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Derivative of Sine

  1. Dec 2, 2008 #1
    I was wondering about this the other day, and it is something that was left over in my head from a thread on Euler's identity from a few weeks ago. It's a bit hard to state, but I'll try to be clear.

    How do you show the relationship between derivatives of sine and cosine?

    Now, obviously, with euler's identity and taylor series, we can show that you can break the real and imaginary parts of e^ix into the taylor series for sine and cosine and then show how they relate to the derivatives of e^ix. But how do we connect the taylor series of sine and cosine with the geometric significance of these functions?

    I have a feeling there is a certain set of geometric properties(*) to which there exists a unique set of functions which provably correspond to the real and imaginary parts of e^ix, but I don't quite know which set of properties that would be.

    (* I'm thinking Pythagorean theorem or something cos^2 x + sin^2 x = 1)

    I'm thinking that it should also be provable without the use of imaginary numbers, as well, as I'm pretty sure their widespread use came after study of trigonometric functions in calculus.
     
  2. jcsd
  3. Dec 2, 2008 #2

    lurflurf

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    When asking questions like this it is important to state the definition that is being used and the methods.
    For example it can be shown that these properties determine sine and cosine (that is there exist a pair of functions that satisfy them and only one such pair)
    for all real x and y
    1)sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
    2)cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
    3)[sin(x)]^2+[cos(x)]^2=1
    4)lim_{x->0} sin(x)/x=1
    other variations are possible
    it should be noticed that these properties can be written using complex numbers as
    1&2) cos(x+y)+i sin(x+y)=[cos(x)+i sin(x)][cos(y)+i sin(y)]
    3)|cos(x)+i sin(x)|=1
    4)lim_{x->0} [cos(x)+i sin(x)-1]/x=i

    so sine and cosine famously act as a homomorphism from modular addition to unit circle multiplition

    a similar formulation of exp is possible

    for all real x and y
    1)exp(x+y)=exp(x)exp(y)
    2)lim_{x->0} [exp(x)-1]/x=1

    this also works on complex numbers

    for all complex x and y
    1)exp(x+y)=exp(x)exp(y)
    2)lim_{x->0} [exp(x)-1]/x=1

    using 1) on exp(x+i y)
    we can deal with (x real) exp(x) and exp(i x) separately exp(x) gives back the real exponential and exp(i x)=cos(x)+i sin(x) since the conditions match.
     
  4. Dec 2, 2008 #3
    Yes, this is exactly what I had in mind.

    Now I just need to figure out how these properties are used in the derivation of the derivative. I also would like to find a geometric proof of (1) and (2). They are given in every trig book and obvious with euler's identity, but I never learned why they were true in terms of triangles.
     
  5. Dec 2, 2008 #4

    HallsofIvy

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    They aren't "true in terms of triangles". "sine" and "cosine" cannot be defined as functions in terms of triangles. For one thing, if x and y are both 50 degrees, then x+y= 100 degrees and it is impossible to have a right triangle with an angle of 100 degrees. You can have an obtuse triangle with an angle of 100 degrees but if you allow that then taking x and y both equal to 100 degrees, we have x+y= 200 degrees and no triangle can have an angle with measure 200 degrees. We want the sin(x) and cos(x) functions defined for all x and the "triangle" definitions simply don't work. That is why lurflurf said how you prove those will depend strongly on how you are defining sine and cosine.
     
  6. Dec 2, 2008 #5
    I'm pretty sure I know this as well as anyone can.

    I'm pretty sure lurflurf gave me what I was looking for. Now I just need to figure out how to put those together to come up with the proof I'm looking for.
     
