
#41




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: 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 unsquaring should help remove remaining imaginary terms.
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http://theoatmeal.com/comics/cat_vs_internet 
#42




Good job for doing
*mumbles something incoherent* *claps
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#43




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 lefthand and righthand 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. 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
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http://theoatmeal.com/comics/cat_vs_internet 
#44




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




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.
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http://theoatmeal.com/comics/cat_vs_internet 
#46




*is too busy coloring the shapes to notce
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#47




The last 24 hours I have made no progress as far as this project here. I was disconnected the entire day because of FatManism 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; */ 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 redefine 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 biconditional statement yet.
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http://theoatmeal.com/comics/cat_vs_internet 
#48




And as for you. How the hell am I going to get you off my back in this thread ...
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.
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http://theoatmeal.com/comics/cat_vs_internet 
#49




Ooh...I love these

#50




The answer is zero
