6. -4 .

Let the coordinates of the common point be $P(\alpha, \sin \alpha)$. Given that $(\sin x)'=\cos x$, the slope of the tangent line to the graph of $y=\sin x$ at point $P$ is $k=\cos \alpha$. Clearly, this tangent line is also a tangent to the circle $C$, with the center of the circle at $M(0,1)$. Therefore, $k \cdot k_{P M}=-1$, which means
$$
\cos \alpha \cdot \frac{\sin \alpha-1}{\alpha}=-1
$$

This simplifies to $\alpha=(1-\sin \alpha) \cos \alpha$. Thus,
$$
\frac{2 \sin 2 \alpha \cos \alpha-4 \cos ^{2} \alpha}{\alpha \cos \alpha}=\frac{2 \sin 2 \alpha \cos \alpha-4 \cos ^{2} \alpha}{(1-\sin \alpha) \cos ^{2} \alpha}=\frac{4(\sin \alpha-1) \cos ^{2} \alpha}{(1-\sin \alpha) \cos ^{2} \alpha}=-4 .
$$
6. Given the circle $C: x^{2}+(y-1)^{2}=r^{2}$ intersects the graph of the function $y=\sin x$ at a unique point, and the x-coordinate of this point is $\alpha$, then the value of $\frac{2 \sin 2 \alpha \cos \alpha-4 \cos ^{2} \alpha}{\alpha \cos \alpha}$ is $\qquad$ (Supplied by An Zhenping)