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Help with sine wave theory

  1. Dec 28, 2011 #1
    Hi. I have been trying to find a way to create a function that I can use to solve sin(x) in my head. I'm familiar with the taylor series but a few weeks ago thought that f(x)= -(x^2)+1 might superimpose well over a sin wave and would be an easy way to approximate.

    e.g. from x= +1 to 0 would represent sin(0) to sin(pi/2) and if you wanted a 45 degree angle you could put .5 into -(x^2)+1.

    It was close but obviously not close enough.

    A few hours ago I was reading about AC current in an electronics book and came up with another related idea. I assumed that if you look at any two x values (that lie between critical points) on a curve (I'll call them a and b), and calculate the slope for the curve along that interval, and then using the derivative find where on that interval the instantaneous slope matches the slope you calculated for the interval (I'll call that c), you could look at the distance a to c divided by distance a to b (along x-axis) and consider that ratio a characteristic of the curve.

    I'm having a hard time putting this in text, hopefully some of the people that read this are still with me.

    e.g. for x^2, from 0 to 3 along the x axis the rise is 9/run=3 so slope=3 a=0 and b=3
    the derivative 2x = 3 at x = 1.5 so c=1.5 (i.e. a tangent line to x^2 at x=1.5 has slope 3)

    a to c = 1.5, a to b = 3 , 1.5/3 = 0.5, and for any interval length along x^2 c-a/b-a = 0.5

    e.g. between 2 and 100 on x rise is 10000-4( or 9996) over run 100-2 (or 98) = 102 = m(slope) from x=2 to 100
    2x=102 x=51
    a line tangent to the curve of x^2 at x=51 has a slope of 102 as well
    (51-2)/(100-2)=0.5

    The reason I tested this curve first was for simplicity, my intentions were to find a (c-a)/(b-a) for a sine wave, in hopes that it would add some insight to the wave and maybe a simpler arithmetic way of explaining and thinking about it.

    so for a=0 (the minimum of the sine curve) b=pi/2 (the maximum of the sine curve) the slope is sin(pi/2)/(pi/2) = 2/pi

    I found a proof that showed the derivative of sin(x) to be cos(x) and used a solve function on my ti-83(because I had no other idea how) to say cos(x) - 2/pi = 0 and solve for x, it returned .8806... which I verified when cos(x) did = 2/pi

    This now being the distance from a aka c, along the x axis, and knowing pi/2 is the distance from a to b, my ratio was x/(pi/2) which = 0.5606641... which was not a fraction and did not divide equally into pi : (

    I redid the steps from 0 to pi/4 and got 0.5733405159, which seemed to prove the ratio concept didn't work for the sine wave

    So I tried pi/1000 and got .57735, not far from pi/4

    pi/1000000 confused my calculator.

    I've considered that the theory could be incorrect, but still intriguing because of how close the ratio is from 0 to pi/2), or that it could be correct and my calculator is giving incorrect answers because of the way it calculates.

    My question is, does anyone have any input that may add insight and or put my mind at ease.

    Hopefully someone finds amusement in this at least.

    Dave
     
  2. jcsd
  3. Dec 30, 2011 #2
    I'm feeling pretty stupid since no one even thought this deserved a reply...haha

    I'm giving it a bump just in case and have a few other questions.

    1. How do I solve Cos(x) = y for x? I can see cos(x)-y=0, or cos(x)/y=1 but I'm lost from there.

    2. I was looking into the procedure for finding arc length, to compare arc length from a to c and arc length from a to b. Looking for a relationship that might stand true for any interval between 0 and pi/2 on a sine wave. I am having trouble comprehending the integral for the arc length calculations though. I followed the algebra to

    [itex] s = \int_a^b \sqrt{1+f '(x)}dx[/itex]

    Using cos(x) as my f '(x) and the concept F(b)-F(a) with a=0 and b=pi/2, I get

    [itex] s = \sqrt{1+cos(pi/2)}dx - \sqrt{1+cos(0)}dx [/itex]

    [itex] = dx - \sqrt{2}dx[/itex]

    Is that right?

    Thanks for any input, Dave
     
  4. Dec 30, 2011 #3
    Forgot to say f prime was squared in the integrals I showed, but oddly the numbers work out the same either way. i.e. 1 squared and 0 squared, then add the one and square root.

    Also forgot to ask what dx represents here. I know this is probably beneath most of you, but I didn't see a better place to ask.

    Thanks.
     
  5. Jan 6, 2012 #4
    Is there a mod than can move this to a forum where I may get some replies?
     
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