\(\sin(\pi\cos\theta) = \cos(\pi\sin\theta) = \sin\left(\frac{\pi}{2} \pm \pi\sin\theta\right)\)

So \(\pi\cos\theta = \frac{\pi}{2} - \pi\sin\theta\) or \(\pi\cos\theta = \frac{\pi}{2} + \pi\sin\theta\) (taking the principal solutions).

• \(\cos\theta + \sin\theta = \frac{1}{2}\): squaring, \(1 + \sin 2\theta = \frac{1}{4}\), so \(\sin 2\theta = -\frac{3}{4}\).
• \(\cos\theta - \sin\theta = \frac{1}{2}\): squaring, \(1 - \sin 2\theta = \frac{1}{4}\), so \(\sin 2\theta = \frac{3}{4}\).

\(\sin 2\theta = \pm\frac{3}{4}\)

Answer: (A)