Angular momentum and rotation symmetry In my book, it is written that for any vector $\mathbf{v}$, we have
$$\{\mathbf{v},\mathbf{L}\cdot \mathbf{n}\}=\mathbf{n}\times\mathbf{v}.\tag{1}$$
For me it is absurd... For example, if we take $\mathbf{v}=(\cos(x),0,0)$
the Poisson bracket, say we consider $L_z=xp_y-yp_x$:
$$(\{\mathbf{v},\mathbf{L}\cdot \mathbf{n}\})_x=\frac{\partial v_1}{\partial x}\frac{\partial L_z}{\partial p_x}-\frac{\partial L_z}{\partial x}\frac{\partial v_1}{\partial p_x}+\frac{\partial v_1}{\partial y}...=-\sin(x)x.$$
I can see that it already fails sincer the left-hand side is $\mathbf{n}\times\mathbf{v}$ can not contain $\sin(x)$...
Or maybe I misunderstood the proposition?
Screenshot of the book's content:

 A: *

*Well, there are different notions of vectors in mathematics and physics. 

*The notion of vectors ${\bf v}$ that seems to be relevant here are maps 
$$\{\text{phase space}\}\times\{\text{time axis}\}\quad\stackrel{\bf v}{\longrightarrow}\quad \mathbb{R}^3$$ 
that transform in the vector representation of the 3D rotation group $SO(3)$. Here the angular momentum $L_x$, $L_y$, and $L_z$ belong to the corresponding $so(3)$ Lie algebra. OP's eq. (1) is the characteristic property of such vectors ${\bf v}$. [Here ${\bf n}$ is a so-called rotation vector, although it is not a vector in above sense].

*Similarly, scalars
$s$ in this context are maps 
$$\{\text{phase space}\}\times\{\text{time axis}\}\quad\stackrel{s}{\longrightarrow}\quad \mathbb{R}$$
that transform in the trivial representation of $SO(3)$.

*OP's example does not fulfill eq. (1) and is hence not a vector in the sense of point 2. 

*Similarly, a constant 3-tuple ${\bf v}=(1,-5,2)$ is not a vector in the sense of point 2 because it stays constant rather than rotate under 3D rotations. 
