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Author Topic: Mathematics Help Thread  (Read 218152 times)

ILikePie

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Re: Mathematics Help Thread
« Reply #555 on: February 19, 2011, 02:02:18 pm »

Code: [Select]
cos x = sin (x - 40)
cos x = cos (90 - (x - 40))
cos x = cos (130 - x)
cos x = cos x*cos 130 + sin x*sin 130
cos x - cos x*cos 130 = sin x*sin 130
cos x*(1 - cos 130) = sin x*sin 130
cos x = (sin x*sin 130) / (1 - cos 130)
cos x = (sin x*cos (90-130)) / (1 - cos 130)
cos x = (sin x*cos (-40)) / (1 - cos 130)
cos x = (sin x*cos 40) / (1 - cos 130)
Now what?
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Christes

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Re: Mathematics Help Thread
« Reply #556 on: February 19, 2011, 02:11:32 pm »

Is your goal to solve for x?

If so, then I'd suggest dividing sin(x) over to get tan(x) = ...  and applying arctan.  Remember that there will be infinitely many solutions, and that arctan's range is (-pi/2,pi/2)
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ILikePie

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Re: Mathematics Help Thread
« Reply #557 on: February 19, 2011, 02:22:43 pm »

That works, thanks.
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Virex

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Re: Mathematics Help Thread
« Reply #558 on: February 19, 2011, 04:55:39 pm »

Code: [Select]
cos x = sin (x - 40)
cos x = cos (90 - (x - 40))
cos x = cos (130 - x)
cos x = cos x*cos 130 + sin x*sin 130
cos x - cos x*cos 130 = sin x*sin 130
cos x*(1 - cos 130) = sin x*sin 130
cos x = (sin x*sin 130) / (1 - cos 130)
cos x = (sin x*cos (90-130)) / (1 - cos 130)
cos x = (sin x*cos (-40)) / (1 - cos 130)
cos x = (sin x*cos 40) / (1 - cos 130)
Now what?
You're doing it the difficult way. After line 3 you've got:
Code: [Select]
cos x = cos (130 - x), and both sides are cosines, so we can equate the arguments of the cosine, using cos(x) = cos(360-x)
x = (130 - x)mod(360) or  x = (360 - (130 - x))mod(360)
Using that you ought to be able to find x.
« Last Edit: February 19, 2011, 05:03:08 pm by Virex »
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Darvi

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Re: Mathematics Help Thread
« Reply #559 on: February 19, 2011, 07:02:39 pm »

Wait, what? Couldn't he just've done

cos(x)=cos(130-x)      l arccos
<=>x=130-x (+k360)  l +x
<=>2x=130 (+k360)   l :2
<=>x=65 (+k180)

?

Edit: ah wait. Whatever.


cos(x)=cos(130-x)
<=>x=130-x (+k360)    or x=x-130(+k360)
<=>2x=130 (+k360)     or 0=-130(+k360) (impossible)
<=>x=65 (+k180)

Still true.

Edit: f*cking radiants
« Last Edit: February 19, 2011, 07:11:33 pm by Darvi »
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Virex

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Re: Mathematics Help Thread
« Reply #560 on: February 19, 2011, 07:06:57 pm »

He's probably working in degrees. Plus, as I pointed out, you'll get 2 sets of intercepts, while your method only finds one. You caught that one
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Darvi

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Re: Mathematics Help Thread
« Reply #561 on: February 19, 2011, 07:09:55 pm »

Yeah, degrees, right. F*cking radiants <_<

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ILikePie

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Re: Mathematics Help Thread
« Reply #562 on: February 20, 2011, 11:21:12 am »

I can get of the cosines if I have them on both sides?
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Virex

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Re: Mathematics Help Thread
« Reply #563 on: February 20, 2011, 11:22:21 am »

Yeah as long as you remember you'll get 2 solutions.
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malimbar04

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Re: Mathematics Help Thread
« Reply #564 on: February 20, 2011, 02:57:57 pm »

Yeah, degrees, right. F*cking radiants <_<
It wouldn't be so bad if degrees were never invented. Seriously, who said "lets divide a circle into some arbitrary 360 pieces". They didn't no jack about circles.
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MonkeyHead

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Re: Mathematics Help Thread
« Reply #565 on: February 20, 2011, 03:33:23 pm »

Yeah, degrees, right. F*cking radiants <_<
It wouldn't be so bad if degrees were never invented. Seriously, who said "lets divide a circle into some arbitrary 360 pieces". They didn't no jack about circles.

Babylonians, approx 3000 BC, based on observation that it took 360 days until Earth returned to same point in orbit (as far as they could tell), which they assumed was circular (not a bad postulation). Thier fault.
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ZetaX

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Re: Mathematics Help Thread
« Reply #566 on: February 20, 2011, 03:50:42 pm »

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MonkeyHead

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Re: Mathematics Help Thread
« Reply #567 on: February 20, 2011, 03:53:19 pm »

Heh yea, nothing is ever simple.
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malimbar04

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Re: Mathematics Help Thread
« Reply #568 on: February 20, 2011, 07:39:12 pm »

At least it's better than the explanation for Fahrenheit. Every explanation I've heard crazy.

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Virex

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Re: Mathematics Help Thread
« Reply #569 on: February 20, 2011, 07:49:24 pm »

You mean that using half the circumference of a circle makes more sense? Anyway, you can always divide the circle in 400 parts if you like that better.
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