根据已知条件$$\left(x+\sin x\cos x=1\right)\land\left(\cos y+2=y-\dfrac{\pi}{2}\right),$$可得$$\left(2x+\sin 2x=2\right)\land\left(\sin\left(\dfrac{\pi}2-y\right)+\dfrac{\pi}2-y=-2\right),$$因此考虑构造函数$$F(x)=x+\sin x,x\in \mathbb R.$$易知 $F(x)$ 为奇函数，且在 $\mathbb R$ 上单调递增．故由$$F(2x)=-F\left(\dfrac{\pi}{2}-y\right)=F\left(y-\dfrac{\pi}{2}\right)$$可得$$2x=y-\dfrac{\pi}2,$$因此$$2x-y=-\dfrac{\pi}2,$$故$$\sin (2x-y)=-1.$$