
sine is 1.44x^2 + 2.44x + 0 = Y
Posted:
Feb 20, 2017 3:18 AM


Now I do not know why I have this mischeviousness within me to never follow instructions until absolutely necessary all frustration out.
So, I employed the tool: The Polynomial Generator is this tool::
For 2 coordinate points, (x0, y0) (x1, y1), we produce the 1st degree polynomial, a straightline or line segment
P(x) = y0(xx1) / (x0x1) + y1(xx0) / (x1x0)
For 3 coordinate points, (x0, y0) (x1, y1) (x2, y2), we produce the 2nd degree polynomial, a compilationcurve
P(x) = y0(xx1)(xx2) / (x0x1)(x0x2) +y1(xx0)(xx2) / (x1x0)(x1x2) +y2(xx0)(xx1) / (x2x0)(x2x1) For 4 coordinate points, (x0, y0) (x1, y1) (x2, y2) (x3, y3), we produce the 3rd degree polynomial, a compilation curve
P(x) = y0(xx1)(xx2)(xx3) / (x0x1)(x0x2)(x0x3) +y1(xx0)(xx2)(xx3) / (x1x0)(x1x2)(x1x3) +y2(xx0)(xx1)(xx3) / (x2x0)(x2x1)(x2x3) +y3(xx0)(xx1)(xx2) / (x3x0)(x3x1)(x3x2)
Obviously I need the tool for 3 coordinate points:
sine (0,0) (.5,.86) (1,1)
And what I end up with is this polynomial
1.44x^2 + 2.44x + 0 = Y
Now, was that so much easier than playing around I think so. For that equation gives me exactly sine at those sought for 3 points. And, the bonus is that it is still quadratic.
And it looks as though the cosine, which I have not yet worked out, but is also quadratic since the last term of y2 is 0 and thus ending as a power of 2 quadratic.
So, let me work that one out, ....
AP

