
can these two formula work Re: Experimenting sin(x)= .4x^2+1.6x .09, cos(x)= .6x^2 .3x +1
Posted:
Feb 19, 2017 11:09 PM


On Sunday, February 19, 2017 at 9:29:08 PM UTC6, Archimedes Plutonium wrote:
Experimenting sin(x)= .4x^2+1.6x .09, cos(x)= .6x^2 .3x +1 > Now i am still hopeful the power of 2, quadratic equations works but the restraints maybe too much > > Sine : (0,0), (.5,.86) (1,1) > > Cosine: starting at (1,1) and continuing (1.5,.86) (2,0) > > Let me see if those two formula work to that end. >
Alright, so I have for sine the function .4x^+1.6x .09
and for cosine the function cos(x)= .6x^2 .3x +1
And what I need to see, is if I can retrieve these points
x y 0 0 .5 .86 1 1 this is for sine and starting cosine i need (1,1) also 1.5 .86 2 0
So what I have here is a single full semicircle from 0 to 2 and half of it is the sine as quartercircle and half is the quartercircle that is cosine.
So does these two equations fulfill that table?
sin(x)= .4x^2+1.6x .09
x y 0 .09 good enough .5 .7 not close enough to .86 1 1.2 not close enough to 1
cos(x)= .6x^2 .3x +1
x y 0 1 exactly what I want when that is pushed over as a horizontal shift .5 .7 not close enough to .86 1 .1 close enough to 0
Judging from the above, I would say there exists two formula that can closely match the desired plotted points of a single semicircle.
AP

