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#41
23rd July 2011, 08:58 PM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

Actually I've just accomplished a couple major things.

First is that I've simplified the squares of both a and b.
The second is that I redefined or rewrote the base definition for the sine of one degree. Not too sure anymore how I managed to arrive to the one I came up with before; I'll sort through my tens of papers I've used up writing math to find out what mistaken factoring I might have done later.

So, going back to the raw conclusion of the sine of one degree:
Quote:
 Originally Posted by Iconoclast
Substituting the variable "a" for the sum of all those radicals and the variable "b" for the imaginary root radical, we begin from the top of this image I rendered.

Not converted to a real number definition yet, but at least the denominator is rationalized. Factorization and squaring followed by un-squaring should help remove remaining imaginary terms.
#42
23rd July 2011, 10:36 PM
 Natch Project Supporter Senior Member Join Date: Jan 2011 Location: Nowhere near you. Posts: 5,078

Good job for doing

*mumbles something incoherent*

*claps
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#43
24th July 2011, 02:24 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

I spent about 4 hours across 3 sheets of paper (both sides) finding the identity, square, and cube of 16 sine 1.

Not the must useful information here but possibly useful in the future. Obviously, plugging in the values for a and b allow for further integer division across the left-hand and right-hand sides of the equations.

I can do that later. It seems I have no choice, regardless, but to finish conducting my operations on binomials defined in terms of a and b.

Quote:
 Originally Posted by natnatertatertot Good job for doing *mumbles something incoherent* *claps
lol, thanks I guess, with this amount of work (as well as in general) positive feedback is kind of useless compared to negative feedback or constructive criticism XD
#44
24th July 2011, 03:48 AM
 Natch Project Supporter Senior Member Join Date: Jan 2011 Location: Nowhere near you. Posts: 5,078

To be honest I haven't really been following you XD Although if you find the sine of one, tell me...

wait that's what you're doing, right?

Also, I think you mean most useful.
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#45
25th July 2011, 06:59 PM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

Yup. My typos pretty much always come in the form of completely different words. It's annoying as Hell.

I will update with some more information I derived a couple nights ago when I eat and get around to it. In the meantime, focusing on the same exact branch of math is kind of boring, so I don't mind if you change the subject to something else about math XD.

Hell, maybe I can use this thread to give you math problems to solve, since you're so good at math from what I've read.
#46
25th July 2011, 07:27 PM
 Natch Project Supporter Senior Member Join Date: Jan 2011 Location: Nowhere near you. Posts: 5,078

*is too busy coloring the shapes to notce
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#47
26th July 2011, 12:07 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

The last 24 hours I have made no progress as far as this project here. I was disconnected the entire day because of FatMan-ism and/or forgetfulness / not caring, but I made use of my time regardless.

Still, useful piece to the puzzle here, is the product of a and b. I've found 2ab, a component used when squaring a binomial in terms of a and b.

And, as usual, proof that I've calculated ab correctly:
Code:
```var sqrt05 =    2.23606797749979;
number     = -512 * Math.sqrt(9 - Math.sqrt(30 + 6 * sqrt05) - sqrt05);
var aper16 =   -0.209343824971776;
var bper16 =    3.9945181390183;
// * i = sqrt(-1)
var abperi = -214.074292772154;
/*number     = aper16 * bper16 * 256;
= -214.074292772154; */```
An approximate decimal value of ab is -214.074292772154i, which is what the expression assigned to the variable number returned, except for the imaginary unit, of course, since Flash doesn't operate on imaginary or complex numbers.

There's only one thing left undone. How can I define b without just taking the square root of b^2 and winding up with the original value of b I took from the original source? a isn't a nested radical like b is, even though their squares are very similar in structure in terms of the radical bases used, so surely there must be a way to re-define b as the sum of several radicals (like the way a is written in that image) and not just one big, messy nested radical? Damn, how do I do that? I have to find out why the two expressions are equal; I haven't finished solving for the bi-conditional statement yet.
#48
26th July 2011, 12:11 AM
 HatCat Alpha Tester Project Supporter Senior Member Join Date: Feb 2007 Location: In my hat. Posts: 16,236

Quote:
 Originally Posted by natnatertatertot *is too busy coloring the shapes to notce
And as for you. How the hell am I going to get you off my back in this thread ...

(x - 5)(x + 3) + 2 = +2.

There, see if you even know what the hell a polynomial function is.
Solve for x.
#49
26th July 2011, 12:31 AM
 ExtremeDude2 Alpha Tester Project Supporter Senior Member Join Date: Apr 2010 Location: USA Posts: 3,118

Quote:
 Originally Posted by Iconoclast Well before I waste any time uploading images of problems, I'll just ask you something algebraic. (x - 5)(x + 3) + 2 = +2. There, see if you even know what the hell a polynomial function is. Solve for x.
Ooh...I love these
__________________
Quote:
 Originally Posted by dsx! are you american or something
#50
26th July 2011, 12:32 AM
 ExtremeDude2 Alpha Tester Project Supporter Senior Member Join Date: Apr 2010 Location: USA Posts: 3,118

__________________
Quote:
 Originally Posted by dsx! are you american or something

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