Question about constraint in lagrangian We know that the lagrangian can be written using constraints.
Suppose I have constraint functions $$f_1=\cos(x) - (x+t)/R=0,\qquad f_2=\sin(x)-y/R=0 .$$
But I know that $1=\cos^2+\sin^2$, so do we use the the $f_1$ and $f_2$ constraints with this one, or only $1=\cos^2+\sin^2$?
In other words, can we combine two constraints into one constraint for the lagrangian formalism?
 A: Calculating the wedge product $df_1\wedge df_2$ gives $$ df_1\wedge df_2=\left(\sin x+\frac{1}{R}\right)\frac{1}{R}dx\wedge dy +\frac{1}{R^2}dt\wedge dy-\frac{\cos x}{R}dt\wedge dx, $$ which is nonzero, hence the constraints $f_1$ and $f_2$ are independent. But then they cannot be replaced by a single constraint function.
Said differently, in the $(t,x,y)$-space the equations $f_1=f_2=0$ determine a curve, while a single constraint function would give a two-dimensional surface.
In fact, unless there is another coordinate (eg. $z$) that does not appear in the constraints, the constraints determine the motion completely and no further degrees of freedom are left.
A: you can  combine the two holonomic constraints into one non- holonomic constraint equation.
from
$$  \mathbf F_c=\left[ \begin {array}{c} \cos \left( x \right) -{\frac {x+t}{R}}
\\ \sin \left( x \right) -{\frac {y}{R}}\end {array}
 \right] 
=\mathbf 0\tag 1$$
thus
$$\mathbf{\dot{F}}_c=
\underbrace{ \left[ \begin {array}{cc} -{\frac {\sin \left( x \right) R+1}{R}}&0
\\  \cos \left( x \right) &-\frac 1R\end {array}
 \right]}_{\mathbf C} \,\begin{bmatrix}
  \dot{x} \\
  \dot{y} \\
\end{bmatrix}-\begin{bmatrix}
  \frac 1R \\
  0 \\
\end{bmatrix}=\mathbf 0\tag 2$$
to solve this equation for $~\dot x~,\dot y~$ the determinate of the matrix $~\mathbf C~$  must be unequal zero this mean also that the constraint equations (Eq. (1) are  independent .
the solution of eq. (2) is:
$$\dot x=-\frac{1}{\sin(x)\,R+1}\quad,
\dot y=-\frac{R\,\cos(x)}{\sin(x)\,R+1}\quad\Rightarrow$$
$$ \dot y-\cos(x)\,R\dot x=0\tag 3$$
Eq. (3) is  non holonomic constraint equation that equivalent to
the two holonomic constraint equations Eq. (1) .
from here you can also see that you obtain from two degrees of freedom $~(x~,y)~$   one generalized coordinate .
