Solution
Given,
$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$
Using,
$\cos x=\sin\left(\frac{\pi}{2}-x\right)$
Therefore,
$\sin(\pi\cos\theta)=\sin\left(\frac{\pi}{2}-\pi\sin\theta\right)$
This gives the following two possible cases.
Case 1:
$\pi\cos\theta=\frac{\pi}{2}-\pi\sin\theta$
$\cos\theta+\sin\theta=\frac{1}{2}$
Squaring both sides,
$(\cos\theta+\sin\theta)^2=\frac{1}{4}$
$\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=\frac{1}{4}$
$1+\sin 2\theta=\frac{1}{4}$
$\sin 2\theta=-\frac{3}{4}$
Case 2:
$\pi\cos\theta=\pi-\left(\frac{\pi}{2}-\pi\sin\theta\right)$
$\cos\theta-\sin\theta=\frac{1}{2}$
Squaring both sides,
$(\cos\theta-\sin\theta)^2=\frac{1}{4}$
$1-\sin 2\theta=\frac{1}{4}$
$\sin 2\theta=\frac{3}{4}$
Hence,
$\sin 2\theta=\pm\frac{3}{4}$
Therefore, the correct answer is:
$\boxed{\text{(a) }\pm\frac{3}{4}}$