how can i divide a circle into 15 equal sectors in matlab?

which code is for dividing a circle into 15 equal sectors?

Answer 1

[Star Strider]

This works:

a = linspace(0, 2*pi, 150);
r = 1;
x = r*cos(a);
y = r*sin(a);

figure(1)
plot(x, y)
hold on
plot([zeros(1,15); x(1:10:end)], [zeros(1,15); y(1:10:end)])
hold off
axis equal

EDIT — (13 Apr 2018 at 19:45 UCT)

To scale it automatically for any desired number of segments, use this code:

N = 15;                                                         % Number Of Segments
a = linspace(0, 2*pi, N*10);
r = 1;
x = r*cos(a);
y = r*sin(a);

figure(1)
plot(x, y)
hold on
plot([zeros(1,N); x(1:10:end)], [zeros(1,N); y(1:10:end)])
hold off
axis equal

[Eman Bany Salameh]

I want to decrease the number of sectors to be 8. I tried to change the number in this statment

    % plot([zeros(1,15); x(1:10:end)], [zeros(1,15); y(1:10:end)]) 
but I got this error 

    % Error using vertcat
Dimensions of matrices being concatenated are not consistent.

    Error in sectors (line 8)
    plot([zeros(1,8); x(1:10:end)], [zeros(1,8); y(1:10:end)])

Could you please tell me what this do?

    % [zeros(1,8); x(1:10:end)], [zeros(1,8); y(1:10:end)]

[Star Strider]

You also have to change ‘a’ to be 10 times the number of segments, so:

a = linspace(0, 2*pi, 80);

for 8 segments.

I tweaked my earlier code to be robust to any number of segments, and added it as an edit. You may want to use it instead.

Answer 2

[Azzi Abdelmalek]

alpha=-pi:0.01:pi;
n_sect=7
sect=2*pi/n_sect
clr='bgrycmk';
for k=1:n_sect
  a0=-pi+(k-1)*sect
  a1=-pi+k*sect
  t=a0:0.01:a1
x=cos(t)
y=sin(t)
fill([ 0 x],[0 y],clr(k))
hold on
end
axis equal

Answer 3

[Image Analyst]

As an example, see my attached colorwheel program.