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Explanation sinus

  1. Dec 16, 2011 #1

    georg gill

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    Here is an article that shows that taylor series for sinus becomes very accurate:


    I wondered if I could prove the sinus values. Explanation for taylor formula have I found here before but the proof for derivative of sinus and cosine relies on values for sinx and cosx which is circular. I might as well upload my attempt to prove cosine to show what I mean. (The same problem goes for sinus):









    The biggest problem of circularity in the proof that I found was that it uses values for sinA and sinB and cosA and cosB to pytagoras theorem in part one for rewriting identity. Does anyone know a way around this to be able to prove values for sinx?
    Last edited: Dec 16, 2011
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  3. Dec 16, 2011 #2

    Simon Bridge

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    That's great because I'm having sinus trouble <sniff>.

    Woah: viewdocsonline is bandwidth-heavy isn't it?

    I don't see your problem - in order to find the derivative of a function you must know the function (I take it you want to prove that the derivative of sin(x) is cos(x)?) How is this circular?
  4. Dec 18, 2011 #3

    georg gill

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    Ok I am not sure if anyone think this is something useful to considerate but I did find one way that could make sense:

    Definition rad:

    [tex]rad=\frac{bowlength}{radius}[/tex] (I)

    then with way to find length of bow from pytagoras and definition of derivative used backwards we get formula for integrating polynoms:

    [tex]\int_a^b \sqrt{1+(\frac{dy}{dx})^2} dx[/tex] so one becomes (II):

    [tex]rad=\frac{\int_a^b \sqrt{1+(\frac{dy}{dx})^2} dx}{radius}[/tex]

    (II) is one way of finding angle from knowing only one side of an triangle that is not hypotenuse.

    Then I tried:

    if one assume an angel and want to find x the cathetus along x-axis one need to integrate the arc length integral. If one start at x=0 then one start at


    so I guess this should go from 0 to 1 for x and from

    [tex]rad=\pi/2[/tex] to [tex]rad=0[/tex]

    I am not quite sure about limits here

    Integrating with taylor polynomials since integral I found for this in my formulas where arccos and is circular for proving how to calculate cathetus from angle. Or does anyone know a other way to integrate this?
  5. Dec 18, 2011 #4

    Simon Bridge

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    There is not enough information here - try drawing a picture.
    What has this to do with the first post?
  6. Dec 18, 2011 #5

    georg gill

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    lets say r=1 as in unity circle



    [tex]rad=\int_a^b \sqrt{1+(\frac{-2x}{2\sqrt{1-x^2}})^2} dx[/tex]

    [tex]rad=\int_a^b \sqrt{\frac{1}{1-x^2}} dx[/tex]

    This is the integral I was thinking to rewrite because of it was defined as answer with arccos in integral book.

    [tex]\int_a^b\sqrt{\frac{1}{1-x^2}} dx[/tex]

    I don't know how to integrate it with substitution or integration by parts but it might work with taylor series. It takes some effort for me to calculate all the derivatives of it to try integrate the taylor series for (a):

    [tex]\sqrt{\frac{1}{1-x^2}} [/tex]

    I think the thing is to say x from x=0 to x and then choose x from angle choosen as an equation to get rad value. Lets say rad=1

    [tex]1=\int_0^x \sqrt{\frac{1}{1-x^2}}[/tex]

    I wonder about if it should be -1 or 1 in numerator in (a) after taking root there. I am not sure how to determine that.

    EDIT: first post was a way of trying to show values for sinx is correct this is a attempt to show how one could calulate sinx from just knowing x i guess. I am not sure if it is possible to prove the first post because taylor series is based on knowing the formula it takes series for
    Last edited: Dec 18, 2011
  7. Dec 18, 2011 #6

    Simon Bridge

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    Lets see if I follow you - you should make the reader guess your intentions at each step.
    I'll tidy up your English as I go - that's so you can see what I understand from what you wrote. Your English is no so much bad as disjointed and archaic.

    We start with a unit circle - defined to have unit radius (r=1).
    [As opposed to a unit circle which has unit circumference or diameter.]

    By definition, the size of an angle is the arclength, about the unit circle, that fits inside the angle. OR: the proportion of the circumference of an arbitrary circle within the angle.

    I guess you mean that, in Cartesian coordinates:

    [tex]y=\pm \sqrt{1-x^2}[/tex]... otherwise it's just a half-circle.
    OK - that's how y changes with x.

    I think you want to put "angle" there instead of "rad", since radiens are a unit for angle.

    So the angle is the arclength about the unit circle - as discussed.

