Blog Space :D Sine of One Degree

[HatCat]

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.

[HatCat]

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:

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.

[Natch]

ya dis is nawt 2 + 2 pls halp

[ExtremeDude2]

Originally Posted by Experiment_150

ya dis is nawt 2 + 2 pls halp

The answer is 5

[HatCat]

Originally Posted by ExtremeDude2

The answer is 5

Originally Posted by Experiment_150

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

[Natch]

The answer is four, unless the kid says it's five, in which case the school can no longer say no without getting sued.

[HatCat]

!! Basically I would like to be a mdoerator and start giving people banns in the mdoerator control pannel !!

"I am the main administrator there and I can ban people !"

lmao, epic reactions ensued from that thread

Anyway, I have a better question. If you know what sine means, how could you not understand what I'm trying to do? You're just pretending like you don't.

[Natch]

because I know what they are, but I have no idea what you're trying to do with them at the moment XD All I see are just a bunch of square roots and such XD

[HatCat]

OMG figure it out.



See this shit, that's a direct value that equates to the sine of one degree.

sin(+1.) = all of that shit in the above image.

If you still don't get what I mean, I'll even link to a "math" page I was working on a year ago but put to rest until recently.
http://dl.dropbox.com/u/16494013/math/trig_ratio.html

That has the exact ratios for the sines of a few popular degree measures. If you evaluate those expressions in a calculator, it will be the same as if you had just looked for the sine of theta degrees.

[Natch]

............................say again? What does theta have to do with this?