Step-by-step Solution:

We need the area between the curves \(y=\sin x\) and \(y=\cos x\) on \(0\le x\le \tfrac{\pi}{2}\). On this interval \(\cos x\ge\sin x\) for \(0\le x\le \tfrac{\pi}{4}\) and \(\sin x\ge\cos x\) for \(\tfrac{\pi}{4}\le x\le \tfrac{\pi}{2}\). Hence \[ \text{Area}=\int_{0}^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx. \] By symmetry the two integrals are equal, so \[ \text{Area}=2\int_{0}^{\pi/4}(\cos x-\sin x)\,dx =2\big[\sin x+\cos x\big]_{0}^{\pi/4}. \] Evaluate: \[ \big(\sin\tfrac{\pi}{4}+\cos\tfrac{\pi}{4}\big)-(\sin 0+\cos 0) =(\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2})-(0+1)=\sqrt2-1. \] Thus \[ \text{Area}=2(\sqrt2-1)=2\sqrt2-2. \] \[ \boxed{\,2\sqrt{2}-2\,} \]