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Skimming The Cosinus Circle - Swan001 Vs The Machines

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posted on Dec, 9 2012 @ 06:46 AM
I was attempting to convert polar coordinates into cartesian coordinates when I realized I needed to compute the cosinus factor.

For those who don't know what cosinus is: You could say it is a factor that tells you the size of a tangent depending on its angle relative to a reference (that is, the size of the tangent when it's at 0°). For instance, let's say you head on a straight road toward south. But your friend decide to head on another road, at the same speed (let's say 100mph), but straight toward south-east instead. Cosinus will let you know that after 1 hour, you will have travelled 100 miles toward the south while your friend will only have traveled 0.707106 of that distance toward the south, meaning, he will have travelled only 70.716 miles toward the south, due to his 45° degree trajectory relative to yours.

Cosinus has applications in astronomy, navigation, home construction, many things.

The problem is, cosinus value for an angle is not computable - only computers can compute cosinus factor. The best mankind can do is make super-long equations when they want to compute cosinus "by hand", and the result is always nothing more than an approximation.

As I often experience power failure, I decided to try and come up with a simple equation to approximatively compute cosinus... using normal current angular degree units (90 degrees units per quarters of a circle; not the ancient radian thing). And, well, after a couple of months of tweaking and comparing to a computer's computations, I came up with a fairly simple equation, which lets you approximate the cosinus with some accuracy... well, its accuracy is not enough for astronomical purpose, though, but anyway, I'll let you guys take a look and decide about its level of accuracy.

So, my equation is:

1-((θ/5q - θ³R)θ)

or 1-((((θ÷5)xq)-(θ³xR))θ)... in which

-θ is the angle, in normal degrees (maximum of 90° - no need to convert to radians)

-q is equal to 0.000 758 496, and

-R is equal to 0.000 000 003 486 71.

And now I am going to show you the computation results which my very simple equation yields over 19 different angles. The left column will show you the cosinus value my equation approximates to, while the right column will show you the true cosinus value as computed by computers. So it's basically my approximation, vs computer's high-tech, foot-long, equations.

swan approx.: computer:

0°: 1.000000 1.000000

5°: 0.996209 0.996194

10°: 0.984864 0.984807

15°: 0.966044 0.965925

20°: 0.939878 0.939692

25°: 0.906549 0.906307

30°: 0.866294 0.866025

35°: 0.819400 0.819152

40°: 0.766207 0.766044

45°: 0.707106 0.707106

50°: 0.642543 0.642787

55°: 0.573015 0.573576

60°: 0.499070 0.500000

65°: 0.421310 0.422618

70°: 0.340389 0.342020

75°: 0.257013 0.258819

80°: 0.171940 0.173648

85°: 0.085981 0.087155

90°: -0.000000 0.000000

For those who would like to see the quantitive difference, between my simplified approximation equation and the truth as computed by computers, you won't have to hunt for a calculator; here it is:

0°: 0.000000

5°: +0.000015

10°: +0.000057

15°: +0.000118

20°: +0.000185

25°: +0.000242

30°: +0.000269

35°: +0.000248

40°: +0.000162

45°: +0.000000

50°: -0.000243

55°: -0.000560

60°: -0.000929

65°: -0.001307

70°: -0.001630

75°: -0.001805

80°: -0.001707

85°: -0.001174

90°: -0.000000

So,... what do you guys think?


posted on Dec, 9 2012 @ 07:01 AM
reply to post by swan001

This makes me tingle like when I read arxiv.


Well done.

posted on Dec, 9 2012 @ 10:35 AM
reply to post by swan001

I can't say I've heard of 'Cosinus', although it appears you are talking about Cosine.

This Sine approximation doesn't use long formula and is used in software where you don't want the overhead of getting an accurate sine.

sin(theta) = (theta * 4/pi) + theta * abs(theta) * -4 * pi^2

the precision can be enhanced to within 0.2%ish by the following:

a = (theta * 4/pi) + theta * abs(theta) * -4 * pi^2
sin(theta) = (a * abs(a) - a) * 0.225 + a

It is trivial to change the formula to return the cosine.

So in effect your formula appears to be a very good one! Well done...

edit on 9-12-2012 by EasyPleaseMe because: (no reason given)

posted on Dec, 9 2012 @ 01:08 PM
reply to post by EasyPleaseMe

Yeah, sorry I meant cosine - I am french, and in french it's "cosinus", so I forgot to translate it up. Thanks for the reminder.

Thanks for the informative post.

posted on Dec, 11 2012 @ 09:54 AM
reply to post by EasyPleaseMe

Wait a min. 0.2%... Does that make it more accurate than mine? I have trouble calculating the accuracy percentage - I'm not exactly the greatest of mathematicians. Last time I checked it seemed as if mine was 1.4% accurate...


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