restart;
with(Physics):
Setup(mathematicalnotation = true);                
Setup(signature = `+---`);
ds2 := ((x^2-y^2)*cos(2*u)+2*x*y*sin(2*u))*(du^2)-2*x*(dv^2)-dx^2-dy^2;
Setup(coordinatesystems=(Z=[u, v, x, y]), metric = ds2);
 

Edit.

[{1, 2}, {2, 3}] minus~ [{2}$2];
                                                      [{1}, {3}]

A more traditional and programmatic way of obtaining the values of a function obtained as a result of solving an equation, etc., is to use  eval  command. See the same example as tomleslie's one:

Download eval.mw

See  question_(2)_new.mw

ode:= diff(y(x),x)=2*x:
plots:-display(DEtools:-DEplot(ode,y(x),x = -2 .. 2,y = -2 .. 2, [[0.1,0]],
               labels=["",""],
               linecolour = red,
               color = blue,
               'arrows' = 'medium',
               axesfont=['Times', 'bold', 12]
               ), axis=[tickmarks=[color=red]]);

For your first example, everything is the same:

restart;
plots:-display(plot(sin(x),x=-Pi..Pi, color=blue),axis=[tickmarks=[color=red]]);
 

Here is an example of solving the problem. The method is of a general nature and can be used for any curves given by parametric equations. In the example, the red curve is a broken line, and the blue one is a straight line.

restart;
d:=(P,Q)->sqrt((P[1]-Q[1])^2+(P[2]-Q[2])^2);
P:=piecewise(s>=0 and s<0.3,[s,s/3+0.3],s>=0.3 and s<0.5,[s,-s+0.7],s>=0.5 and s<0.7,[s,7/2*s-1.55],s>=0.7 and s<=1,[s,-s/3+17/15]);
Q:=[t,t];
A:=plot(piecewise(seq(op([op(P)[2*n-1],op(P)[2*n][2]]), n=1..nops([op(P)])/2)), s=0..1,color=red, thickness=3):
B:=plot(t, t=0..1, color=blue, thickness=3):
epsilon:=0.3;
C:=plots:-inequal({d(P,Q)<=epsilon}, s=0..1, t=0..1, optionsfeasible = [color = "WhiteSmoke"], optionsexcluded = [color = "DarkGrey"], view=[0..1,0..1]):
plots:-display(<plots:-display(A,B) | C>, scaling=constrained); # This is "free space diagram"

If the curve is given by a conventional equation, for example  y=f(x) , so you must specify it as the list  [x, f(x)]

FD.mw

Edit.

Try  
A.C;  # dot product 
# or  
expand(A.C); 

Type this ( . ) as an ordinary dot on the keyboard. I use 1d-math input. In 2d-math input (which you use) , this will look like a fat dot in the middle.

For real domain use  discont  command instead of  singular  (the output is empty set):

discont(ln(y^2+1), y);
                                           { }

You are not right. Maple calculates everything correctly. Should be

combinat:-numbcomb(n, m) = binomial(n, m)  

See help on this command.

For example  combinat:-numbcomb(7, 3) = (7*6*5)/3! = 35

The mapping of the square  [0,1] x [0,1]  with 2d-grid:

restart;
f:=(r,t)->[r*exp(t),r*exp(-t)];
S:=plot([seq([x,t,t=0..1],x=0..1,0.1),seq([t,y,t=0..1],y=0..1,0.1)], color=red, size=[200,200]):
F:=plottools:-transform(f):
T:=plots:-display(F(S), size=[400,400]):
plots:-display(<S | T>, scaling=constrained);

                       

Edit.
                        

Try use inert multiplication for this:

A:=x%*y=y%*x + 2*z;
expand(subs(A, w*x%*y));
                              

Sum is an inert form for sum, when the summation is simply shown on the display, but not calculated in any closed form. See the following example:

Sum(1/n^2, n=1..infinity) = sum(1/n^2, n=1..infinity);

Addition. To avoid possible errors, end each line of code with delimiters (;  or  : ).

In the beginning: 

restart; 
# and so on

Here is another solution. Perhaps for a not very experienced user it will be clearer:

F:=proc(M::Matrix,N::Matrix,a::nonnegint)
local m1, m2, n1, n2, k, i, j, S;
if not a in convert(M, set) then error cat(a, " is not an element of M") fi;
m1, m2 :=op([1,1],M), op([1,1],N);  # Row dimensions of M and N
n1, n2 :=op([1,2],M), op([1,2],N);  # Column dimensions of M and N
if m1<>m2 or n1<>n2 then error "Matrices M and N should be the same size" fi;
k:=0;
for i from 1 to m1 do
for j from 1 to n1 do
if M[i,j]=a then k:=k+1; S[k]:=N[i,j] fi;
od; od;
convert(S, list);
end proc:

I did not understand what you are trying to do with this bunch of parameters. If the parameter values are known from the very beginning, then you can simply assign them these values. Then simplify the resulting equation and solve it (separately for positive and negative  x ):

Download Polynom_Calculation_new.mw

The loop variable  m  starts to take values starting from 1. But as soon as m = 2, the loop exits.