Jacobian linearization of trigonometric functions If I have to linearize a nonlinear trigonometric system around the origin $(0,0)$:
$$\dot{x_1} = x_2$$
$$\dot{x_2} = \cos(x_1)$$
I can apply the small angle approximation to find the matrix A:
$$ A = \pmatrix{0&1\\1&0} $$
However, if I apply Jacobian linearization and take the partial derivatives of $\dot{x}$, I get:
$$ A =  \pmatrix{0&1\\-\sin(x_1)&0}$$
Evaluated at the origin, I get:
$$ A = \pmatrix{0&1\\0&0}  $$
Seeing as how the two different linearization techniques yield different results, is one more valid than the other?

"Approximation" in our case means Taylor's theorem, i.e.
$$
f(x) = f(0) + Df(0)x
$$
where $f(x) := (x_2, \cos(x_1))$. This is where $Df(0)$, i.e. your second matrix appears. So this is a linearization in the normal sense. This type of linearization is actually useful when we want to determine stability of an equilibrium point.
The first approach I would call incoherent, especially since you leave $x_2$ untouched. This is not how linearization in analysis is understood. Linear maps have the property that they vanish in $0$. So what we do, is to always approximate
$$
x \mapsto f(x) - f(0)
$$
by a linear map since the former vanishes in $0$, too. This is the idea behind Taylor's theorem. We just accept, that this is actually an affine approximation of $f(x)$ since it simply has no major ramifications in practice.
But what you did, is to find a linear map that approximates $f(x)$ only. But since $f(0) \neq 0$ and $A0 = 0$, this might yield huge errors. To my knowledge (which is of course limited), the first one has no application whatsoever, but I would love to be corrected if I were wrong.
