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Why does Euler's identity work only in Radians?

  1. Oct 25, 2011 #1
    e^iA = cosA + i*sinA is true iff A is expressed in Radians. Why that particular unit?
    (I'm not sure this rubric is the right one for this question, but since it didn't seem to fit any of the other rubrics, I put it here.)
     
  2. jcsd
  3. Oct 25, 2011 #2

    D H

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    One way to prove Euler's formula is in terms of power series. First express exp(ix) as a power series. Then split the even and odd terms of this to form two series. Next rewrite the powers of i in terms of powers of -1 (and a residual i for the odd series). Those two power series are exactly those of cosine(x) and i*sine(x) where x is in radians.
     
  4. Oct 25, 2011 #3

    lurflurf

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    It would work fine with another scaling, except it would have another scaling.

    given small x we have approximately that
    exp(i x)~1+i x
    cos(x)+i sin(x)~1+i x

    which combined with the addition laws gives us the identity

    if we had for some reason wanted to use other scaling we have
    otherexp(i x/a)=exp(i x)~1+i x
    othercos(x/b)+i othersin(x/b)=cos(x)+i sin(x)~1+i x
    we would have the identity

    otherexp(i x/a)=othercos(x/b)+i othersin(x/b)

    for suitable a and b

    in calculus we would say
    a=otherexp'(0)
    b=othersin'(0)

    and of course when a=b=1 we have the usual functions
     
    Last edited: Oct 25, 2011
  5. Oct 25, 2011 #4
    Ah, that makes sense. Thank you both, D H and lurflurf
     
  6. Oct 29, 2011 #5
    Ah, one detail. In doing the MacLauren series for sin and cos, in the expansion one only needs that sin(0) = 0 and cos(0) = 1. This is true whether or not your units are in degrees or in Radians. Therefore the answer from D H seems to be begging the question. There's something else to this.... What?
     
  7. Oct 29, 2011 #6
    It all has to do with the following limit:
    [tex]
    \lim_{x \rightarrow 0}{\frac{\sin{x}}{x}} = 1
    [/tex]
    which works only in radians.
     
  8. Oct 29, 2011 #7

    D H

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    Not true. Here is the series for sine(x) when x is in radians:

    [tex]\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]

    This is not valid when x is in degrees. The reason is simple: [itex]d/dx(\sin x) = \cos x[/itex] is only valid when x is in radians. When x is expressed in degrees,

    [tex]\frac{d}{dx}\sin x^{\circ} = \frac{\pi}{180} \cos x^{\circ}[/tex]

    and thus the Maclaurin series for [itex]\sin x^{\circ}[/itex] is

    [tex]\sin x^{\circ} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{180}\right)^{2n+1}x^{2n+1}[/tex]
     
  9. Oct 29, 2011 #8

    symbolipoint

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    All of those terms are circular functions. Radians are based on pi. 360 Degrees is arbitrary.
     
  10. Oct 29, 2011 #9

    HallsofIvy

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    You may be thinking that, because (sin x)'= cos x, (sin x)''= -sin x, (sin x)'''= -cos x, etc. and similarly for the derivatives of cos x, in order to find the MacLauren series, we only need to evaluate those at x= 0.

    But the point is that those derivative formula are only true if x is in radians.

    In particular, the derivative of sin x is given by
    [tex]\lim_{h\to 0}\frac{sin(x+h)- sin(x)}{h}[/tex]
    Using the trig identity [itex]sin(x+h)= sin(x)cos(h)+ cos(x)sin(h)[/itex] that becomes
    [tex]\lim_{h\to 0}\frac{sin(x)cos(h)- cos(x)sin(h)- sin(x)}{h}[/tex]
    [tex]= \lim_{h\to 0} sin(x)\frac{cos(h)- 1}{h}+ cos(x)\frac{sin(h)}{h}[/tex]
    so to have the derivative of sin(x) be cos(x) we must have
    [tex]\lim_{h\to 0}\frac{cos(h)- 1}{h}= 0[/tex]
    and
    [tex]\lim_{h\to 0}\frac{sin(h)}{h}= 1[/tex]
    and, as Dickfore noted, those are only true if h is in radian.

    In fact, if you are going to use the trig functions as general functions and not just as ways to "solve triangles", you should stop thinking of the arguments as being angles at all- and just think of them as number, without any "degrees" or "radians". One way of doing that is to use the "circle" definition: Given a unit circle (circle with center at the origin, radius 1, in some coordinate system), to find sin(t) or cos(t), start from the point with coordinates (1, 0) and measure counter-clockwise (clockwise if t is negative) a distance t around the circumference of the circle. What ever (x,y) point you end at gives you cos(t)= x, sin(t)= y. No angles at all there! (Yes, you have measured a distance but the "units" are whatever "units" you used to construct the coordinate system.)
     
  11. Oct 31, 2011 #10
    thanks, HallsofIvy. There are two curious points that I do not quite understand.
    (1) First, nowhere in your proof is it explicitly shown that x and h need to be in radians. I presume it is implicit in the use of the limit of sin(h)/h = 1 which is derived from using the arc formula arc length = rx, where x is in radians, and r is the radius. Right?
    (2) Assuming this to be the case, although the proof that sin'(x) = cos x and so forth rests on x being in Radians, nonetheless sin'(x) = cos x works perfectly well for x in degrees, since one just substitutes from the radians to the equivalent number of degrees and back again. The derivative of sin x at x=60 degrees is indeed cos(60 degrees). Therefore I would appreciate some clarification to your statement that these derivatives are only valid if x is in Radians.
     
  12. Oct 31, 2011 #11

    lurflurf

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    sin'(0)=1 is only true in radians, it can be taken as the definition of radians.
    sin'(0)=pi/Arccos(-1)
    As I said above the sine and cosine in another scaling are easily expressible in terms of the usual one, but that does not make them equal. Some confusion might be resulting from thinking of degrees as a way of expressing a number versus as being a different number. Like if we said
    60 degrees=π/3
    or
    degrees=π/180
    by sine in degrees we mean a function where
    (60,√3/2)
    not
    (π/3,√3/2)
     
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