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I am working on my own math project of movement around a square. (I am stuck) I have never been taught this stuff before(I'm only 15). So to start with I made a 4 unit by 4 unit square. I started by knowing that the rotation of a point in a circle is: $(sin(x_1),cos(x_1))$
But then I figured out that the radius of the rotation needs to be larger. So then I made another variable for radius, so the point became: $((r)sin(x_1),(r)cos(x_1)$ (oh also I am using desmos graphing calculator) So once I started playing around with the radius size to where the point was on the edge of the square I came up with the sine wave: $$\frac{d-2}{2}\sin\left(\frac{\left(x-\frac{\pi}{8}\right)}{\frac{1}{4}}\right)+\sqrt{2}+1$$ The only problem with this wave is that when I use it in connection with $x_1$ it only works for the corners and in the center of the sides. I am trying to make a wave that works all the way around the square. So I went back to the rotation around a circle formula again and made the y value side of the point equal to 2 (we are trying to find the radius of the rotation when it is on the square): $\left(r\right)\sin\left(x_{1}\right)=1$
So then I made it so that r was on its own side of the equals sign , which I got: $$r=2\frac{1}{\cos\left(x_{1}\right)}$$
The only issue with this one is that it only works for the top of the square and it doesn't move down. I'd like to continue on with my sine wave I was needing to improve but Im not sure how to write an equation of a sine wave that doesn't have a smooth peak. All I need is a wave that looks like this: Sine Wave

I would be very grateful to whoever could help me :)

Byron E
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Solution

The main problem you appear to be struggling with seems to be the use of 'x' as the variable in polar coordinates, x measures the x axis in a cartesian graph. When working with polar coordinates dependent on angles, the "theta" variable is used.

I've compiled an explanation in Desmos graphing calculator here: https://www.desmos.com/calculator/kieymy69a0

Equation in Desmos: $$r=\min\left(\frac{1}{\sin\left(\operatorname{mod}\left(\theta,\frac{\pi}{2}\right)\right)},\frac{1}{\cos\left(\operatorname{mod}\left(\theta,\frac{\pi}{2}\right)\right)}\right) $$

Andrei
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