Find x,y coordinates of a square given an angle $\alpha$ To find the $x,y$ coordinates of a circle where it's center is located at $(0,0)$ we use $x = r \cdot \sin(\alpha)$ and $y = r \cdot \cos(\alpha)$ 
How can I do the same for a square? Finding $x,y$ coordinates given an angle $\alpha$ ?
E.g. If the square is of length 2 then for: 


*

*$\alpha = 0°$ I should have $x = 0$ and $y = 1$ 

*$\alpha = 90°$ I should have $x = 1$ and $y = 0$


this for every possible angle?
 A: Use 
$$
x(t) = \frac{s}{M(t)} \cdot \cos(t)\\
y(t) = \frac{s}{M(t)} \cdot \sin(t)
$$
where 
$$
M(t) = \max\{ |\cos t | , |\sin t |\}
$$
and $s$ is the half-length of the square's side. 
Note that in the formulas you gave, you had $x$ tied to $\sin \alpha$, which uses angles measured counterclockwise from the vertical. I swapped that to $\cos$, following the rule that angles are generally measured (in math) counterclockwise from the positive $x$-axis. If you really want the one you said, just swap the "sin" and "cos" in the formulas for $x$ and $y$. 
Here's matlab code to show this in action. 
N = 105;
t = linspace(0, 2*pi, N); 

x = zeros(1, N);
y = x; 
s = 14; % a randomly chosen radius

x = cos(t);
y = sin(t); 
M = max(abs(x), abs(y)); 
x = s*x ./ M; 
y = s*y ./ M;

plot(x, y, 'b-', x, y, 'ro');
set(gca, 'XLim', [-(s+1), s+1], 'YLim', [-(s+1), s+1], 'DataAspectRatio', [1,1,1]);
figure(gcf);

and the resulting figure:
  
Notice the spacing of the red markers is not uniform, because the radial projection from the uniform spacing on the circle leads to nonuniform spacing on the square. 
