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Trig. Derivatives

  1. Nov 13, 2008 #1
    So its pretty much common knowledge that:

    [tex]\frac{d}{dx}\sin (x) = \cos (x)[/tex]

    But how does one go about to actually prove it?

    This came up in my Math Methods class while we were talking about Bessel Functions.
     
  2. jcsd
  3. Nov 13, 2008 #2
    Hia,

    The way I would do it would be to write sin(x) in terms of exponentials, then differentiate that. You should end up with cos(x) in exponential form, but im not sure if that can be considered a proof.
     
  4. Nov 13, 2008 #3

    tiny-tim

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    Hi castusalbuscor! :smile:

    Either use the lim{(f(x+h)-f(x))/h} definition, and sinA - sinB = 2.sin(A-B)/2.cos(A+B)/2,

    or use 2isinx = eix - e-ix :smile:
     
  5. Nov 13, 2008 #4
    This was actually brought up in class the the professor shot it down, he also shot down expanding it in a Taylor series (since we need to know the derivative for it to work).

    The limit definition might be more appropriate. ^_^ Thanks!

    I wonder is there a geometric way to prove it?
     
  6. Nov 13, 2008 #5

    tiny-tim

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    Yup … just draw two very close triangles inside a circle, and the tiny triangle at the tip will have hypotenuse r∆θ and height r∆(sinθ), so you get the equation r∆(sinθ) = r∆θcosθ. :smile:
     
  7. Nov 13, 2008 #6
  8. Nov 13, 2008 #7

    lurflurf

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    Ones proof would depend upon ones definition. What definition are you using?
    If your definition of cos is
    cos(x):=sin'(x)
    It is an easy proof.
     
  9. Nov 14, 2008 #8

    HallsofIvy

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    I had typed up a long response and it has disappeared! Must have hit the wrong key.

    As lurflurf said, how you prove the derivatives of sin(x) and cos(x) depends upon how you define them!

    It is perfectly valid to define sin(x) to be
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]
    and cos(x) to be
    [tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}[/tex]
    Once you have done that and proven technical details like that fact that these converge absolutely for all x, you can differentiate term by term.
    Of course, your teacher might ask to prove that these "sine" and "cosine" functions have the various properties of the trig functions- for example being periodic of period [itex]2\pi[/itex]

    My personal favorite is to define y= sin(x) as "the function satisfying the initial value problem y"= -y with y(0)= 0, y'(0)= 1" and y= cos(x) as "the function satisfying the initial value problem y"= -y with y(0)= 1, y'(0)= 0". It is not too difficult to show that those do, in fact, has all the properties of the "usual" sine and cosine, including periodicity. In fact, if I were asked to prove that sine and cosine, as defined by the infinite sums above, were periodic, I would first use term by term by term differentiation to show that they satisfy the above differential equations and then use that.

    The proof Desh627 links to defines sine and cosine in terms of "circular functions": given non-negative t, measure counterclockwise around the unit circle from (1, 0) a distance t. Cos(t) and sin(t) are defined as the y and x coordinates of the endpoint, respectively. If t is negative, measure clockwise instead of counterclockwise. That proof uses the fact that
    [tex]\lim_{x\rightarrow 0}\frac{sin(x)}{x}= 1[/tex]

    The proof of that (given in most Calculus texts) is:

    Assume t is some small, positive value and draw the line from (0,0) to (cos(t), sin(t)) on the unit circle. Drop a perpendicular from (cos(t), sin(t)) to x-axis. We have a right triangle with legs of length cos(t) and sin(t) and so area (1/2)sin(t)cos(t). Draw a vertical line from (1, 0) to the extended radius through (cos(t), sin(t)). That forms a right triangle with one leg of length 1 and similar to the smaller right triangle: cos(t)/1= sin(t)/h where h is the vertical height, so h= sin(t)/cos(t) and the area of this larger triangle is (1/2)sin(t)/cos(t). Finally, the area of the circular sector between them is (1/2)t. Looking at those 3 areas, [itex](1/2)sin(t)cos(t)\le (1/2)t \le (1/2) sin(t)/cos(t)[/itex]. Multiplying through by the positive number 2/sin(t) we have [itex]cos(t)\le t/sin(t)\le 1/cos(t)[/itexs]. Inverting each part, which also reverses the inequalities, [itex]1/cos(t)\ge sin(t)/t\ge cos(t)[/itex]. By the "sandwiching theorem" we have
    [tex] 1\le \lim_{t\rightarrow 0%+} \frac{sin(x)}{x}\le 1[/tex]

    Of course, we can argue by symmetry that the same thing is true for t approaching 0 from below.
     
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