{\pi}{2}-\pi \cos x\right)=\cos (\pi \sin x)$

$\Rightarrow \frac{\pi}{2}-\pi \cos x=\pi \sin x$

$\Rightarrow \pi \sin x+\pi \cos x=\frac{\pi}{2}$

$\Rightarrow \sin x+\cos x=\frac{1}{2}$

Squaring both sides we get,

$\Rightarrow \sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=\frac{1}{4}$

$\Rightarrow 1+\sin 2 x=\frac{1}{4}$

$\Rightarrow \sin 2 x=\frac{1}{3}$

Therefore, $\sin 2 x=\pm \frac{1}{3}$