If sin (π cos x) = cos (π sin x), then sin 2 x =
(a) $\pm \frac{3}{4}$
(b) $\pm \frac{4}{3}$
(c) $\pm \frac{1}{3}$
(d) none of these
(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, $\sin 2 x=\pm \frac{1}{3}$