MHB Is simplification of sin(x) * (1 - sqrt(cos(x)) possible?

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The discussion revolves around the simplification of the expression w = (1 - sqrt(cos(απ)) * sin(απ) / 2. Participants note that the angle απ must be in the first or fourth quadrants for w to be real, and there are no known identities to eliminate the square root from the cosine function. An alternative formulation is suggested, expressing w in terms of a squared relation, but it is concluded that further simplification may depend on specific properties of α. The conversation highlights the complexity of the equation and the challenges in finding a more efficient representation. Ultimately, the participants acknowledge the intriguing nature of the equation and its connections to theoretical physics.
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hi folks I've looked on List of trigonometric identities - Wikipedia, the free encyclopedia
and the equation below (expressed identically as both latex and python) i don't
see on the list. can anyone think of a way in which this equation may be
re-factored so that the square-root is no longer part of it?

many many thanks,
l.

(update: let me try some {math} brackets round the latex... it worked! yay! thanks mark for the hint)
[math]
w = \frac{
\left( 1 - \sqrt{\cos{\alpha \pi}} \right)
\left(sin{\alpha \pi} \right)
}
{2}
[/math]
python:

cs = 1 - pow(cos(fsc * pi), 0.5)
si = sin(fsc * pi)

cs * si / 2
 
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I have moved this thread here to our trigonometry subforum as it is a better fit. For easier reading, the expression is equivalently:

$$w = \frac{\left(1-\sqrt{\cos(\alpha\pi)} \right)\sin(\alpha\pi)}{2}$$

This first thing I notice is we need to assume the angle $\alpha\pi$ is in the first or fourth quadrants if $w$ is to be real. But, I don't know of any identities that would allow you to express that cosine function as a square.
 
MarkFL said:
I have moved this thread here to our trigonometry subforum as it is a better fit. For easier reading, the expression is equivalently:

$$w = \frac{\left(1-\sqrt{\cos(\alpha\pi)} \right)\sin(\alpha\pi)}{2}$$

This first thing I notice is we need to assume the angle $\alpha\pi$ is in the first or fourth quadrants if $w$ is to be real. But, I don't know of any identities that would allow you to express that cosine function as a square.

thank you mark - yes w is real. i thought about squaring the equation (x^2 + 2xy + y^2) but doh! that just leaves sqrt cos in the middle. this is a very strange equation but i cannot argue with it: it was discovered phenomenologically (big word, i know *grin*) by mrob's program "ries". alpha is the fine structure constant and this is something fascinating and exciting based on qiu-hong hu's paper about the electron prescribing a path of a hubius helix. in case any of these help in any way, or are generally of interest:

[1206.0620] On the Wave Character of the Electron
[physics/0512265] The nature of the electron
http://members.optushome.com.au/walshjj/toroid2.jpg
 
Well, you can rewrite it as:

$$\left [ 1 - \frac{2w}{\sin{\pi \alpha}} \right ]^2 = \cos{\pi \alpha}$$

Which is an aesthetically pleasing relation, but I don't think you can get it any simpler than what you've got (in order to calculate $w$) unless $\alpha$ has some special properties. It will certainly be more efficient than expressing $w$ as a real root of a fourth order polynomial, as can be seen by setting $\cos \pi \alpha = \sqrt{1 - \sin^2 \pi \alpha}$ and squaring both sides.
 
Bacterius said:
Well, you can rewrite it as:

$$\left [ 1 - \frac{2w}{\sin{\pi \alpha}} \right ]^2 = \cos{\pi \alpha}$$

Which is an aesthetically pleasing relation, but I don't think you can get it any simpler than what you've got (in order to calculate $w$) unless $\alpha$ has some special properties.

$\alpha$ does actually have some special properties, but they're expressed as an iterative algorithm:

Chip Architect: An exact formula for the Electro Magnetic coupling constant ( fine structure constant )
It will certainly be more efficient than expressing $w$ as a real root of a fourth order polynomial, as can be seen by setting $\cos \pi \alpha = \sqrt{1 - \sin^2 \pi \alpha}$ and squaring both sides.

oh! that would seem to do the trick, wouldn't it. awwwesome. the reason i say that is because there is a near-identical equation of similar layout that I'm presently dealing with in the form:

[math]
\left( \frac{ 1 + 2 x}{y} \right) ^4 = 1 - y^2
[/math]

or something to that effect, which would make this a lot easier. so, thank you!
 
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