    [tex]rad=\int_a^b \sqrt{\frac{1}{1-x^2}} dx[/tex]

    This is the integral I was thinking to rewrite because of it was defined as answer with arccos in integral book.[/quote]
    Well yes, that's what arccos means.

    I think I see what you mean now.

    You were talking about differentiation of sine before, which process does not refer back to the definitions to procede. The way to do this is to treat an integration as an anti-differentiation. You need to find the difference equation that satisfies the integrand.

    I think the thing is to say x from x=0 to x and then choose x from angle choosen as an equation to get rad value. Lets say [θ=1radien]

    [tex]1=\int_0^x \sqrt{\frac{1}{1-x^2}}[/tex][/quote]
    This would be if you orient your axis so that the x-axis lines up with one side of the angle.
    Good call.

    If you started with a -1, then of course. You have to recover the original expression.
    [itex]y=-x; \; z=y^2; \; \text{so} \; z=x^2; \; \text{and} \; \sqrt{z} = -x[/itex]
    But only because the mapping for y has been defined.

    We would usually write: [itex]\sqrt{z}=\pm y[/itex] since we cannot tell, looking at [itex]z[/itex] which sign to give to [itex]y[/itex]

    But bear in mind you also neglected the "±" in the root earlier.

    Taylor series is also only valid for a range about a point ... so you'd need it to converge as the order gets very big.

    But if I have understood you correctly, you are basically concerned that any derivation for trig functions will end up referring to the definitions of the functions. You want a description that does not.

    If so then I don't think that can be done. See next post.
  8. Dec 18, 2011 #7

    Simon Bridge

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    It can be helpful to think about it like this:

    A line drawn so as to intersect a circle at only one point is called a tangent.
    A line-segment so drawn is called "tangent to" the circle at that point.
    A line segment drawn with it's end-points on the circle is called a chord.
    A chord that also intersects the center of the circle is called a diameter.
    (The length of any diameter is called the diameter of the circle.)
    Half the diameter is called the radius.
    Half the length of a chord is called a sine.

    We make new words as a shorthand for these concepts - it means we can talk about them without having to go on and on.

    In the unit-radius circle, the radius to be 1 (of course).
    When we do this, the name of the unit is radien - just like the name of the unit of force is "Newton" - the word just means "in units of the radius". Thus, "radien" is a dimensionless unit for proportional length (ust as "Newton" is a dimentioned unit for rate of change of momentum).

    Thus the diameter is 2 radiens and the circumference is 2π radiens.

    It is also common to define a circle to have unit circumference ... people don't like doing fractions so the unit gets divided, by convention, into 360 sub-units called "degrees". Here I will focus on the unit-radius circle.

    If we draw two diameters and a chord, so that the first of the diameters is perpendicular to the chord, and the second intersects the sine where the sine intersects the circle, then you have a triangle. The angle at the center is called "the angle".

    The size of the angle is defined as the arclength between the diameters.
    So the angle is just the proportional arc-length - when we write the formula for an arc-length [itex]s=R\theta[/itex] we are just repeating the definition of the angle.

    The length of the half-chord, then, is related to the angle. It is the sine - that's it's name.

    Since they [the trig relations] are just definitions of words, don't expect to be able to derive or prove them without referring to them. It would be like deriving your name from your personality - it becomes circular.

    Instead we use the words as labels to talk about other relationships.

    So when we say x=cos(A) and y=sin(A), we mean that x is to y what the sine is to the cosine. In this way we can understand the behavior of other curves in terms of the behavior of circles.

    That integral that turns into an arccos? - well that is to be expected because you put x equal to the distance from the origin to the chord: i.e. the cosine. So you just found that the angle was equal to the angle. Well done.

    The integral is more useful when the curve is not a circle.

    Taking the taylor series for the trig functions will just repeat this definition.

    I suspect that's why the derivations look circular to you.

    But what you can do, is find values for sine and so on algebraically - using the sum of a number sequence. Just like pi has a definition as the ratio of circumference to diameter, and most derivations involving pi come back to that, but you can compute pi algebraically from a long sum.

    Perhaps that is what you are trying to do here?
  9. Dec 21, 2011 #8

    georg gill

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    I have tried to find out if the taylor approximation of the integral converges. This is what I ended up with:


    I have problem fully grasping all the problems. Especially that since I have not calculated that many terms of taylor approxiamtion and I dont know if it converges because of that. Anyway if anyone could he3lp sort things out it would be great!
    Last edited: Dec 21, 2011
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