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Sine and cosine

  1. Apr 27, 2009 #1
    Hello.

    I know the definition of sine and cosine, but how were these formulas originally invented? I mean, how did people derive the power series for sine and cosine for the first time?
     
  2. jcsd
  3. Apr 27, 2009 #2

    HallsofIvy

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    Be careful when you talk of "the definition". Of course, historically, the original definitions of sine and cosine were not in terms of series but in terms of right triangles. From those we can show that the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x). From those follow all the derivatives of sine and cosine and then the standard series forms for sine and cosine are given by the Taylor's series about x= 0.

    You can also define sine and cosine by:
    "y= sin(x) is the function satisfying y"= -y for all x, with y(0)= 0, y'(0)= 1 and y= cos(x) is the function satisfying y"= -y for all x, with y(0)= 1, y'(0)= 0" and, personally, I prefer that definition. From that the Taylor's series about x= 0 immediately follow.
     
  4. Apr 27, 2009 #3
    Can I find the derivative like this:

    (cos(x+dx)-cos(x))/dx = (cos(x)cos(dx)-sin(x)sin(dx)-cos(x))/dx = sin(x)sin(dx)/dx --> sin(x)

    ? (I'm sorry this is so ugly, I hope you understand what I mean. Is TeX broken?)

    The only problem is that I have to argue that sin(dx) goes to zero linearly, but that seems to make sense geometrically.
     
  5. Apr 27, 2009 #4
    I quite like the geometry of the unit-circle definitions of sin and cos myself. I know they aren't the most useful or practical versions, but they provide a very visual interpretation that give (at least to me) a good first meaning to the otherwise arbitrary trig functions :)

    Edit:
    Yes, LaTeX is broken at the moment :(
     
  6. Apr 27, 2009 #5

    HallsofIvy

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    sin(x) does NOT "go to 0 linearly" but it is close. The 'standard' proof involve dropping a vertical line from the point (cos(t), sin(t)) to the x-axis as well as extending the line from (0,0) to the tangent line to the circle at (1, 0). The smaller triangle has area (1/2)xy= (1/2)cos(t) sin(t), the circular sector has area (1/2)t, and the larger triangle has area (1/2)(1)(tan(t)= (1/2)(sin(t)/cos(t)).

    So we have (1/2)cos(t)sin(t)<= (1/2)t<= (1/2)sin(t)/cos(t). Multiplying through by 2/sin(x), we have cos(t)<= t/sin(t)<= 1/cos(t) which, inverting, gives 1/cos(t)<= sin(t)/sin(t)<= cos(t). Since cos(t) is continuous and cos(0)= 1, taking the limit as t goes to 0, lim sin(t)/t= 1.

    We also will need sin(x+y)= sin(x)cos(y)+ cos(x)sin(y) and sin(x-y)= sin(x)cos(y)- cos(x)sin(y). Adding those sin(x+y)- sin(x- y)= 2cos(x)sin(y). In particular, if we take A= x+y and B= x-y, then x= (A+ B)/2 and y= (A-B)/2 so that sin(A)- sin(B)= 2 cos((A+B)/2)sin((A-B)/2).

    Now we can write sin(x+h)- sin(x)= 2cos((2x+h)/2)sin(h/2)= 2 cos(x+ h/2)sin(h/2)
    Then [sin(x+h)- sin(x)]/h= 2 cos(x+h/2)sin(h/2)/h= cos(x+h/2)sin(h/2)/(h/2).

    lim(h->0) [sin(x+h)- sin(x)]/y= lim(h->0) cos(x+ h/2) lim(h->0) sin(h/2)/(h/2). cos(x+ h/2) goes to cos(x) and sin(h/2)/(h/2) goes to 1 so that limit is cos(x): the derivative of sin(x) is cos(x).
     
  7. Apr 28, 2009 #6
    I know this is circular reasoning, but if you look at the taylor series for sine

    sin(x)=x-(x^3)/3+...

    does this not go to zero linearly ? I.e. lim(x->0) sin(x) = lim(x->0) x.

    Anyway, it is a nice proof and a nice way to derive sine and cosine that you have told me. What about before people knew Taylor series? I just tried to do some geometric approximations (approximating an arc by a straight line), and I got

    cos(x)=1-1/2*x^2

    but I could not get much further because the equations got too complicated ...
     
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