Visualisation of a sine wave

Instructions

So... (SCROLL DOWN-only if youre interested in incredibly simple trigonometry) WIP. I messed up the clones, just need a bit of time to fix the actual wave. Keep scrolling. I wanted to show you how sine works using a circle The circle drawn in the project is using the equation: (x-h)^2+(y-k)^2=r^2 x and y being the coordinates of any point on the circle h and k being the x and y coordinates of the origin (centre) of the circle r is the radius ^2 means 'squared' or 'to the power of 2' I will make a better explanation later but for now, this is what you get. The wave amplitude (distance from wave crest to the x axis) is equal to the distance from the x axis of a point along a circle at a specific angle from the origin of the circle. Now the thing I'm missing here is properly representing the wave length of the sine wave. One full distance from one crest (peak) to the other should represent the result of the sine wave after a complete 360 rotation. I will need to make the sine wave clones stationery and build the sine wave from the back This will be done by setting the x coordinate of each clone to the angle of the point from the origin which I can calculate using simple right angle trigonometry using sin rule and doing the inverse sine of (distance of the point from the x axis/the distance of the point from the origin). This is represented as: θ=sin^-1(distance of point from x axis/distance from origin) θ is an unknown angle pretty much btw Since I don't want to cheat and use any sort of "distance from" blocks, I'll just use pythagoras to find the distance of the point to the origin of the circle by doing the distance of the square root of the ((distance of x position from origin x position)^2+(the y position of the point)^2) It;s literally just a^2+b^2=c^2, except instead of using triangle lengths, we're imagining the triangle and making the x positions the lengths. The equation looks like this: distance from origin=sqrt((x position distance from origin x position)^2+(y position of point)^2) This x position distance from origin x position is just the difference between the two, which I can just do subtraction for: x position of point-origin x position Combined ALL TOGETHER, the equation for getting the x position of the clones (which is equivalent to the angle of the point from the origin) is... θ=sin^-1(y position/sqrt((x position of point-origin x position)^2+(y position of point)^2)) This is actually really really easy, it's just on text it looks complicated. But it's not. Anyway, that's the next step. Hopefully you didn't get bored out of your mind. Correct me if I'm wrong. See you later.