Can cos(x) or sin(x) be the function of stationary action? Is there a way to express $\cos(x(t))$ (or $\sin(x(t))$) as the solution to the Euler-Lagrange equation, in other words is there a sense in which this function is the path of stationary action?
 A: If x(t) solves the Euler Lagrange equation for $L(x,\dot{x})$, then f(x(t)) is stationary for $L(f^{-1}x, \dot{f^{-1}x(t)})$. The reason is that the action evaluates to the same thing, so that a small perturbation in either coordinate gives zero action change.
This means that in the domain where the function sin(x) has an inverse, you can freely change $x(t)$ to $\sin(x(t))$, so long as you change the action as above. This is a coordinate change on the configuration space, and the change in action automatically extends it to be a symplectic transformation on phase space (once you define the new momentum from the new Lagrangian).
A: Most naturally, both sine and cosine – I suppose you meant simply $x=\sin(t)$ and $x=\cos(t)$ because $\cos(x(t))$ isn't a particular path, it's a functional of a path – are solutions to the differential equation
$$ \frac{d^2}{dt^2} x(t) = -x(t) $$
which is the equation for a harmonic oscillator (with a unit spring constant, in this case). This equation may also be derived as the Euler-Lagrange equation from the action for the harmonic oscillator,
$$ S = \int dt \left[ \frac 12 \left(\frac{dx(t)}{dt}\right)^2 - \frac {x^2}{2} \right] $$
which is the difference between the kinetic and potential energy (the Lagrangian) integrated over  time (the action).
