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Topic: improvement on sin(x) = −0.4176x^2 +1.3122x -
0.0504 ;; we need semicircle sine, not the fakery of sinusoi
d sine

Replies: 15   Last Post: Feb 25, 2017 3:31 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
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 UTC-6, 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