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Old 30th June 2011, 11:20 PM
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Talking Blog Space :D Sine of One Degree

Storing further text notes on my Flash drive just doesn't render my notes out in the eminent way I may very well need, so I'm just going to take notes via spam on a vBulletin board.

So method here is that I've based off the original expression for the sine of one degree:


Which this web page explains how to derive using basic trigonometric properties,
http://www.efnet-math.org/Meta/sine1.htm


Now that expression looks huge as Hell, but the only two problems I can see are that a) its definition is in terms of the imaginary unit i = sqrt(-1) and that b) tediousness in writing, in spite of the fact that two very large components of that expression are repeatedly used throughout the expression and can be replaced.

After hours of working with imaginary or complex numbers, I've [possibly mistakenly] simplified that expression down to:
[(a + b)^2 - 64] / 512(a + b)

Where a is that frequently used component, stemming from [4 * sqrt(2)] to [4 * sqrt(5) * sqrt(25 + 11(sqrt5))], and b is 16i * sqrt(8 + ... - 2 * sqrt(2) * sqrt(25 + 11 * sqrt(5)).

I've been working several days now on simplifying both expressions (mostly analytic factorization) to a clearer real-number solution for the exact value of the sine of one degree.
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Old 30th June 2011, 11:40 PM
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Today's accomplishment, simplified the expression that I shorten using the variable name "a".



Well, maybe not "simplified". It's only shorter because of factorization, but it does save having to write the same symbols over and over.

I've proven this by plugging in the original expression and the factorized expression I've created into the Flash environment with the following ActinScript:
Code:
var srt2 = Math.sqrt(2);
var srt3 = Math.sqrt(3);
var srt5 = Math.sqrt(5);
var srt6 = Math.sqrt(6);
var disc = Math.sqrt(25 + (11 * srt5));
number = 4 * srt2 - 4 * srt2 * srt5 + 4 * srt3 * srt2 - 4 * srt5 * srt3 * srt2 + 12 * disc * srt3 - 12 * disc - 4 * srt5 * disc * srt3 + 4 * srt5 * disc;
//number = (4 * disc * (srt3 - 1) * (3 - srt5)) + (4 * (srt6 + srt2) * (1 - srt5));
The line that is commented out uses the factorized expression I've created instead of the original as shown in the image. Both statements evaluate to the same constant, so I've at least successfully done this much so far.
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Old 1st July 2011, 12:46 AM
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ya dis is nawt 2 + 2 pls halp
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Old 1st July 2011, 01:22 AM
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Quote:
Originally Posted by Experiment_150 View Post
ya dis is nawt 2 + 2 pls halp
The answer is 5
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Old 1st July 2011, 02:36 AM
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Quote:
Originally Posted by ExtremeDude2 View Post
The answer is 5


Quote:
Originally Posted by Experiment_150 View Post
ya dis is nawt 2 + 2 pls halp
Depends what you mean by "not".

If you mean the additive inverse, then the evaluation is as follows:
[-(2 + 2)] = (-2 + -2) = -4.


If you mean the rational or multiplicative inverse, then the evaluation is more:
1:(2 + 2) = 4^-1 = +0.25
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Old 1st July 2011, 02:42 AM
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The answer is four, unless the kid says it's five, in which case the school can no longer say no without getting sued.
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Old 1st July 2011, 02:43 AM
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You said it's NOT (2 + 2) though, which means it's the inverse of 4, not 4.
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Old 1st July 2011, 09:37 PM
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Not really an accomplishment, more an update.

I don't like looking at the huge amount of ink flooding my past few pages, so I transferred what little work I could do on the expression assigned to variable named "b" to an image of its own.

So for convenient reference, the definitions of both a and b:


And once again, proof that I've rewritten the expression accurately:
Code:
var srt2 = Math.sqrt(2);
var srt3 = Math.sqrt(3);
var srt5 = Math.sqrt(5);
var disc = Math.sqrt(25 + (11 * srt5));
//number   = 8 + srt3 + srt5 * srt3 + srt5 * srt2 * disc - 2 * srt2 * disc;
number   = (1 + srt5) * srt3 - (2 - srt5) * srt2 * disc + 8;
The line that is commented out defines the Flash text variable number to the original expression in the image, before I factored it down to the expression assigned via the last statement.

Both of them returned the floating-point constant: 15.9561751629462

So just take the square root of that, multiply it by 16 and i = sqrt(-1.) (gulp, this issue again), then you get b.

I tried to take care of simplifying the top statement there rewriting the sine of one degree accurately, but I can't verify it yet obviously if b is a complex or imaginary number without temporarily assuming it's correct for now. Which means that the next step is to add a + b, square that shit, then divide by (512a + 512b). Stay tuned for the next exciting episode, of DRAGON, MATH, Z!!1!1!!1
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Old 4th July 2011, 04:28 AM
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Hi episode incomplete, sorry I don't have any pretty pictures to flood this post with as usual, just a mid-update that the above approach is still in progress lmao

I am still finding the square of the sum of a and b, because I am still finding the square of a.

(a + b)^2
(a + b)(a + b)
a(a + b) + b(a + b)
a^2 + ab + ba + b^2
a^2 + 2ab ^ b^2

(b)^2 is easy to find since the entire expression in of itself is already a square root, so squaring that is easy.

I'm still writing out the last "few" (relatively speaking) bits of the square of a, which so far has already taken up three full pages of paper XD.

Have to admit, didn't get anything accomplished after yesterday's break because of a sort of ache in my head area I've found myself wound up into this afternoon. It's extremely tedious, but I'll never give up !

Last edited by HatCat; 4th July 2011 at 04:30 AM.
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Old 4th July 2011, 12:30 PM
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Quote:
Originally Posted by Lil Cool b Clast View Post
It's extremely tedious, but I'll never give up !
What if
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