What is being represented by this 2 images? image 1

image 2

It's possible that image 1 is showing some kind of methods for building polygons out of trigonometric functions ?
It's also possible that image 2 is a quadratic bezier curve ?
 A: Here's how we can produce the figures in image 1.  The graphs are showing the $ \ y-$ coordinate of the point being "swept along" on each curve. 
The circle is simple because familiar trigonometric functions can be defined in terms of a circle.  For a circle of radius $ \ a \ , $ the coordinates of the point are just $ \ (a \cos \theta \ , \ a \sin \theta) \ , $ with $ \ \theta \ $ being the angle measured (counter-clockwise) from the positive $ \ x-$  direction.  [Keep in mind that, in the animations, the graphs are being drawn "right to left".]
The square of side $ \ 2a \ $ , having four straight sides lying along the lines $ \ x \ = \ \pm a \ $ and $ \ y \ = \ \pm a \ , $ requires a piecewise function with four "branches":
$$ \sigma(\theta) \ = \ \left\{ \begin{array}{cc}( \ a \ , \ a \ \tan \theta \ ) \ ,&
-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}\\(  \ a \ \cot \theta \ , \ a ) \ ,&
\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}\\( \ a \ , \ -a \ \tan \theta \ ) \ ,&
\frac{3\pi}{4} \le \theta \le \frac{5\pi}{4}\\(  \ -a \ \cot \theta \ , \ a ) \ ,&
\frac{5\pi}{4} \le \theta \le \frac{7\pi}{4}\end{array}\right.  $$
The hexagon depicted is correspondingly more complicated and I won't do all of it here.  We will call the lengths of the sides $ \ a \ . $  Except for the vertical sides, the rest make 30º angles to the "horizontal", so their slopes are $ \ \pm \frac{\sqrt{3}}{3} \ . $  Since the altitude of an equilateral triangle is $ \ \frac{\sqrt{3}}{2}  a \ ,  $ the right vertical side lies on the line $ \ x \ = \ \frac{\sqrt{3}}{2}  a \ , $ so that branch of the piecewise function is 
$$ ( \ \frac{\sqrt{3}}{2}  a \ , \ \frac{\sqrt{3}}{2}  a \ \tan \theta \ ) \ , \  -\frac{\pi}{6} \ \le \ \theta \ \le \ \frac{\pi}{6} \ \ .  $$
The next side in the counter-clockwise direction has a slope of $ \ -\frac{\sqrt{3}}{3} \ $ and a $ \ y-$ intercept of $ \ a \ $ , so it lies on the line $ \ y \ = \ a \ - \ \frac{\sqrt{3}}{3} x \ . $  In polar coordinates, this is 
$$ r \ = \ \frac{a}{\frac{\sqrt{3}}{3}  \cos \theta \ + \ \sin \theta} \ \ . $$
The branch of the piecewise function definition covering this side is then
$$ ( \ \frac{a \ \cos \theta}{\frac{\sqrt{3}}{3}  \cos \theta \ + \ \sin \theta} \ , \ \frac{a \ \sin \theta}{\frac{\sqrt{3}}{3}  \cos \theta \ + \ \sin \theta} \ ) \ , \  \frac{\pi}{6} \ \le \ \theta \ \le \ \frac{\pi}{2} \ \ .  $$
The branch definitions for the other four sides of the hexagon can be determined analogously.
A: I just figured out this formula:
$\theta \in [0,2\pi] $
$r = \sqrt{1 + tan^2[|mod(\theta,\frac{2\pi}{n})-\frac{\pi}{n}|-\frac{\pi}{n}]} = sec(|mod(\theta,\frac{2\pi}{n})-\frac{\pi}{n}|-\frac{\pi}{n}) $
For example, in matlab:
n = 5;
theta = linspace(0,2*pi,1000);
r = sec(abs(mod(theta,2*pi/n)-pi/n)-pi/n);
plot(r.*sin(theta),-r.*cos(theta),'linewidth',2);
axis equal;

Displays this:

