Find the choice of c_0,c_1 and c_2 that gives the best fit of a
function of the form f(t) = c_0 + c_1*sin(t) + c_2*cos(t)
to the data points (0,0),(1,1), (2,2), (3,3).  Substituting:

f(0) = c_0              + c_2        = 0
f(1) = c_0 + c_1*sin(1) + c_2*cos(1) = 1
f(2) = c_0 + c_1*sin(2) + c_2*sin(2) = 2
f(3) = c_0 + c_1*sin(3) + c_2*sin(3) = 3

Which gives us a linear system in the unknowns c_0,c_1 and c_2.
To compute the matrix of this system in MATLAB and set up a vector for 
the right-hand side:

>> A = [1 0 1; 1 sin(1) cos(1); 1 sin(2) cos(2); 1 sin(3) cos(3)]
>> b = [0;1;2;3]

   [ 1.0000         0    1.0000] [c_0]   [0]
   [ 1.0000    0.8415    0.5403]*[c_1] = [1]
   [ 1.0000    0.9093   -0.4161] [c_2]   [2]
   [ 1.0000    0.1411   -0.9900]         [4]

To find the least squares solution we multiply both sides by the
transpose of A.  (Remember, in MATLAB the transpose of A is A')

    [4.0000    1.8919    0.1342] [c_0] [ 6.0000]
    [1.8919    1.5548   -0.0635]*[c_1]=[ 3.0834]
    [0.1342   -0.0635    2.4452] [c_2] [-3.2620]

If the columns of the original matrix A were linearly independent
then the 3x3 matrix on the left will be invertible.  In MATLAB we
can find the inverse of a matrix M using inv(M):

Thus the best choice of coefficients c = inv(A'*A)*A'*b giving

c0 = 1.5000, c1 = 0.1003, and c2 = -1.4137

We can plot these functions to see how well the trigonometric function

    f(t) = 1.5 + 0.1003*sin(t) - 1.4137*cos(t) 

fits the linear function g(t) = t.

>> t = 0:0.1:3;       % pick a vector of t values between 0 and 3
                      % the graph of g(t) = t can be obtained using
                      % plot(t,t)

Next graph f(t) =  1.5 + 0.1003*sin(t) -1.4137*cos(t)
First compute the y-values

>> y = 1.5 + 0.1003*sin(t) -1.4137*cos(t);  

Now plot(t,y) will produce the graph of f on [0,3]

We can also include the data points.

>> x = [0,1,2,3];     % plot(x,x,'*') will plot the data points as '*'

Putting it all together:

>> plot(t,t,t,y,x,x,'*')