  7. Dec 2, 2008 #6

    lurflurf

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    Derivation of the derivative of what?
    If it is sine the newton quotient is
    [sin(x+h)-sin(x)]/h=cos(x+h/2)sin(h/2)/(h/2)
    (show using addition formulas or difference formula that can be derived from it)
    (sin(y)-sin(x)=cos((y+x)/2)sin((y-x)/2))
    cos(x+h/2)->cos(x) as h->0 (continuity of cosine)
    sin(h/2)/(h/2)->1 as h->0 (property 4)
    thus sin'(x)=cos(x)

    Properties 1) and 2) do not need to be proven you can start with them
    If you want a geometric proof try this
    let
    [cos(x),sin(x)] and [-cos(y),sin(y)] be two points on the unit circle
    use the distance formula with d^2 the square of the distance
    d^2([cos(x),sin(x)],[-cos(y),sin(y)])=[cos(x)+cos(y)]^2+[sin(x)-sin(y)]^2
    =2+2[cos(x)cos(y)-sin(x)sin(y)]
    not consider a different coordinate system in which the coordinates of the points change
    [cos(x),sin(x)]->[1,0]
    [-cos(y),sin(y)]->[-cos(x+y),sin(x+y)]
    d^2([cos(x+y),sin(x+y)],[1,0])=[-cos(x+y)-1]^2+[sin(x+y)]^2
    =2+2cos(x+y)
    thus
    cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
    similar arguments give
    sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
     
  8. Dec 2, 2008 #7
    There's a proof in my calculus book that shows that [tex]\int^b_a(sinx)dx = -(cos(b) - cos(a))[/tex]. I will post the proof. Then by the FTC we will have the derivative of -cosx being sinx.

    Proof:

    We will prove this using Riemann sums. Let [tex]h = \frac{b - a}{n}[/tex]. Then the integral

    [tex]\int^b_a(sinx)dx[/tex]

    clearly is the limit of the sum [tex]S_h = h[sin(a+h) + sin(a+2h) + \cdots + sin(a+nh)][/tex],

    arising from division of the interval of integration into cells of size [tex]h = \frac{b - a}{n}[/tex]. We multiply the right-hand expression by [tex]\frac{2sinh}{2}[/tex] and recall the well-known trigonometrical formula:

    [tex]2sin(u)sin(v) = cos(u-v) - cos(u + v).[/tex]

    Provided h is not a multiple of [tex]2\pi[/tex], we obtain the formula

    [tex] S_h = \frac{h}{2sin(\frac{h}{2})}[ cos(a+\frac{h}{2}) - cos(a+\frac{3h}{2}) + cos(a + \frac{3h}{2}) - cos(a + \frac{5h}{2}) + \cdots + cos(a + \frac{(2n-1)h}{2}) - cos(a + \frac{(2n+1)h}{2})[/tex]


    [tex] = \frac{h}{2sin(\frac{h}{2})}[cos(a + \frac{h}{2}) - cos(a + \frac{(2n+1)h}{2}][/tex].

    Since [tex] a + nh = b[/tex], the integral becomes the limit of [tex]\frac{h}{2sin(\frac{h}{2})}[cos(a+\frac{h}{2}) - cos(b + \frac{h}{2})][/tex] as [tex]h \rightarrow 0[/tex].

    Now, we know that for [tex] h \rightarrow 0 [/tex], the expression [tex]\frac{\frac{h}{2}}{sin(\frac{h}{2})}[/tex] apprroaches the limit one. The desired limit is then simply [tex]cos(a) - cos(b)[/tex], and we arrive at the integral [tex] \int^b_a(sinx)dx = -(cosb - cosa)[/tex]. QED

    As an exercise in the back it says using a similar method, prove that [tex]\int^b_a(cosx)dx = sinb - sina[/tex]. So, that's how you show the relationships between the derivatives of the sin and cosine.
     
  9. Dec 2, 2008 #8
    You can define sine and cosine as the x and y coordinates of a point a given arc-length (angle) along the unit circle on the complex plane. Then the geometric significance of the connection between sin, cos, and exp( i x ) is that exp( i x ) is a unit speed parameterization of the unit circle (like we have seen in the previous thread)!

    Furthermore, if you know exp( i pi) = -1 but not necessairily the full euler formula, then there is also a deep connection between exp( i x ) and rotations. First define sin(x) and cos(x) as the x and y coordinates of the point (1,0) rotated by angle x, and note that composing n rotations is the same as rotating by n-times the original angle. Then with a bit of thinking you should be able to convince yourself that rotating by an arbitrary angle is the same as raising a known rotation matrix to a some power. If you actually compute this matrix power via diagonalization, on the diagonal you get get exp( i x ) and exp( -i x ), and multiplying it out gets the complex exponential definitions of sine and cosine.
     
    Last edited: Dec 2, 2008
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